text:
\section{A section}\lipsum
\section{ Introduction}

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret.

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$. \\

Alice begins by taking her qubit (secret), which is in the state
$\alpha|0\rangle_A + \beta|1\rangle_A$, combines it with her 3-qubit GHZ resource, and after measuring the pair in the Bell basis, she now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.
Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret. Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
\noindent{Verstraete et al \cite{verstraete} propose four qubits which can be entangled in nine different ways.}
In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Figure 1).

\begin{figure}[h]

    \caption{}
    \includegraphics[width=.575\textwidth]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.
\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state:
\begin{eqnarray}
|G_{abcd}\rangle=\frac{a+d}{2}(|0000\rangle+|1111\rangle) + {}\nonumber \\ \frac{a-d}{2}(|1100\rangle+|0011\rangle) + {}\nonumber \\  \frac{b+c}{2}(|0101\rangle+|1010\rangle) + {}\nonumber \\ \frac{b-c}{2}(|0110\rangle+|1001\rangle) {}&&
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
={}\nonumber\\&&\{\alpha|0\rangle+\beta|1\rangle\}\otimes \frac{a+d}{2}(|0000\rangle+|1111\rangle) + {}\nonumber\\&& \frac{a-d}{2}(|1100\rangle+|0011\rangle) +  \frac{b+c}{2}(|0101\rangle+|1010\rangle) + {}\nonumber\\&&\frac{b-c}{2} (|0110\rangle+|1001\rangle) \
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
={}\nonumber\\&&(|\phi^+\rangle)[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle) + {}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)] + {}\nonumber\\&&(|\phi^-\rangle)[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle) + {}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)] + {}\nonumber\\&&(|\psi^+\rangle)[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle) + {}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)] + {}\nonumber\\&&(|\psi^-\rangle)[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle) + {}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)]\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.
\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\begin{center}Alice's Measurement Outcomes\end{center}} & \parbox{1.75in}{\begin{center}Alice, Bob and Charlie's Combined State\end{center}}\\
\hline $\lvert\phi^\pm \rangle$  &
$\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]$+ \\ & $\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]$+ \\ & $\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]$+ \\ & $\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]$  \\
\hline $\lvert\psi^\pm\rangle$ & $\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]$+\\ & $\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]$+ \\ & $\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]$+ \\ &
$\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]$   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\
\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+{}\nonumber\\&&\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+{}\nonumber\\&&\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+{}\nonumber\\&&\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
&&|Y_{234}^{\phi^\pm}\rangle=(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B}(|+\rangle+|-\rangle)_{C}]\nonumber\\&&\pm{}\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}{}\nonumber\\&&+(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]{}\nonumber\\&&\pm{}\beta[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}]\}{}\nonumber\\&&+(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}{}\nonumber\\&&\pm{}\beta[|0\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] {}\nonumber\\&&+(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\nonumber\\&&\pm{}\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie send their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\begin{figure}[h]
  \caption{}

  \includegraphics[width=.75\textwidth]{alice-recon-final.pdf}
\end{figure}
\noindent{The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.}

\\ \\
\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.
\\
\\
\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
&&|Y_{234}^{\phi^{\pm}}\rangle=(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]{}\nonumber\\&&\pm\beta[(|+\rangle-|-\rangle)_{A}(|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}{}\nonumber\\&&+(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]{}\nonumber\\&&\pm\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}{}\nonumber\\&&+(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}{}\nonumber\\&&\pm\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] {}\nonumber\\&&+(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]{}\nonumber\\&&\pm\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}
\end{eqnarray}
The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}
\\
\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $\lvert\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha \pm (a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha \pm (a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha \pm (a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha \pm (-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha \pm  (-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha \pm (-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha \pm (a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha \pm (a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha \pm (a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha \pm (a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha \pm (a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha \pm (-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha \pm (-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha \pm (-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha \pm (a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha \pm (a+b-c+d)\beta)]|1\rangle]. $\\

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\begin{figure}[h]
  \caption{}

  \includegraphics[width=.75\textwidth]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.
\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\

\section{ Introduction}

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states.
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocol ensures security only if some of the share holders are \textcolor{blue}{semi-honest} but however fail to address the security when the dealer, after sharing finds all the share holders to be \textcolor{blue}{semi-honest}. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently \textcolor{blue}{retrieve} the secret when she finds both of them to be \textcolor{blue}{semi-honest}.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with four qubit pure entangled states as a resource.\\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in 3-party, 3-qubit protocol and  why do we need a 4-qubit resource in a protocol where revocation of the secret is also possible along with the reconstruction at the share holder's site. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie \textcolor{blue}{(the two parties with whom Alice wants to share the secret)} are sharing a three qubit resource among them with each \textcolor{blue} {possessing} one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is {\textcolor{blue}{semi-honest}, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are \textcolor{blue}{semi-honest} they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if \textcolor{blue}{semi-honest}
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource (see FIG. 1 (a)), and measures the pair in the Bell basis (see FIG. 1 (b)).

\\
\begin{figure}[ht]
  \subfloat[The figure above indicates a three qubit GHZ state shared among Alice (A), Bob (B) and Charlie (C). Alice is also having the secret (depicted by an isolated dot) with herself.]{
	\begin{minipage}[c][1\width]{
	   0.4\textwidth}
	   \centering
	   \includegraphics[width=1\textwidth]{FIGNEW1.png}
	\end{minipage}}
 \hfill
  \subfloat[The figure above describes the situation after Alice's measurement in the Bell basis, where Bob and Charlie are sharing the secret entangled qubit between them.]{
	\begin{minipage}[c][1\width]{
	   0.4\textwidth}
	   \centering
	   \includegraphics[width=1.1\textwidth]{FIGNEW2.png}
	\end{minipage}}
\caption{}
\end{figure}

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is \textcolor{blue}{semi-honest} or Charlie is \textcolor{blue}{semi-honest} or both are \textcolor{blue}{semi-honest}, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is \textcolor{blue}{semi-honest} or Charlie is \textcolor{blue}{semi-honest} or both are \textcolor{blue}{semi-honest} after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be \textcolor{blue} {semi-honest}, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties.\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob
or Charlie is \textcolor{blue}{semi-honest} or both are \textcolor{blue}{semi-honest}, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret when both of other parties are \textcolor{blue}{semi-honest} and consequently bring the entire secret back to herself without revealing it even after sharing .
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\begin{figure}[ht]

	   \centering
	   \includegraphics[width=.575\textwidth,left]{Recon-Comb.pdf}

 \end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is \textcolor{blue}{revoking} back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  \textcolor{blue}{Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a
subroutine.}
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)]\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when A\bflice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when A\bf lice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $|+\rangle \rightarrow |0\rangle$, $|-\rangle \rightarrow |1\rangle$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following lemma, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{lemma}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen to be real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$ (b)$c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (c)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units (d) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:\\
     (a) $a^2 < \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (b) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (c)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (d)$|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:\\
     (a) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (b)$c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (d)$|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:\\
     (a) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (b) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (d) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

\end{enumerate}
\end{lemma}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not \textcolor{blue}{semi-honest}. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $|+\rangle \rightarrow |0\rangle$, $|-\rangle \rightarrow |1\rangle$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{lemma}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e)$|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b)$b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = d$ and $a = c$}
     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e)|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -c$ and $b = -d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:
     (a) {$a = -c$ and $b = -d$}
     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b)  $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e)$|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = d$ and $a = c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c)$d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

\end{enumerate}
\end{lemma}
\\

\noindent For the entire protocol to be successful there will be common range of the input parameters $a,b,c,d$ for which both revocation and reconstruction will be possible. We give this range in form of the following theorem:

\begin{theorem}
\label{QSS}
Quantum Secret Sharing protocol can be implemented using the ${G}_{abcd}$ state, where $a$, $b$, $c$ and $d$ are complex numbers, such that for Alice to retrieve the secret, and either Bob or Charlie to reconstruct the secret if and only if the following conditions are met:

\begin{enumerate}

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely imaginary and $(a+b+c-d)$ is purely real such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:}
     (a) {$b = c$ and $a = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a+b-c+d)$ is purely real and $(a+b+c-d)$ is purely imaginary such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = d$ and $a = c$}
     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely real and $(-a+b+c+d)$ is purely imaginary such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -c$ and $b = -d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d)$|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -c$ and $b = -d$}
     (b)$a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c)$c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$
     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d)$|b|^2 = \frac{1}{2} - a^2$ which means $b$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $c^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < c < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|d|^2 = \frac{1}{2} - c^2$ which means $d$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

     \item If $(a-b+c+d)$ is purely imaginary and $(-a+b+c+d)$ is purely real such that $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a){$b = d$ and $a = c$}

     (b) $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
     (c) $d^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < d < \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$ which means $a$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units
     (e) $|c|^2 = \frac{1}{2} - d^2$ which means $c$ lies within the circle $x^2 + y^2 < 1/2$ with centre at origin and radius less than $\frac{1}{\sqrt2}$ units

\end{theorem}

\\

\\

\section{Conclusion:} In this article we  addresses the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, the dealer also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that share, when she finds other shareholders to be \textcolor{blue}{semi-honest}. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are \textcolor{blue}{semi-honest}.

\noindent \textit{\bf Note:} For proof of \textbf{Lemma 1} and \textbf{Lemma 2}, please see the Appendix.\\

\noindent\textit{\bf Acknowledgment:} This work is done at  Center for Security, Theory and Algorithmic Research (CSTAR), IIIT, Hyderabad.

\section{Introduction}
Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource

\\

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Fig).
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)].\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}.
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\} .
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.
\end{theorem}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}.
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle]. $\\

\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{alice-recon-final.pdf}
\end{figure}

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\\
\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.

\\

\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\\
\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\

\section{ Introduction}

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource

\\

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Fig).
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)].\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}.
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\} .
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.
\end{theorem}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}.
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle]. $\\

\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{alice-recon-final.pdf}
\end{figure}

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\\
\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.

\\

\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\\
\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\

\section{Introduction}
Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource

\\

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Fig).
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)].\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}.
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\} .
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.
\end{theorem}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}.
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle]. $\\

\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{alice-recon-final.pdf}
\end{figure}

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\\
\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.

\\

\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\\
\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\

\section{ Introduction}

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource

\\

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Fig).
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)].\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}.
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\} .
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.
\end{theorem}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}.
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle]. $\\

\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{alice-recon-final.pdf}
\end{figure}

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\\
\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.

\\

\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\\
\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\

\section{Appendix 1: (Proof of Theorem 1)}

For Alice to successfully retrieve the secret, Bob and Charlie must measure their respective qubits in the Hadamard Basis. In TABLE II we show how Alice can retrieve back her message by enlisting down the respective local operations corresponding to Bob's and Charlie's measurement outcomes. Consider Alice's resultant state outcome in Hadamard Basis $\ket{+}\ket{+}$:
\begin{eqnarray}
\sqrt{2}\{(a\alpha + b\beta)|0\rangle + (b\alpha + a\beta)|1\rangle\} {}\nonumber
\end{eqnarray}
Condition of normalization of the above Alice's resultant state is as follows:
\begin{eqnarray}
|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} {}\nonumber&&
\end{eqnarray}
In order to find Alice's Local Operations state for retrieving the secret, we proceed by taking a Unitary Matrix as follows:
\begin{equation*}
U =
\begin{bmatrix}
a_{1} + ib_{1} & c_{1} - id_{1} \\
c_{1} + id_{1} & -a_{1} + ib_{1} \\
\end{bmatrix}
\end{equation*}
such that det(U) = -1
\\
\\
We need to find $a_{1}, b_{1}, c_{1}, d_{1}$ by the solving the following equation for Alice to retrieve the secret:

\[
\begin{bmatrix}
    a_{1} + ib_{1} & c_{1} - id_{1} \\
    c_{1} + id_{1} & -a_{1} + ib_{1}      \\

\end{bmatrix}
\begin{bmatrix}
    \sqrt{2}(a\alpha + b\beta) \\
    \sqrt{2}(b\alpha + a\beta)    \\

\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
\\
As $a, b$, $\alpha$ and $\beta$ are complex numbers, we write $a = x_{1} + iy_{1}, b = x_{2} + iy_{2}, \alpha = x_{3} + iy_{3}, \beta = x_{4} + iy_{4}$.
\\
The above equation transforms as follows:
\[
\begin{bmatrix}
    a_{1} + ib_{1} & c_{1} - id_{1} \\
    c_{1} + id_{1} & -a_{1} + ib_{1}      \\

\end{bmatrix}
\begin{bmatrix}
    \sqrt{2}(\lambda_{1} + i\lambda_{2}) \\
    \sqrt{2}(\lambda_{3} + i\lambda_{4})    \\

\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
\\
Where we have:\\
\begin{eqnarray}
\lambda_{1} = x_{1}x_{3} + x_{2}x_{4} - y_{1}y_{3} - y_{2}y_{4}{}\nonumber\\
\lambda_{2} = x_{3}y_{1} + x_{1}y_{3} + x_{2}y_{4} + x_{4}y_{2}{}\nonumber\\
\lambda_{3} = x_{2}x_{3} + x_{1}x_{4} - y_{1}y_{4} - y_{2}y_{3}{}\nonumber\\
\lambda_{4} = x_{3}y_{2} + x_{4}y_{1} + x_{2}y_{3} + x_{1}y_{4}{}\nonumber
\end{eqnarray}
\\
Such that:\\
\begin{eqnarray}
\lambda_{1}^2 + \lambda_{2}^2 + \lambda_{3}^2 + \lambda_{4}^2 = |a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta) =  \frac{1}{2}
\end{eqnarray}
Solving the above equation by comparing real and imaginary parts on both sides to get:
\begin{eqnarray}
a_{1} = \frac{\lambda_{1}x_{3} + \lambda_{2}y_{3} - \lambda_{3}x_{4} - \lambda_{4}y{4}}{\sqrt{2}{(\lambda_{1}^2 + \lambda_{2}^2 + \lambda_{3}^2 + \lambda_{4}^2)}} = \sqrt{2}(\lambda_{1}x_{3} + \lambda_{2}y_{3} - \lambda_{3}x_{4} - \lambda_{4}y_{4})} {}\nonumber\\
b_{1} = \frac{\lambda_{1}y_{3} - \lambda_{2}x_{3} + \lambda_{3}y_{4} - \lambda_{4}x_{4}}{\sqrt{2}{(\lambda_{1}^2 + \lambda_{2}^2 + \lambda_{3}^2 + \lambda_{4}^2)}} = \sqrt{2}(\lambda_{1}y_{3} - \lambda_{2}x_{3} + \lambda_{3}y_{4} - \lambda_{4}x_{4})} {}\nonumber\\
c_{1} = \frac{\lambda_{1}x_{4} + \lambda_{2}y_{4} + \lambda_{3}x_{3} + \lambda_{4}y_{3}}{\sqrt{2}{(\lambda_{1}^2 + \lambda_{2}^2 + \lambda_{3}^2 + \lambda_{4}^2)}} = \sqrt{2}(\lambda_{1}x_{4} + \lambda_{2}y_{4} + \lambda_{3}x_{3} + \lambda_{4}y_{3})} {}\nonumber\\
d_{1} = \frac{\lambda_{4}x_{3} - \lambda_{3}y_{3} + \lambda_{1}y_{4} - \lambda_{2}x_{4}}{\sqrt{2}{(\lambda_{1}^2 + \lambda_{2}^2 + \lambda_{3}^2 + \lambda_{4}^2)}} = \sqrt{2}(\lambda_{4}x_{3} - \lambda_{3}y_{3} + \lambda_{1}y_{4} - \lambda_{2}x_{4})} {}\nonumber
\end{eqnarray}
\begin{eqnarray}
a_{1} + ib_{1} = \sqrt{2}\{(\lambda_{1} - i\lambda_{2})\alpha - (\lambda_{3} + i\lambda_{4})\tilde{\beta}\} {}\nonumber\\
= \sqrt{2}\{(\tilde{a}\tilde{\alpha} + \tilde{b}\tilde{\beta})\alpha - (b\alpha + a\beta)\tilde{\beta}\} {}\nonumber\\
= \sqrt{2}\{(\tilde{a} - (\tilde{a} + a)|\beta|^2 + (\tilde{b} - b)\alpha\tilde{\beta})\}
\end{eqnarray}
\begin{eqnarray}
-a_{1} + ib_{1} = \sqrt{2}\{-(\lambda_{1} + i\lambda_{2})\tilde{\alpha} + (\lambda_{3} - i\lambda_{4})\beta\} {}\nonumber\\
= \sqrt{2}\{-(a\alpha + b\beta)\tilde{\alpha} + (\tilde{a}\tilde{\beta} + \tilde{b}\tilde{\alpha})\beta\} {}\nonumber\\
= \sqrt{2}\{-a + (\tilde{a} + a)|\beta|^2 + (\tilde{b} - b)\tilde{\alpha}\beta\}
\end{eqnarray}
\begin{eqnarray}
c_{1} + id_{1} = \sqrt{2}\{(\lambda_{1} - i\lambda_{2})\beta + (\lambda_{3} + i\lambda_{4})\tilde{\alpha}\} {}\nonumber\\
= \sqrt{2}\{(\tilde{a}\tilde{\alpha} + \tilde{b}\tilde{\beta})\beta + (b\alpha + a\beta)\tilde{\alpha}\} {}\nonumber\\
= \sqrt{2}\{\tilde{b} + (\tilde{a} + a)\tilde{\alpha}\beta - (\tilde{b} - b)|\alpha|^2\}
\end{eqnarray}
\begin{eqnarray}
c_{1} - id_{1} = \sqrt{2}\{(\lambda_{1} + i\lambda_{2})\tilde{\beta} + (\lambda_{3} - i\lambda_{4})\alpha\} {}\nonumber\\
= \sqrt{2}\{(a\alpha + b\beta)\tilde{\beta} + (\tilde{a}\tilde{\beta} + \tilde{b}\tilde{\alpha})\alpha\} {}\nonumber\\
= \sqrt{2}\{b + (\tilde{a} + a)\alpha\tilde{\beta} + (\tilde{b} - b)|\alpha|^2\}
\end{eqnarray}
Using equations (31), (32), (33) and (34), we obtain the Unitary matrix as follows:\\
\begin{equation*}
\begin{bmatrix}
    \sqrt{2}\{(\tilde{a} - (\tilde{a} + a)|\beta|^2 + (\tilde{b} - b)\alpha\tilde{\beta})\} & \sqrt{2}\{b + (\tilde{a} + a)\alpha\tilde{\beta} + (\tilde{b} - b)|\alpha|^2\} \\
    \sqrt{2}\{\tilde{b} + (\tilde{a} + a)\tilde{\alpha}\beta - (\tilde{b} - b)|\alpha|^2\} & \sqrt{2}\{-a + (\tilde{a} + a)|\beta|^2 + (\tilde{b} - b)\tilde{\alpha}\beta\}      \\

\end{bmatrix}
\end{equation*}
\\
\\
Since the elements of the matrix need to be independent of $\alpha$ and $\beta$, we have the following conditions:\\
\begin{eqnarray}
\tilde{a} + a = 0
\end{eqnarray}
\begin{eqnarray}
\tilde{b} - b = 0
\end{eqnarray}
From eq (14) we have $a$ to be purely imaginary and from eq (15) we have $b$ to be purely real.
After the imposition of the above conditions eq (14) and (15), the Unitary Matrix reduces to:\\
\begin{equation*}
U =
\begin{bmatrix}
    \sqrt{2}\tilde{a} & \sqrt{2}b \\
    \sqrt{2}\tilde{b} & -\sqrt{2}a      \\

\end{bmatrix}
= \begin{bmatrix}
    -\sqrt{2}a & \sqrt{2}b \\
    \sqrt{2}b & -\sqrt{2}a      \\

\end{bmatrix}
\end{equation*}
Where we made use of eq (14) and eq (15) above, $\tilde{a} + a = 0$ or $\tilde{a} = -a$ and $\tilde{b} - b = 0$ or $\tilde{b} = b$
\\
\\
det(U) = $2a^2 - 2b^2 = -1$ or $2b^2 - 2a^2 = 1$\\
\\
Using the above matrix, Alice successfully retrieves the secret as follows:
\[
\begin{bmatrix}
    -\sqrt{2}a & \sqrt{2}b \\
    \sqrt{2}b & -\sqrt{2}a      \\

\end{bmatrix}
\begin{bmatrix}
    \sqrt{2}}(a\alpha + b\beta) \\
    \sqrt{2}}(b\alpha + a\beta)    \\

\end{bmatrix}
=
\begin{bmatrix}
    2(-a^2\alpha - ab\beta + b^2\alpha + ab\beta)      \\
    2(ab\alpha + b^2\beta -ab\alpha - a^2\beta )
\end{bmatrix}
=
\begin{bmatrix}
    2(b^2 - a^2)\alpha      \\
    2(b^2 - a^2)\beta
\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
Expressing the above matrix in terms of Pauli Matrices, we have U = $\sqrt{2}(-aI + b\sigma_x)$}
\\
\\
Now as $2b^2 - 2a^2 = 1$, we have $a^2 = b^2 - \frac{1}{2}$
\\
\\
Since $a$ is purely imaginary, we must have $b^2 < \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} < b < \frac{1}{\sqrt{2}}$
Now from the condition of normalization of Alice's resultant state, we have:
\begin{eqnarray}
|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} {}\nonumber&&
\end{eqnarray}
As $b = \tilde{b}$, the above equation reduces to:
\begin{eqnarray}
|a|^2+|b|^2+b(\tilde{a}+a)(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} {}\nonumber
\end{eqnarray}
And as $\tilde{a}+a$ = 0, the above equation further reduces to:
\begin{eqnarray}
|a|^2+|b|^2 = \frac{1}{2} {}\nonumber&&
\end{eqnarray}
As $b$ is a real number, $b^2 = |b|^2$ and so we have:
\begin{eqnarray}
|a|^2+b^2 =  \frac{1}{2}{}\nonumber &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 =  \frac{1}{2} - b^2 {}\nonumber&&
\end{eqnarray}
\\
\\
\\
\\
\newpage
\noindent\textit{\bf Appendix 2:} (Proof of Theorem 2)\\
\item
For Charlie to successfully reconstruct the secret, Alice and Bob must measure their respective qubits in the Hadamard Basis. In TABLE II we show how Charlie can reconstruct the secret by enlisting down the respective local operations corresponding to Alice's and Bob's measurement outcomes. Consider Charlie's resultant state outcome in Hadamard Basis $\ket{+}\ket{+}$:
\begin{eqnarray}
\frac{1}{\sqrt{2}}\{[(a+b-c+d)\alpha+(a+b+c-d)\beta]|0\rangle\}+\frac{1}{\sqrt{2}}\{[(a+b+c-d)\alpha$ + $(a+b-c+d)\beta]|1\rangle\} {}\nonumber&
\end{eqnarray}
Denoting $(a+b-c+d)$ as $\lambda_1$ and $(a+b+c-d)$ as $\lambda_2$, the above resultant state can be written as:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(\lambda_1\alpha+\lambda_2\beta)|0\rangle + \frac{1}{\sqrt{2}}(\lambda_2\alpha+\lambda_1\beta)|1\rangle {}\nonumber
\end{eqnarray}
Condition of normalization of the above Charlie's resultant state is as follows:
\begin{eqnarray}
\frac{1}{2}\{|\lambda_1\alpha+\lambda_2\beta|^2 + |\lambda_2\alpha+\lambda_1\beta|^2 \} = 1 {}\nonumber&&
\end{eqnarray}
Simplifying, we get:
\begin{eqnarray}
|\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta) = 2 {}\nunumber&&
\end{eqnarray}
\\
In order to find Charlie's Local Operations state for retrieving the secret, we proceed by taking a Unitary Matrix as follows:
\begin{equation*}
U =
\begin{bmatrix}
a_{1} + ib_{1} & c_{1} - id_{1} \\
c_{1} + id_{1} & -a_{1} + ib_{1} \\
\end{bmatrix}
\end{equation*}
such that det(U) = -1
\\
\\
We need to find $a_{1}, b_{1}, c_{1}, d_{1}$ by the solving the following equation for Alice to retrieve the secret:
\[
\begin{bmatrix}
    a_{1} + ib_{1} & c_{1} - id_{1} \\
    c_{1} + id_{1} & -a_{1} + ib_{1}      \\

\end{bmatrix}
\begin{bmatrix}
    \frac{1}{\sqrt{2}}(\lambda_{1}\alpha + \lambda_{2}\beta) \\
    \frac{1}{\sqrt{2}}(\lambda_{2}\alpha + \lambda_{1}\beta)    \\

\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
\\
As $a$, $b$, $c$, $d$, $\alpha$ and $\beta$ are complex numbers, we write $\lambda_{1} = x_{1} + iy_{1}, \lambda_{2} = x_{2} + iy_{2}, \alpha = x_{3} + iy_{3}, \beta = x_{4} + iy_{4}$.
\\
\\
The above equation transforms as follows:
\[
\begin{bmatrix}
    a_{1} + ib_{1} & c_{1} - id_{1} \\
    c_{1} + id_{1} & -a_{1} + ib_{1}      \\

\end{bmatrix}
\begin{bmatrix}
    \frac{1}{\sqrt{2}}(\nu_{1} + i\nu_{2}) \\
    \frac{1}{\sqrt{2}}(\nu_{3} + i\nu_{4})    \\

\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
\\
Where we have:\\
\begin{eqnarray}
\nu_{1} = x_{1}x_{3} + x_{2}x_{4} - y_{1}y_{3} - y_{2}y_{4}{}\nonumber\\
\nu_{2} = x_{3}y_{1} + x_{1}y_{3} + x_{2}y_{4} + x_{4}y_{2}{}\nonumber\\
\nu_{3} = x_{2}x_{3} + x_{1}x_{4} - y_{1}y_{4} - y_{2}y_{3}{}\nonumber\\
\nu_{4} = x_{3}y_{2} + x_{4}y_{1} + x_{2}y_{3} + x_{1}y_{4}{}\nonumber
\end{eqnarray}
\\
Such that:\\
\begin{eqnarray}
\nu_{1}^2 + \nu_{2}^2 + \nu_{3}^2 + \nu_{4}^2 = |\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta) =  2
\end{eqnarray}
Solving the above equation by comparing real and imaginary parts on both sides to get:
\begin{eqnarray}
a_{1} = \frac{\sqrt{2}(\nu_{1}x_{3} + \nu_{2}y_{3} - \nu_{3}x_{4} - \nu_{4}y_{4})}{\nu_{1}^2 + \nu_{2}^2 + \nu_{3}^2 + \nu_{4}^2}} = \frac{ \nu_{1}x_{3} + \nu_{2}y_{3} - \nu_{3}x_{4} - \nu_{4}y_{4}}{\sqrt{2}} {}\nonumber\\
b_{1} = \frac{\sqrt{2}(\nu_{1}y_{3} - \nu_{2}x_{3} + \nu_{3}y_{4} - \nu_{4}x_{4})}{\nu_{1}^2 + \nu_{2}^2 + \nu_{3}^2 + \nu_{4}^2}} = \frac{ \nu_{1}y_{3} - \nu_{2}x_{3} + \nu_{3}y_{4} - \nu_{4}x_{4}}{\sqrt{2}} {}\nonumber\\
c_{1} = \frac{{\sqrt{2}}(\nu_{1}x_{4} + \nu_{2}y_{4} + \nu_{3}x_{3} + \nu_{4}y_{3})}{(\nu_{1}^2 + \nu_{2}^2 + \nu_{3}^2 + \nu_{4}^2)}} = \frac{ \nu_{1}x_{4} + \nu_{2}y_{4} + \nu_{3}x_{3} + \nu_{4}y_{3}}{\sqrt{2}} {}\nonumber\\
d_{1} = \frac{{\sqrt{2}}(\nu_{4}x_{3} - \nu_{3}y_{3} + \nu_{1}y_{4} - \nu_{2}x_{4})}{(\nu_{1}^2 + \nu_{2}^2 + \nu_{3}^2 + \nu_{4}^2)}} = \frac{ \nu_{4}x_{3} - \nu_{3}y_{3} + \nu_{1}y_{4} - \nu_{2}x_{4}}{\sqrt{2}} {}\nonumber
\end{eqnarray}
\begin{eqnarray}
a_{1} + ib_{1} = \frac{1}{\sqrt{2}}\{(\nu_{1} - i\nu_{2})\alpha - (\nu_{3} + i\nu_{4})\tilde{\beta}\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{(\tilde{\lambda_1}\tilde{\alpha} + \tilde{\lambda_1}\tilde{\beta})\alpha - (\lambda_2\alpha + \lambda_1\beta)\tilde{\beta}\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{(\tilde{\lambda_1} - (\tilde{\lambda_1} + \lambda_1)|\beta|^2 + (\tilde{\lambda_2} - \lambda_2)\alpha\tilde{\beta})\}
\end{eqnarray}
\begin{eqnarray}
-a_{1} + ib_{1} = \frac{1}{\sqrt{2}}\{-(\nu_{1} + i\nu_{2})\tilde{\alpha} + (\nu_{3} - i\nu_{4})\beta\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{-(\lambda_1\alpha + \lambda_2\beta)\tilde{\alpha} + (\tilde{\lambda_1}\tilde{\beta} + \tilde{\lambda_2}\tilde{\alpha})\beta\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{-\lambda_1 + (\tilde{\lambda_1} + \lambda_1)|\beta|^2 + (\tilde{\lambda_2} - \lambda_2)\tilde{\alpha}\beta\}
\end{eqnarray}
\begin{eqnarray}
c_{1} + id_{1} = \frac{1}{\sqrt{2}}\{(\nu_{1} - i\nu_{2})\beta + (\nu_{3} + i\nu_{4})\tilde{\alpha}\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{(\tilde{\lambda_1}\tilde{\alpha} + \tilde{\lambda_2}\tilde{\beta})\beta + (\lambda_2\alpha + \lambda_1\beta)\tilde{\alpha}\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{\tilde{\lambda_2} + (\tilde{\lambda_1} + \lambda_1)\tilde{\alpha}\beta - (\tilde{\lambda_2} - \lambda_2)|\alpha|^2\}
\end{eqnarray}
\begin{eqnarray}
c_{1} - id_{1} = \frac{1}{\sqrt{2}}\{(\nu_{1} + i\nu_{2})\tilde{\beta} + (\nu_{3} - i\nu_{4})\alpha\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{(\lambda_1\alpha + \lambda_2\beta)\tilde{\beta} + (\tilde{\lambda_1}\tilde{\beta} + \tilde{\lambda_2}\tilde{\alpha})\alpha\} {}\nonumber\\
= \frac{1}{\sqrt{2}}\{\lambda_2 + (\tilde{\lambda_1} + \lambda_1)\alpha\tilde{\beta} + (\tilde{\lambda_2} - \lambda_2)|\alpha|^2\}
\end{eqnarray}
Using equations (18), (19), (20) and (21), we obtain the Unitary matrix as follows:\\
\begin{equation*}
\begin{bmatrix}
    \frac{1}{\sqrt{2}}\{(\tilde{\lambda_1} - (\tilde{\lambda_1} + \lambda_1)|\beta|^2 + (\tilde{\lambda_2} - \lambda_2)\alpha\tilde{\beta})\} & \frac{1}{\sqrt{2}}\{\lambda_2 + (\tilde{\lambda_1} + \lambda_1)\alpha\tilde{\beta} + (\tilde{\lambda_2} - \lambda_2)|\alpha|^2\} \\
    \frac{1}{\sqrt{2}}\{\tilde{\lambda_2} + (\tilde{\lambda_1} + \lambda_1)\tilde{\alpha}\beta - (\tilde{\lambda_2} - \lambda_2)|\alpha|^2\} & \frac{1}{\sqrt{2}}\{-\lambda_1 + (\tilde{\lambda_1} + \lambda_1)|\beta|^2 + (\tilde{\lambda_2} - \lambda_2)\tilde{\alpha}\beta\}      \\

\end{bmatrix}
\end{equation*}
\\
\\
Since the elements of the matrix need to be independent of $\alpha$ and $\beta$, we have the following conditions:\\
\begin{eqnarray}
\tilde{\lambda_1} + \lambda_1 = 0
\end{eqnarray}
\begin{eqnarray}
\tilde{\lambda_2} - \lambda_2 = 0
\end{eqnarray}
From eq (22) we have $\lambda_1$ or $(a+b-c+d)$ to be purely imaginary and from eq (23) we have $\lambda_2$ or $(a+b+c-d)$ to be purely real.
After the imposition of the above conditions eq (22) and (23), the Unitary Matrix reduces to:\\
\begin{equation*}
U =
\begin{bmatrix}
    \frac{1}{\sqrt{2}}\tilde{\lambda_1} & \frac{1}{\sqrt{2}}\lambda_2 \\
    \frac{1}{\sqrt{2}}\tilde{\lambda_2} & -\frac{1}{\sqrt{2}}\lambda_1      \\

\end{bmatrix}
= \begin{bmatrix}
    -\frac{1}{\sqrt{2}}\lambda_1 & \frac{1}{\sqrt{2}}\lambda_2 \\
    \frac{1}{\sqrt{2}}\lambda_2 & -\frac{1}{\sqrt{2}}\lambda_1      \\

\end{bmatrix}
\end{equation*}
Where we made use of eq (22) and eq (23) above, $\tilde{\lambda_1} + \lambda_1 = 0$ or $\tilde{\lambda_1} = -\lambda_1$ and $\tilde{\lambda_2} - \lambda_2 = 0$ or $\tilde{\lambda_2} = \lambda_2$
\\
\\
det(U) = $\frac{1}{2}\lambda_1^2 - \frac{1}{2}\lambda_2^2 = -1$ or $\frac{1}{2}\lambda_2^2 - \frac{1}{2}\lambda_1^2 = 1$\\
\\
Using the above matrix, Alice successfully retrieves the secret as follows:
\[
\begin{bmatrix}
    -\frac{1}{\sqrt{2}}\lambda_1 & \frac{1}{\sqrt{2}}\lambda_2 \\
    \frac{1}{\sqrt{2}}\lambda_2 & -\frac{1}{\sqrt{2}}\lambda_1      \\

\end{bmatrix}
\begin{bmatrix}
    \frac{1}{\sqrt{2}}(\lambda_1\alpha + \lambda_2\beta) \\
    \frac{1}{\sqrt{2}}(\lambda_2\alpha + \lambda_1\beta)    \\

\end{bmatrix}
=
\begin{bmatrix}
    \frac{1}{2}(-\lambda_1^2\alpha - \lambda_1\lambda_2\beta + \lambda_2^2\alpha + \lambda_1\lambda_2\beta)      \\
    \frac{1}{2}(\lambda_1\lambda_2\alpha + \lambda_2^2\beta -\lambda_1\lambda_2\alpha - \lambda_1^2\beta )
\end{bmatrix}
=
\begin{bmatrix}
    \frac{1}{2}(\lambda_2^2 - \lambda_1^2)\alpha      \\
    \frac{1}{2}(\lambda_2^2 - \lambda_1^2)\beta
\end{bmatrix}
=
\begin{bmatrix}
    \alpha      \\
    \beta
\end{bmatrix}
\]
Expressing the above matrix in terms of Pauli Matrices, we have U = $\frac{1}{\sqrt{2}}(-\lambda_1I + \lambda_2\sigma_x)$ or U = $\frac{1}{\sqrt{2}}[-(a+b-c+d)I + (a+b+c-d)\sigma_x]$
\\
\\
From eq (22) we derived $\lambda_1$ or $(a+b-c+d)$ to be purely imaginary, so we have the following conditions of $a$, $b$, $c$ and $d$ from Alice's retrieval derived in Appendix 1:\\
\\
\textbf{Case 1:} $(a+b-c+d)$ is purely imaginary\\
If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have $a$ = $c$\\
If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have $a$ = $-d$\\
If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have $b$ = $c$\\
If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have $b$ = $-d$\\
\\
From eq (23) we derived $\lambda_1$ or $(a+b+c-d)$ to be purely real, so we have the following conditions of $a$, $b$, $c$ and $d$ from Alice's retrieval derived in Appendix 1:\\
\\
\textbf{Case 2:} $(a+b+c-d)$ is purely real\\
If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have $b$ = $d$\\
If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have $b$ = $-c$\\
If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have $a$ = $d$\\
If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have $a$ = $-c$\\
\\

\\
\\
\\
\\

\end{widetext}
\end{document}

\begin{appendix}
\begin{widetext}
\noindent\textit{\bf Appendix 1:} (Proof of Lemma 1)\\
\\
Alice's resultant state obtained in the Hadamard Basis $\ket{+}\ket{+}$, when Alice's initial measurement outcome is $\ket{\phi}$ is:
$\sqrt{2}\{(a\alpha + b\beta)|0\rangle + (b\alpha + a\beta)|1\rangle\}$ with the  normalization condition as,
$|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2}$
For the QSS protocol to exist, the condition of normalization obtained  above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde{a}b + a\tilde{b} = 0$, which means that either $a=0$ or $b=0$, but both $a$ and $b$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{+}$ and $\ket{-}\ket{-}$.
\\
\\
\textbf{Case 1:}\\
If $a = 0$, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}b\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 2:}\\
If $b$ = 0, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}a\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow \ I$ &&
\end{eqnarray}
\\
From Alice's resultant state obtained in the Hadamard Basis $\ket{+}\ket{-}$, the condition of normalization is as follows:
\begin{eqnarray}
2\{|d\alpha + c\beta|^2+|c\alpha + d\beta|^2\} = 1 &&
\end{eqnarray}
or
\begin{eqnarray}
|c|^2+|d|^2+(\tilde{c}d+c\tilde{d})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} &&
\end{eqnarray}
For the QSS protocol to exist, the condition of normalization obtained in (41) above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde{c}d + c\tilde{d} = 0$, which means that either $c=0$ or $d=0$, but both $c$ and $d$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{-}$ and $\ket{-}\ket{+}$.
\\
\\
\\
\textbf{Case 3:}\\
If $c = 0$, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{-}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}d\{\alpha|0\rangle - \beta|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|d|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle - \beta|1\rangle \rightarrow \sigma_z$ &&
\end{eqnarray}
\\
\textbf{Case 4:}\\
If $d$ = 0, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{-}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}c\{\beta|0\rangle - \alpha|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|c|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle - \alpha|1\rangle \rightarrow \ \sigma_x\sigma_z$ &&
\end{eqnarray}
\\
From Cases 1, 2, 3 and 4 we have the following conditions:\\
\begin{eqnarray}
1.\ $a = 0$, \  $c = 0$, \  $b \ne 0$, \  $d \ne 0$, \  $|b|^2 = \frac{1}{2}$, \  $|d|^2 = \frac{1}{2}$ &&
\\
2.\ $a = 0$, \  $d = 0$, \  $b \ne 0$, \  $c \ne 0$, \  $|b|^2 = \frac{1}{2}$, \  $|c|^2 = \frac{1}{2}$ &&
\\
3.\ $b = 0$, \  $c = 0$, \  $a \ne 0$, \  $c \ne 0$, \  $|a|^2 = \frac{1}{2}$, \  $|d|^2 = \frac{1}{2}$ &&
\\
4.\ $b = 0$, \  $d = 0$, \  $a \ne 0$, \  $d \ne 0$, \  $|a|^2 = \frac{1}{2}$, \  $|c|^2 = \frac{1}{2}$ &&
\end{eqnarray}
\\
\noindent\textit{\bf Appendix 2:} (Proof of Lemma 2)\\
\\
Charlie's resultant state obtained in the Hadamard Basis $\ket{+}\ket{+}$, when Alice's initial measurement outcome is $\ket{\phi}$ is:
\begin{eqnarray}
\frac{1}{\sqrt{2}}\{[(a+b-c+d)\alpha+(a+b+c-d)\beta]|0\rangle\}+\frac{1}{\sqrt{2}}\{[(a+b+c-d)\alpha$ + $(a+b-c+d)\beta]|1\rangle\} &
\end{eqnarray}
Denoting $(a+b-c+d)$ as $\lambda_1$ and $(a+b+c-d)$ as $\lambda_2$, (51) can be written as:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(\lambda_1\alpha+\lambda_2\beta)|0\rangle + \frac{1}{\sqrt{2}}(\lambda_2\alpha+\lambda_1\beta)|1\rangle
\end{eqnarray}
Condition of normalization of (57) is as follows:
\begin{eqnarray}
\frac{1}{2}\{|\lambda_1\alpha+\lambda_2\beta|^2 + |\lambda_2\alpha+\lambda_1\beta|^2 \} = 1 &&
\end{eqnarray}
or
\begin{eqnarray}
|\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta) = 2 &&
\end{eqnarray}
\\
For the QSS protocol to exist, the condition of normalization obtained in (59) above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2 = 0$, which means that either $\lambda_1 = 0$ or $\lambda_2 = 0$, but both $\lambda_1$ and $\lambda_2$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{+}$ and $\ket{-}\ket{-}$ So it means either $(a+b-c+d) = 0$ or $(a+b+c-d) = 0$.
\\
\\
\textbf{Case 1:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (57) above if we take $a = 0$ and $c = 0$, we have $b + d = 0$ or $b = -d$ or $-b = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $c = 0$ and $b = -d$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $c = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
If we take $b = -d$, $a + b + c - d = 2b$ or $a + b + c - d = -2d$ or $\lambda_2 = 2b$ or $\lambda_2= -2d$\\
\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\sqrt{2}b\{\beta|0\rangle+\alpha|1\rangle\} \rightarrow \sigma_x$ &&
\end{eqnarray}
or
\begin{eqnarray}
$-\sqrt{2}d\{\beta|0\rangle+\alpha|1\rangle\} \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 2:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (57) above if we take $a = 0$ and $c = 0$, we have $b - d = 0$ or $b = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $c = 0$ and $b = d$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $c = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 3:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (58) above if we take $a = 0$ and $d = 0$, we have $b - c = 0$ or $b = c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $d = 0$ and $b = c$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $d = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 4:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (58) above if we take $a = 0$ and $d = 0$, we have $b + c = 0$ or $b = -c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $d = 0$ and $b = -c$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $d = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 5:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (59) above if we take $b = 0$ and $c = 0$, we have $a + d = 0$ or $a = -d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $c = 0$ and $a = -d$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $c = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 6:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (59) above if we take $b = 0$ and $c = 0$, we have $a - d = 0$ or $a = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $c = 0$ and $a = d$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 7:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (60) above if we take $b = 0$ and $d = 0$, we have $a - c = 0$ or $a = c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $d = 0$ and $a = c$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 8:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (60) above if we take $b = 0$ and $d = 0$, we have $a + c = 0$ or $a = -c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $d = 0$ and $a = -c$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\\
\noindent\textit{\bf Appendix 3:} (Proof of Theorem 1)\\
\\
Combining the results of Lemma 1 and Lemma 2, we establish Theorem 3
\\

\begin{table}[h]
\caption{\bf Reconstruction of Quantum Information by Charlie after the measurement of Alice and Bob in the Hadamard Basis
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Bob and Charlie's Outcomes\\ \ \\} &\parbox{.75in}{\ \\$a$\\ \ \\} &\parbox{.75in}{\ \\$b$\\ \ \\} &\parbox{.75in}{\ \\$c$\\ \ \\} &\parbox{.75in}{\ \\$d$\\ \ \\} &\parbox{.75in}{\ \\Charlie's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Charlie's Local Operations\ \\ \ \\}\\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|0\rangle-\beta|0\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}

\end{widetext}

\end{document}
\newpage
\begin{appendix}
\begin{widetext}
\noindent\textit{\bf Appendix 1:}\\
Consider the 4-qubit ${G}_{abcd}$ state:\\
\\
 $= \frac{a + d}{{2}}(|0000\rangle+|1111\rangle) + \frac{a - d}{{2}}(|1100\rangle+|0011\rangle) + \frac{b + c}{{2}}(|0101\rangle+|1010\rangle) + \frac{b - c}{{2}}(|0110\rangle+|1001\rangle)$\\
\\
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie and let $|\psi\rangle = \alpha|0\rangle+\beta|1\rangle$ be the message with Alice. We first consider the case where a, b, c and d are real numbers.\\
\end{widetext}
\begin{table}[h]
\caption{\bf Retrieving Quantum Information after re-normalization of Alice's resultant state when ($a = 0$ and $c = 0$) and ($b =\frac{1}{\sqrt{2}}$ and $d = {\frac{1}{\sqrt{2}}}$)}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Bob and Charlie's Outcomes\\ \ \\} &\parbox{.75in}{\ \\Alice's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Alice's Local Operations\ \\ \ \\}\\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $\ket{\phi^{+}}$  & $\ket{-}\ket{+}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $\ket{\phi^{+}}$  & $\ket{-}\ket{-}$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$  \\
\hline $\ket{\phi^{-}}$  & $\ket{+}\ket{+}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $\ket{\phi^{-}}$  & $\ket{+}\ket{-}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$ \\
\hline $\ket{\phi^{-}}$  & $\ket{-}\ket{+}$ & $\alpha|0 \rangle+\beta|1\rangle$ & $I$ \\
\hline $\ket{\phi^{-}}$  & $\ket{-}\ket{-}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$ \\
\hline $\ket{\psi^{+}}$  & $\ket{+}\ket{+}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\psi^{+}}$  & $\ket{+}\ket{-}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $\ket{\psi^{+}}$  & $\ket{-}\ket{+}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\psi^{+}}$  & $\ket{-}\ket{-}$ & $-\alpha|0\rangle-\beta|1\rangle$ & $I$    \\
\hline $\ket{\psi^{-}}$  & $\ket{+}\ket{+}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$    \\
\hline $\ket{\psi^{-}}$  & $\ket{+}\ket{-}$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$  \\
\hline $\ket{\psi^{-}}$  & $\ket{-}\ket{+}$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\psi^{-}}$  & $\ket{-}\ket{-}$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{table}[h]
\caption{\bf Retrieving Quantum Information after re-normalization of Charlie's resultant state when ($a = 0$ and $c = 0$) and ($b =\frac{1}{\sqrt{2}}$ and $d =\frac{1}{\sqrt{2}}$)}.
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Alice and Bob's Outcomes\\ \ \\} &\parbox{.75in}{\ \\Charlie's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Charlie's Local Operations\ \\ \ \\}\\
\hline $|\phi^+\rangle$  & $|++\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $|\phi^+\rangle$  & $|+-\rangle$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $|\phi^+\rangle$  & $|-+\rangle$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $|\phi^+\rangle$  & $|--\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$  \\
\hline $|\phi^-\rangle$  & $|++\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $|\phi^-\rangle$  & $|+-\rangle$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$ \\
\hline $|\phi^-\rangle$  & $|-+\rangle$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$ \\
\hline $|\phi^-\rangle$  & $|--\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$ \\
\hline $|\psi^+\rangle$  & $|++\rangle$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $|\psi^+\rangle$  & $|+-\rangle$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$  \\
\hline $|\psi^+\rangle$  & $|-+\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$    \\
\hline $|\psi^+\rangle$  & $|--\rangle$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $|\psi^-\rangle$  & $|++\rangle$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $|\psi^-\rangle$  & $|+-\rangle$ & $-\alpha|0\rangle-\beta|1\rangle$ & $I$  \\
\hline $|\psi^-\rangle$  & $|-+\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $|\psi^-\rangle$  & $|--\rangle$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}
\newpage

\begin{table}[H]
\caption{\bf Retrieving Quantum Information after re-normalization of Alice's resultant state when ($b = 0$ and $d = 0$) and ($a =\frac{1}{\sqrt{2}}$ and $c = {\frac{1}{\sqrt{2}}}$)}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Bob and Charlie's Outcomes\\ \ \\} &\parbox{.75in}{\ \\Alice's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Alice's Local Operations\ \\ \ \\}\\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $\ket{\phi^{+}}$  & $\ket{-}\ket{+}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $\ket{\phi^{+}}$  & $\ket{-}\ket{-}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$  \\
\hline $\ket{\phi^{-}}$  & $\ket{+}\ket{+}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $\ket{\phi^{-}}$  & $\ket{+}\ket{-}$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$ \\
\hline $\ket{\phi^{-}}$  & $\ket{-}\ket{+}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$ \\
\hline $\ket{\phi^{-}}$  & $\ket{-}\ket{-}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$ \\
\hline $\ket{\psi^{+}}$  & $\ket{+}\ket{+}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\psi^{+}}$  & $\ket{+}\ket{-}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $\ket{\psi^{+}}$  & $\ket{-}\ket{+}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\psi^{+}}$  & $\ket{-}\ket{-}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\psi^{-}}$  & $\ket{+}\ket{+}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\psi^{-}}$  & $\ket{+}\ket{-}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$  \\
\hline $\ket{\psi^{-}}$  & $\ket{-}\ket{+}$ & $-\alpha|0\rangle-\beta|1\rangle$ & $I$    \\
\hline $\ket{\psi^{-}}$  & $\ket{-}\ket{-}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}[H]
\caption{\bf Retrieving Quantum Information after re-normalization of Charlie's resultant state when ($b = 0$ and $d = 0$) and ($a =\frac{1}{\sqrt{2}}$ and $c =\frac{1}{\sqrt{2}}$)}.
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Alice and Bob's Outcomes\\ \ \\} &\parbox{.75in}{\ \\Charlie's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Charlie's Local Operations\ \\ \ \\}\\
\hline $|\phi^+\rangle$  & $|++\rangle$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $|\phi^+\rangle$  & $|+-\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $|\phi^+\rangle$  & $|-+\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$  \\
\hline $|\phi^+\rangle$  & $|--\rangle$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$  \\
\hline $|\phi^-\rangle$  & $|++\rangle$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $|\phi^-\rangle$  & $|+-\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$ \\
\hline $|\phi^-\rangle$  & $|-+\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$ \\
\hline $|\phi^-\rangle$  & $|--\rangle$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$ \\
\hline $|\psi^+\rangle$  & $|++\rangle$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $|\psi^+\rangle$  & $|+-\rangle$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$  \\
\hline $|\psi^+\rangle$  & $|-+\rangle$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $|\psi^+\rangle$  & $|--\rangle$ & $-\alpha|0\rangle-\beta|1\rangle$ & $I$    \\
\hline $|\psi^-\rangle$  & $|++\rangle$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$    \\
\hline $|\psi^-\rangle$  & $|+-\rangle$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$  \\
\hline $|\psi^-\rangle$  & $|-+\rangle$ & $-\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x$    \\
\hline $|\psi^-\rangle$  & $|--\rangle$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{table}[H]
\caption{\bf List of some examples of the $G_{abcd}$ state using values of $a$, $b$, $c$, $d$ from Table 3.11 above for which there is a possible Quantum Secret Sharing Protocol. All the resource states are normalized.}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline \parbox{.75in}{\ \\$a$\\ \ \\} & \parbox{.75in}{\ \\$b$\\ \ \\} &\parbox{.75in}{\ \\$c$\\ \ \\} & \parbox{.75in}{\ \\$d$\ \\ \ \\} & \parbox{1.75in}{\ \\$G_{abcd}$ State\ \\ \ \\} \\
\hline $\frac{1}{\sqrt{2}}$  & $0$ & $\frac{1}{\sqrt{2}}$ & $0$ & $\{\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]+ \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle] - \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] + \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $\frac{1}{\sqrt{2}}$  & $0$ & $-\frac{1}{\sqrt{2}}$ & $0$ & $\{\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]+ \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]+ \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] - \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $-\frac{1}{\sqrt{2}}$  & $0$ & $\frac{1}{\sqrt{2}}$ & $0$ & $\{\(-\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]- \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]- \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] + \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $-\frac{1}{\sqrt{2}}$  & $0$ & $-\frac{1}{\sqrt{2}}$ & $0$ & $\{\(-\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]- \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]+ \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] - \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $\frac{1}{\sqrt{2}}$  & $0$ & $0$ & $\frac{1}{\sqrt{2}}$ & $\{\(\frac{1}{\sqrt{2}}[|0000\rangle+|1111\rangle]\}$  \\
\hline $\frac{1}{\sqrt{2}}$  & $0$ & $0$ & $-\frac{1}{\sqrt{2}}$ & $\{\(\frac{1}{\sqrt{2}}[|0011\rangle+|1100\rangle]\}$  \\
\hline $-\frac{1}{\sqrt{2}}$  & $0$ & $0$ & $\frac{1}{\sqrt{2}}$ & $\{\(-\frac{1}{\sqrt{2}}[|0011\rangle+|1100\rangle]\}$  \\
\hline $-\frac{1}{\sqrt{2}}$  & $0$ & $0$ & $-\frac{1}{\sqrt{2}}$ & $\{\(-\frac{1}{\sqrt{2}}[|0000\rangle+|1111\rangle]\}$  \\
\hline $0$  & $\frac{1}{\sqrt{2}}$ & $\frac{1}{\sqrt{2}}$ & $0$ & $\{\(\frac{1}{\sqrt{2}}[|0101\rangle+|1010\rangle]\}$  \\
\hline $0$  & $\frac{1}{\sqrt{2}}$ & $-\frac{1}{\sqrt{2}}$ & $0$ & $\{\(\frac{1}{\sqrt{2}}[|0110\rangle+|1001\rangle]\}$  \\
\hline $0$  & $-\frac{1}{\sqrt{2}}$ & $\frac{1}{\sqrt{2}}$ & $0$ & $\{\(-\frac{1}{\sqrt{2}}[|0101\rangle+|1010\rangle]\}$  \\
\hline $0$  & $-\frac{1}{\sqrt{2}}$ & $-\frac{1}{\sqrt{2}}$ & $0$ & $\{\(-\frac{1}{\sqrt{2}}[|0110\rangle+|1001\rangle]\}$  \\
\hline $0$  & $\frac{1}{\sqrt{2}}$ & $0$ & $\frac{1}{\sqrt{2}}$ & $\{\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]- \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]+ \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] + \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $0$  & $\frac{1}{\sqrt{2}}$ & $0$ & $-\frac{1}{\sqrt{2}}$ & $\{-\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]+ \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]+ \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] + \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $0$  & $-\frac{1}{\sqrt{2}}$ & $0$ & $\frac{1}{\sqrt{2}}$ & $\{\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]- \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]- \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] - \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline $0$  & $-\frac{1}{\sqrt{2}}$ & $0$ & $-\frac{1}{\sqrt{2}}$ & $\{-\(\frac{1}{2\sqrt{2}}[|0000\rangle+|1111\rangle]+ \(\frac{1}{2\sqrt{2}}[|1100\rangle+|0011\rangle]- \\ & & & & \(\frac{1}{2\sqrt{2}}[|0110\rangle+\beta|1001\rangle] - \(\frac{1}{2\sqrt{2}}[\alpha|0101\rangle+\beta|1010\rangle]\}$    \\
\hline
\end{tabular}
\end{center}
\end{table}

\newpage
\begin{widetext}
\noindent\textit{\bf Appendix 2:}\\
Consider the 4-qubit ${G}_{abcd}$ state:\\
\\
 $= \frac{a + d}{{2}}(|0000\rangle+|1111\rangle) + \frac{a - d}{{2}}(|1100\rangle+|0011\rangle) + \frac{b + c}{{2}}(|0101\rangle+|1010\rangle) + \frac{b - c}{{2}}(|0110\rangle+|1001\rangle)$\\
\\
Where the first two qubits are with Alice, the third one with Bob and the last one with Charlie and let $|\psi\rangle = \alpha|0\rangle+\beta|1\rangle$ be the message with Alice. We now consider the case where $a$, $b$, $c$ and $d$ are complex numbers. \\
Normalization condition of the ${G}_{abcd}$ state yields: \\
$\lvert\frac{a+d}{2}$\rvert^2 + \lvert\frac{a+d}{2}\rvert^2 + \lvert\frac{a-d}{2}\rvert^2 + \lvert\frac{a-d}{2}\rvert^2 + \lvert\frac{b+c}{2}\rvert^2 + \lvert\frac{b+c}{2}\rvert^2 + \lvert\frac{b-c}{2}\rvert^2 + \lvert\frac{b-c}{2}\rvert^2 = 1$
\\
Simplifying we get:  $|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1 $
\\
\begin{eqnarray}
&&|\psi\rangle\otimes\{\frac{a + d}{2}(|0000\rangle+|1111\rangle) + \frac{a - d}{2}(|1100\rangle+|0011\rangle) + {}\nonumber\\&&\frac{b + c}{2}(|0101\rangle+|1010\rangle) + \frac{b - c}{2}(|0110\rangle+|1001\rangle)\} {}\nonumber\\&&
= \{\alpha|0\rangle+\beta|1\rangle\}\otimes\{\frac{a + d}{2}(|0000\rangle+|1111\rangle) + \frac{a - d}{2}(|1100\rangle+|0011\rangle) + {}\nonumber\\&&\frac{b + c}{2}(|0101\rangle+|1010\rangle) + \frac{b - c}{2}(|0110\rangle+|1001\rangle)\}{}\nonumber\\&&
= \frac{a + d}{2}(\alpha|00000\rangle+\alpha|01111\rangle) + \frac{a - d}{2}(\alpha|01100\rangle+\alpha|00011\rangle) + {}\nonumber\\&&\frac{b + c}{2}(\alpha|00101\rangle + \alpha|01010\rangle) + \frac{b - c}{2}(\alpha|00110\rangle+\alpha|01001\rangle) + {}\nonumber\\&&\frac{a + d}{2}(\beta|10000\rangle+\beta|11111\rangle) + \frac{a - d}{2}(\beta|11100\rangle+\beta|10011\rangle) + {}\nonumber\\&&\frac{b + c}{2}(\beta|10101\rangle+\beta|11010\rangle) + \frac{b - c}{2}(\beta|10110\rangle+\beta|11001\rangle)\\{}\nonumber\\&&
= \frac{a+d}{2}(|00\rangle\alpha|000\rangle+|01\rangle\alpha|111\rangle ) + \frac{a-d}{2}(|01\rangle\alpha|100\rangle+|00\rangle\alpha|011\rangle) + \nonumber\\&&\frac{b+c}{2}(|00\rangle\alpha|101\rangle+|01\rangle\alpha|010\rangle) + \frac{b-c}{2}(|00\rangle\alpha|110\rangle+|01\rangle\alpha|001\rangle) + \nonumber\\&&\frac{a+d}{2}(|10\rangle\beta|000\rangle+|11\rangle\beta|111\rangle) + \frac{a-d}{2}(|11\rangle\beta|100\rangle+|10\rangle\beta|011\rangle) + \nonumber\\&&\frac{b+c}{2}(|10\rangle\beta|101\rangle+|11\rangle\beta|010\rangle) + \frac{b-c}{2}(|10\rangle\beta|110\rangle+|11\rangle\beta|001\rangle)\\{}\nonumber\\&&
= \frac{a+d}{2\sqrt{2}}([|\phi^+\rangle + |\phi^-\rangle]\alpha|000\rangle+[|\psi^+\rangle + |\psi^-\rangle]\alpha|111\rangle)+{}\nonumber\\&&\frac{a-d}{2\sqrt{2}}([|\psi^+\rangle + |\psi^-\rangle]\alpha|100\rangle+[|\phi^+\rangle +
|\phi^-\rangle]\alpha|011\rangle)+{}\nonumber\\&&
\frac{b+c}{2\sqrt{2}}([|\phi^+\rangle + |\phi^-\rangle]\alpha|101\rangle
+[|\psi^+\rangle + |\psi^-\rangle]\alpha|010\rangle)+{}\nonumber\\&&\frac{b-c}{2\sqrt{2}}([|\phi^+\rangle +
|\phi^-\rangle]\alpha|110\rangle+[|\psi^+\rangle +
|\psi^-\rangle]\alpha|001\rangle)+{}\nonumber\\&&
\frac{a+d}{2\sqrt{2}}([|\psi^+\rangle - |\psi^-\rangle]\beta|000\rangle+[|\phi^+\rangle - |\phi^-\rangle]\beta|111\rangle)+{}\nonumber\\&&\frac{a-d}{2\sqrt{2}}([|\phi^+\rangle - |\phi^-\rangle]\beta|100\rangle+[|\psi^+\rangle - |\psi^-\rangle]\beta|011\rangle)+{}\nonumber\\&&
\frac{b+c}{2\sqrt{2}}([|\psi^+\rangle - |\psi^-\rangle]\beta|101\rangle
+[|\phi^+\rangle -
|\phi^-\rangle]\beta|010\rangle)+{}\nonumber\\&&\frac{b-c}{2\sqrt{2}}([|\psi^+\rangle - |\psi^-\rangle]\beta|110\rangle+[|\phi^+\rangle -
|\phi^-\rangle]\beta|001\rangle)\\{}\nonumber
\end{eqnarray}
\newpage
\begin{eqnarray}
= &&(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)]\\{}\nonumber
\end{eqnarray}

Suppose Alice measures in $|\phi^+\rangle$ Bell Basis. So the secret already being shared between Alice, Bob and Charlie is:
\begin{eqnarray}
&&\{\frac{a+d}{2\sqrt{2}}[\alpha|000\rangle+\beta|111\rangle]+\frac{a-d}{2\sqrt{2}}[\alpha|011\rangle+\beta|100\rangle]+{}\nonumber\\&&\frac{b+c}{2\sqrt{2}}[\alpha|101\rangle+\beta|010\rangle]+\frac{b-c}{2\sqrt{2}}[\alpha|110\rangle+\beta|001\rangle]\}_{ABC}\\{}\nonumber
\end{eqnarray}
Re-normalized Alice, Bob and Charlie's resultant state after the measurement of Alice is as follows:

\begin{eqnarray}
&&\{\frac{a+d}{\sqrt{2}}[\alpha|000\rangle+\beta|111\rangle]+\frac{a-d}{\sqrt{2}}[\alpha|011\rangle+\beta|100\rangle]+{}\nonumber\\&&\frac{b+c}{\sqrt{2}}[\alpha|101\rangle+\beta|010\rangle]+\frac{b-c}{\sqrt{2}}[\alpha|110\rangle+\beta|001\rangle]\}_{ABC} \\{}\nonumber
\end{eqnarray}
\begin{eqnarray}
&&= \{\frac{a+d}{\sqrt{2}}[\alpha|0\rangle|00\rangle+\beta|1\rangle|11\rangle]+\frac{a-d}{\sqrt{2}}[\alpha|0\rangle|11\rangle+\beta|1\rangle|00\rangle]+\nonumber\\&&\frac{b+c}{\sqrt{2}}[\alpha|1\rangle|01\rangle+\beta|0\rangle|10\rangle]+\frac{b-c}{\sqrt{2}}[\alpha|1\rangle|10\rangle+\beta|0\rangle|01\rangle]\}_{ABC} \\{}\nonumber
\end{eqnarray}
\begin{eqnarray}
&&=\{(\frac{a+d}{2\sqrt{2}})[\alpha|0\rangle[(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]+\beta|1\rangle[(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]]+\nonumber\\&&(\frac{a-d}{2\sqrt{2}})[\alpha|0\rangle[(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]+\beta|1\rangle[(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]]+\nonumber\\&&(\frac{b+c}{2\sqrt{2}})[\alpha|1\rangle[(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]+\beta|0\rangle[(|+\rangle-|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]]+\nonumber\\&&(\frac{b-c}{2\sqrt{2}})[\alpha|1\rangle[(|+\rangle-|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]+\beta|0\rangle[(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]]\} \\{}\nonumber
\end{eqnarray}
\begin{eqnarray}
&&= \{(\frac{1}{\sqrt{2}}(a\alpha+b\beta)|0\rangle+\frac{1}{\sqrt{2}}(b\alpha+a\beta)|1\rangle\}[|+\rangle_{B}|+\rangle_{C}]+\nonumber\\&&\{(\frac{1}{\sqrt{2}}(d\alpha+c\beta)|0\rangle-\frac{1}{\sqrt{2}}(c\alpha+d\beta)|1\rangle\}[|+\rangle_{B}|-\rangle_{C}]+\nonumber\\&&\{(\frac{1}{\sqrt{2}}(d\alpha-c\beta)|0\rangle+\frac{1}{\sqrt{2}}(c\alpha-d\beta)|1\rangle\}[|-\rangle_{B}|+\rangle_{C}]+\nonumber\\&&\{(\frac{1}{\sqrt{2}}(a\alpha-b\beta)|0\rangle-\frac{1}{\sqrt{2}}(b\alpha-a\beta)|1\rangle\}[|-\rangle_{B}|-\rangle_{C}]\\{}\nonumber
\end{eqnarray}
When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(a\alpha+b\beta)|0\rangle+\frac{1}{\sqrt{2}}(b\alpha+a\beta)|1\rangle \\{}\nonumber&&
\end{eqnarray}
When Bob and Charlie measure in the Hadamard Basis $|+\rangle|-\rangle$, Alice's resultant state is as follows:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(d\alpha+c\beta)|0\rangle-\frac{1}{\sqrt{2}}(c\alpha+d\beta)|1\rangle \\{}\nonumber&&
\end{eqnarray}
When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(d\alpha-c\beta)|0\rangle+\frac{1}{\sqrt{2}}(c\alpha-d\beta)|1\rangle \\{}\nonumber&&
\end{eqnarray}
When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(a\alpha-b\beta)|0\rangle-\frac{1}{\sqrt{2}}(b\alpha-a\beta)|1\rangle \\{}\nonumber&&
\end{eqnarray}

\noindent\textit{\bf Appendix 6:}\nonumber\\
\\
Re-normalized Alice's resultant state after the measurement of Bob and Charlie in the Hadamard Basis $|+\rangle|+\rangle$ is as follows:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{\([|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)]}}(a\alpha+b\beta)|0\rangle+\(\frac{1}{\sqrt{[|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)]}}(b\alpha+a\beta)|1\rangle&&
\end{eqnarray}
From the above re-normalized state of Alice, for the QSS protocol to be effective, then Alice's resultant state has to be independent of $\alpha$ and $\beta$. For Alice's resultant state to be independent of $\alpha$ and $\beta$, the necessary condition is $ab = 0$, which means that either $a=0$ or $b=0$.
\\
\\
If $a = 0$, then Alice's resultant state reduces to:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{\(|b|^2}}(b\beta)|0\rangle+\(\frac{1}{\sqrt{|b|^2}}(b\alpha)|1\rangle =  \(\frac{b}{\(|b|}[\beta|0\rangle+\alpha|1\rangle] &&
\end{eqnarray}
\\
If $b$ = 0, then Alice's resultant state reduces to:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{\(|a|^2}}(a\alpha)|0\rangle+\(\frac{1}{\sqrt{|a|^2}}(a\beta)|1\rangle =  \(\frac{a}{\(|a|}[\alpha|0\rangle+\beta|1\rangle]&&
\end{eqnarray}
\\
\\
\noindent\textit{\bf Appendix 7:}
\\
Re-normalized Alice's resultant state after the measurement of Bob and Charlie in the Hadamard Basis $|+\rangle|-\rangle$ is as follows:\\

\begin{eqnarray}
\(\frac{1}{\sqrt{[|c|^2+|d|^2+(\tilde{c}d+c\tilde{d})(\alpha\tilde\beta+\tilde\alpha\beta)]}}(d\alpha+c\beta)|0\rangle-\(\frac{1}{\sqrt{[|c|^2+|d|^2+(\tilde{c}d+c\tilde{d})(\alpha\tilde\beta+\tilde\alpha\beta)]}}(c\alpha+d\beta)|1\rangle&&
\end{eqnarray}
\\
From the above re-normalized state of Alice, for the QSS protocol to be effective, then Alice's resultant state has to be independent of $\alpha$ and $\beta$. For Alice's resultant state to be independent of $\alpha$ and $\beta$, we infer that $cd = 0$, which means that either $c=0$ or $d=0$.\\
\\
\textbf{Case 1:}\\
Consider the re-normalized state of Alice in the Hadamard Basis $|+\rangle|-\rangle$ (6.24). If $c = 0$, then Alice's resultant state reduces to:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{\(|d|^2}}(d\alpha)|0\rangle-\(\frac{1}{\sqrt{\(|d|^2}}(d\beta)|1\rangle = \(\frac{d}{\(|d|}[\alpha|0\rangle-\beta|1\rangle]&&
\end{eqnarray}
\\
If we choose $d = 0$, the QSS protocol will fail in the Hadamard Basis $|+\rangle|-\rangle$ and $|-\rangle|+\rangle$ (6.12 and 6.13). From the re-normalized state of Alice in the Hadamard Basis $|+\rangle|+\rangle$ above in (6.18), for QSS protocol to be effective, we obtained the condition either $a = 0$ or $b = 0$.\\
\\
If we choose $a = 0$, from the condition of normalization, we have \\
\begin{eqnarray}
|a|^2+|b|^2+|c|^2+|d|^2 = 1
\end{eqnarray}
Since we chose  $b = 0$, and $c = 0$, the above equation reduces to:
\\
\begin{eqnarray}
|a|^2 + |d|^2 = 1
\end{eqnarray}
\\
However, if we choose $d = \pm1$, then from equation (6.27) above, we get $a$ = 0. Since we already chose $b$ = 0 earlier, we cannot have $a$ to be equal to 0, or else the QSS protocol will fail in the Hadamard Basis $|+\rangle|+\rangle$ and $|-\rangle|-\rangle$ (6.11 and 6.14). Hence $d \neq \pm1$.
So we have from 6.27 above:\\
\begin{eqnarray}
|a| = \pm\sqrt{1 - |d|^2}
\end{eqnarray}
\\
\\
Now as modulus of a complex number is always positive, we have:\\
\begin{eqnarray}
0 < |d|^2 < 1
\end{eqnarray}
\\
And $|d|^2$ cannot be greater than 1 because then $|a|$ would be a complex number which is not possible.\\
Notice that the modulus of a complex number is always a real number and in fact it will never be negative since square roots always return a positive number or zero depending on what is under the radical.\\

Similarly, if we choose $a = \pm1$, then from equation (6.27) above, we get $d$ = 0. Since we already chose $c$ = 0 earlier, we cannot have $d$ to be equal to 0, or else the QSS protocol will now fail in the Hadamard Basis $|+\rangle|-\rangle$ and $|-\rangle|+\rangle$ (6.12 and 6.13). Hence $a \neq \pm1$. So we have from 6.27 above:\\
\begin{eqnarray}
|d| = \pm\sqrt{1 - |a|^2}
\end{eqnarray}
\\
\newpage
\noindent\textit{\bf Appendix 8:}\\
\\
From Appendix 1, when Alice combines her secret with the $G_{abcd}$ state, and does a two-qubit measurement in the $|\phi^+\rangle$ Bell basis, we have Alice, Bob and Charlie's resultant re-normalized state as follows:
\begin{eqnarray}
&&\{\frac{a+d}{\sqrt{2}}[\alpha|000\rangle+\beta|111\rangle]+\frac{a-d}{\sqrt{2}}[\alpha|011\rangle+\beta|100\rangle]+{}\nonumber\\&&\frac{b+c}{\sqrt{2}}[\alpha|101\rangle+\beta|010\rangle]+\frac{b-c}{\sqrt{2}}[\alpha|110\rangle+\beta|001\rangle]\}_{ABC} {}\nonumber
\end{eqnarray}
Now if Alice and Bob measure their qubits in the Hadamard basis, it is Charlie who will be able to retrieve the secret.
\begin{eqnarray}
&&= \{\frac{a+d}{\sqrt{2}}[\alpha|00\rangle|0\rangle+\beta|11\rangle|1\rangle]+\frac{a-d}{\sqrt{2}}[\alpha|01\rangle|1\rangle+\beta|10\rangle|0\rangle]+{}\nonumber\\&&\frac{b+c}{\sqrt{2}}[\alpha|10\rangle|1\rangle+\beta|01\rangle|0\rangle]+\frac{b-c}{\sqrt{2}}[\alpha|11\rangle|0\rangle+\beta|00\rangle|1\rangle]\}_{ABC} {}\nonumber
\end{eqnarray}
\begin{eqnarray}
&&=\{(\frac{a+d}{2\sqrt{2}})[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|0\rangle+\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|1\rangle]+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|1\rangle+\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|0\rangle]+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle+\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|0\rangle]+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})[\alpha[(|+\rangle-|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]|1\rangle+\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle]\} {}\nonumber
\end{eqnarray}
\begin{eqnarray}
&&= \frac{1}{2\sqrt{2}}[[(a+b-c+d)\alpha+(a+b+c-d)\beta)]|0\rangle+{} \nonumber\\&&[(a+b+c-d)\alpha+(a+b-c+d)\beta)]|1\rangle][|+\rangle_{A}|+\rangle_{B}]+\nonumber\\&&\frac{1}{2\sqrt{2}}[[(a-b+c+d)\alpha+(a-b-c-d)\beta)]|0\rangle+{} \nonumber\\&&[(-a+b+c+d)\alpha+(-a+b-c-d)\beta)]|1\rangle][|+\rangle_{A}|-\rangle_{B}]+\nonumber\\&&\frac{1}{2\sqrt{2}}[[(a-b+c+d)\alpha+(-a+b+c+d)\beta)]|0\rangle+{} \nonumber\\&&[(a-b-c-d)\alpha+(-a+b-c-d)\beta)]|1\rangle][|-\rangle_{A}|+\rangle_{B}]+\nonumber\\&&\frac{1}{2\sqrt{2}}[[(a+b-c+d)\alpha+(a+b+c-d)\beta)]|0\rangle+{} \nonumber\\&&[(-a-b-c+d)\alpha+(a+b-c+d)\beta)]|1\rangle][|-\rangle_{A}|-\rangle_{B}] {}\nonumber &&
\end{eqnarray}
\\
When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows:
\begin{eqnarray}
&&\frac{1}{2\sqrt{2}}[[(a+b-c+d)\alpha+(a+b+c-d)\beta)]|0\rangle+{} \nonumber\\&&[(a+b+c-d)\alpha+(a+b-c+d)\beta)]|1\rangle]{}\nonumber&&
\end{eqnarray}
When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows:
\begin{eqnarray}
&&\frac{1}{2\sqrt{2}}[[(a-b+c+d)\alpha+(a-b-c-d)\beta)]|0\rangle+{} \nonumber\\&&[(-a+b+c+d)\alpha+(-a+b-c-d)\beta)]|1\rangle] {}\nonumber&&
\end{eqnarray}
When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
\begin{eqnarray}
&&\frac{1}{2\sqrt{2}}[[(a-b+c+d)\alpha+(-a+b+c+d)\beta)]|0\rangle+{} \nonumber\\&&[(a-b-c-d)\alpha+(-a+b-c-d)\beta)]|1\rangle] {}\nonumber&&
\end{eqnarray}
When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
\begin{eqnarray}
&&\frac{1}{2\sqrt{2}}[[(a+b-c+d)\alpha+(a+b+c-d)\beta)]|0\rangle+{} \nonumber\\&&[(-a-b-c+d)\alpha+(a+b-c+d)\beta)]|1\rangle] {}\nonumber&&
\end{eqnarray}
\\
Denoting $(a+b-c+d)$ as $\lambda_1$ and $(a+b+c-d)$ as $\lambda_2$, the re-normalized Charlie's resultant state after the measurement of Bob and Charlie in the Hadamard Basis $|+\rangle|+\rangle$ is as follows:\\
The square root of the sum of squares of all coefficients gives:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{|\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta)}}(\lambda_1\alpha+\lambda_2\beta)|0\rangle+\(\frac{1}{\sqrt{|\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta)}}(\lambda_2\alpha+\lambda_1\beta)|1\rangle\nonumber&&
\end{eqnarray}
\\
From the above re-normalized state of Charlie, for the QSS protocol to be effective, then Charlie's resultant state has to be independent of $\alpha$ and $\beta$. For Charlie's resultant state to be independent of $\alpha$ and $\beta$, the necessary condition is $\lambda_1\lambda_2 = 0$, which means that either $\lambda_1=0$ or $\lambda_2=0$. So it means either $(a+b-c+d) = 0$ or $(a+b+c-d) = 0$.
\\ \\
\textbf{Case 1:}\\
If $(a+b+c-d) = 0$ or $(\lambda_2 = 0)$, then Charlie's resultant state reduces to:\\
\begin{eqnarray}
\(\frac{1}{\sqrt{\(|\lambda_1|^2}}(\lambda_1\alpha)|0\rangle+\(\frac{1}{\sqrt{|\lambda_1|^2}}(\lambda_1\beta)|1\rangle =  \(\frac{\lambda_1}{\(|\lambda_1|}[\alpha|0\rangle+\beta|1\rangle] &&
\end{eqnarray}
\\
From Appendix 1, when Bob and Charlie measured in the Hadamard Basis, and Alice recovers the secret, we had the following constraints on the values of $a$, $b$, $c$ and $d$ for the QSS protocol to work: \\
\\
\\
\textbf{Case 1.1}\\
From Appendix 7 above, we chose $b = 0$ and $c = 0$\\
So the equation $(a+b+c-d = 0)$ reduces to $a-d = 0$ or we have $a = d$\\
For the above values of $a$, $b$, $c$ and $d$, from the condition of normalization, we have \\
\begin{eqnarray}
|a|^2+|b|^2+|c|^2+|d|^2 = 1 {}\nonumber
\end{eqnarray}
Since we chose  $b = 0$, and $c = 0$, the above equation reduces to:
\\
\begin{eqnarray}
|a|^2 + |d|^2 = 1 {}\nonumber
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 + |a|^2 = 1 {}\nonumber
\end{eqnarray}
or
\begin{eqnarray}
2|a|^2 = 1 {}\nonumber
\end{eqnarray}
so we have:\\
\begin{eqnarray}
|a| = |d| = \frac{1}{\sqrt{2}} {}\nonumber
\end{eqnarray}
Since we already have the following condition:
\begin{eqnarray}
0 < |d|^2 < 1
\end{eqnarray}
and
\begin{eqnarray}
0 < |a|^2 < 1
\end{eqnarray}
We choose $a = d = \(\frac{1}{2}(1+i)$\\
\\
$G_{abcd}$ = $\(\frac{1}{2}(1+i)[|0000\rangle + |1111\rangle]$
\\

So Charlie's resultant state in $|+\rangle|+\rangle$ is as follows:\\
\begin{eqnarray}
\(\frac{1+i}{\sqrt{2}}[\alpha|0\rangle + \beta|1\rangle]
\end{eqnarray}
\\
The Unitary Matrix to retrieve the secret is as follows:
\begin{equation*}
U =
\begin{bmatrix}
\frac{1-i}{\sqrt{2}} & 0 \\
0 & \frac{1-i}{\sqrt{2}} \\

\end{bmatrix}
\end{equation*}

\end{widetext}

\begin{appendix}
\begin{widetext}
\noindent\textit{\bf Appendix 1:} (Proof of Lemma 1)\\
\\
Alice's resultant state obtained in the Hadamard Basis $\ket{+}\ket{+}$, , when Alice's initial measurement outcome is $\ket{\phi}$ is:
\begin{eqnarray}
\sqrt{2}\{(a\alpha + b\beta)|0\rangle + (b\alpha + a\beta)|1\rangle\}
\end{eqnarray}
Condition of normalization of (40) is as follows:
\begin{eqnarray}
|a|^2+|b|^2+(\tilde{a}b+a\tilde{b})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} &&
\end{eqnarray}
For the QSS protocol to exist, the condition of normalization obtained in (41) above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde{a}b + a\tilde{b} = 0$, which means that either $a=0$ or $b=0$, but both $a$ and $b$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{+}$ and $\ket{-}\ket{-}$.
\\
\\
\textbf{Case 1:}\\
If $a = 0$, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}b\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 2:}\\
If $b$ = 0, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}a\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow \ I$ &&
\end{eqnarray}
\\
From Alice's resultant state obtained in the Hadamard Basis $\ket{+}\ket{-}$, the condition of normalization is as follows:
\begin{eqnarray}
2\{|d\alpha + c\beta|^2+|c\alpha + d\beta|^2\} = 1 &&
\end{eqnarray}
or
\begin{eqnarray}
|c|^2+|d|^2+(\tilde{c}d+c\tilde{d})(\alpha\tilde\beta+\tilde\alpha\beta)
 =  \frac{1}{2} &&
\end{eqnarray}
For the QSS protocol to exist, the condition of normalization obtained in (41) above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde{c}d + c\tilde{d} = 0$, which means that either $c=0$ or $d=0$, but both $c$ and $d$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{-}$ and $\ket{-}\ket{+}$.
\\
\\
\\
\textbf{Case 3:}\\
If $c = 0$, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{-}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}d\{\alpha|0\rangle - \beta|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|d|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle - \beta|1\rangle \rightarrow \sigma_z$ &&
\end{eqnarray}
\\
\textbf{Case 4:}\\
If $d$ = 0, then Alice's resultant state in the Hadamard Basis $\ket{+}\ket{-}$ reduces to:\\
\begin{eqnarray}
\sqrt{2}c\{\beta|0\rangle - \alpha|1\rangle\} &&
\end{eqnarray}
And from (41) above, we have:
\begin{eqnarray}
|c|^2 = \frac{1}{2} &&
\end{eqnarray}
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle - \alpha|1\rangle \rightarrow \ \sigma_x\sigma_z$ &&
\end{eqnarray}
\\
From Cases 1, 2, 3 and 4 we have the following conditions:\\
\begin{eqnarray}
1.\ $a = 0$, \  $c = 0$, \  $b \ne 0$, \  $d \ne 0$, \  $|b|^2 = \frac{1}{2}$, \  $|d|^2 = \frac{1}{2}$ &&
\\
2.\ $a = 0$, \  $d = 0$, \  $b \ne 0$, \  $c \ne 0$, \  $|b|^2 = \frac{1}{2}$, \  $|c|^2 = \frac{1}{2}$ &&
\\
3.\ $b = 0$, \  $c = 0$, \  $a \ne 0$, \  $c \ne 0$, \  $|a|^2 = \frac{1}{2}$, \  $|d|^2 = \frac{1}{2}$ &&
\\
4.\ $b = 0$, \  $d = 0$, \  $a \ne 0$, \  $d \ne 0$, \  $|a|^2 = \frac{1}{2}$, \  $|c|^2 = \frac{1}{2}$ &&
\end{eqnarray}
\\
\noindent\textit{\bf Appendix 2:} (Proof of Lemma 2)\\
\\
Charlie's resultant state obtained in the Hadamard Basis $\ket{+}\ket{+}$, when Alice's initial measurement outcome is $\ket{\phi}$ is:
\begin{eqnarray}
\frac{1}{\sqrt{2}}\{[(a+b-c+d)\alpha+(a+b+c-d)\beta]|0\rangle\}+\frac{1}{\sqrt{2}}\{[(a+b+c-d)\alpha$ + $(a+b-c+d)\beta]|1\rangle\} &
\end{eqnarray}
Denoting $(a+b-c+d)$ as $\lambda_1$ and $(a+b+c-d)$ as $\lambda_2$, (51) can be written as:
\begin{eqnarray}
\frac{1}{\sqrt{2}}(\lambda_1\alpha+\lambda_2\beta)|0\rangle + \frac{1}{\sqrt{2}}(\lambda_2\alpha+\lambda_1\beta)|1\rangle
\end{eqnarray}
Condition of normalization of (57) is as follows:
\begin{eqnarray}
\frac{1}{2}\{|\lambda_1\alpha+\lambda_2\beta|^2 + |\lambda_2\alpha+\lambda_1\beta|^2 \} = 1 &&
\end{eqnarray}
or
\begin{eqnarray}
|\lambda_1|^2+|\lambda_2|^2+(\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2)(\alpha\tilde\beta+\tilde\alpha\beta) = 2 &&
\end{eqnarray}
\\
For the QSS protocol to exist, the condition of normalization obtained in (59) above should be independent of $\alpha$ and $\beta$. This can be possible only if $\tilde\lambda_1\lambda_2+\lambda_1\tilde\lambda_2 = 0$, which means that either $\lambda_1 = 0$ or $\lambda_2 = 0$, but both $\lambda_1$ and $\lambda_2$ cannot be 0 simultaneously, otherwise, the protocol will not exist in the Hadamard Basis $\ket{+}\ket{+}$ and $\ket{-}\ket{-}$ So it means either $(a+b-c+d) = 0$ or $(a+b+c-d) = 0$.
\\
\\
\textbf{Case 1:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (57) above if we take $a = 0$ and $c = 0$, we have $b + d = 0$ or $b = -d$ or $-b = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $c = 0$ and $b = -d$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $c = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
If we take $b = -d$, $a + b + c - d = 2b$ or $a + b + c - d = -2d$ or $\lambda_2 = 2b$ or $\lambda_2= -2d$\\
\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\sqrt{2}b\{\beta|0\rangle+\alpha|1\rangle\} \rightarrow \sigma_x$ &&
\end{eqnarray}
or
\begin{eqnarray}
$-\sqrt{2}d\{\beta|0\rangle+\alpha|1\rangle\} \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 2:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (57) above if we take $a = 0$ and $c = 0$, we have $b - d = 0$ or $b = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $c = 0$ and $b = d$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $c = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 3:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (58) above if we take $a = 0$ and $d = 0$, we have $b - c = 0$ or $b = c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $d = 0$ and $b = c$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $d = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 4:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (58) above if we take $a = 0$ and $d = 0$, we have $b + c = 0$ or $b = -c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $a = 0$ and $d = 0$ and $b = -c$, eq (67) reduces to:
\begin{eqnarray}
|2b|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|b|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $a = 0$ and $d = 0$ and $|b|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 5:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (59) above if we take $b = 0$ and $c = 0$, we have $a + d = 0$ or $a = -d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $c = 0$ and $a = -d$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $c = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|d|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 6:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (59) above if we take $b = 0$ and $c = 0$, we have $a - d = 0$ or $a = d$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $c = 0$ and $a = d$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\textbf{Case 7:}\\
If $a + b - c + d = 0$ or $\lambda_1 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_2\{\beta|0\rangle + \alpha|1\rangle\} &&
\end{eqnarray}
From (60) above if we take $b = 0$ and $d = 0$, we have $a - c = 0$ or $a = c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_2|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b+c-d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $d = 0$ and $a = c$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\beta|0\rangle+\alpha|1\rangle \rightarrow \sigma_x$ &&
\end{eqnarray}
\\
\textbf{Case 8:}\\
If $a + b + c - d = 0$ or $\lambda_2 = 0$, then Charlie's resultant state in the Hadamard Basis $\ket{+}\ket{+}$ reduces to:\\
\begin{eqnarray}
\frac{1}{\sqrt{2}}\lambda_1\{\alpha|0\rangle + \beta|1\rangle\} &&
\end{eqnarray}
From (60) above if we take $b = 0$ and $d = 0$, we have $a + c = 0$ or $a = -c$
And from (64) above, we have:
\begin{eqnarray}
|\lambda_1|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a+b-c+d|^2 = 2 &&
\end{eqnarray}
Since $b = 0$ and $d = 0$ and $a = -c$, eq (67) reduces to:
\begin{eqnarray}
|2a|^2 = 2 &&
\end{eqnarray}
or
\begin{eqnarray}
|a|^2 = \frac{1}{2} &&
\end{eqnarray}
From the condition of normalization of the $G_{abcd}$ state, we have:
$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$\\
As $b = 0$ and $d = 0$ and $|a|^2 = \frac{1}{2}$, we also have $|c|^2 = \frac{1}{2}$\\
The transformation which Alice should perform in order to obtain the secret is (ignoring external phase):
\begin{eqnarray}
$\alpha|0\rangle+\beta|1\rangle \rightarrow I$ &&
\end{eqnarray}
\\
\\
\noindent\textit{\bf Appendix 3:} (Proof of Theorem 1)\\
\\
Combining the results of Lemma 1 and Lemma 2, we establish Theorem 3
\\

\begin{table}[h]
\caption{\bf Reconstruction of Quantum Information by Charlie after the measurement of Alice and Bob in the Hadamard Basis
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline \parbox{.75in}{\ \\Alice's  Outcomes\\ \ \\} & \parbox{.75in}{\ \\Bob and Charlie's Outcomes\\ \ \\} &\parbox{.75in}{\ \\$a$\\ \ \\} &\parbox{.75in}{\ \\$b$\\ \ \\} &\parbox{.75in}{\ \\$c$\\ \ \\} &\parbox{.75in}{\ \\$d$\\ \ \\} &\parbox{.75in}{\ \\Charlie's Resultant state\\ \ \\} & \parbox{.75in}{\ \\Charlie's Local Operations\ \\ \ \\}\\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $\alpha|0\rangle+\beta|1\rangle$ & $I$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $\alpha|0\rangle-\beta|1\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{-}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|0\rangle-\beta|0\rangle$ & $\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $-\alpha|1\rangle+\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $\frac{1}{2} + \frac{i}{2}$ & $0$ & $\alpha|1\rangle-\beta|0\rangle$ & $\sigma_x\sigma_z$    \\
\hline $\ket{\phi^{+}}$  & $\ket{+}\ket{+}$ & $0$ & $\frac{1}{2} + \frac{i}{2}$ & $-\frac{1}{2} - \frac{i}{2}$ & $0$ & $-\alpha|0\rangle+\beta|1\rangle$ & $\sigma_z$    \\
\hline
\end{tabular}
\end{center}
\end{table}

\end{widetext}

\end{document}

\section{ Introduction}

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more systems have to be described with reference to each other, even in the case when the systems have to be spatially separated \cite{ent}. This can also be viewed as correlations between observable physical properties of the systems. It not only gives us insight in understanding the deepest nature of reality but also has applications in the emerging technologies \cite{tech}. In fact, entanglement is widely considered as a key resource to the areas of quantum computing, quantum information processing  and quantum cryptography \cite{gisin}.

In classical cryptography, secret sharing is a protocol where some sensitive classical data is distributed among number of people such that  sufficiently large subsets of those people can collectively reconstruct the data, but they can gain no information about the sensitive data from their respective individual shares. For instance, a possible application is to share the key for a accessing a joint  account controlled by many people.  No individual is able to withdraw money, but sufficiently large groups can withdraw the money or access the account. One particular symmetric variety of secret sharing scheme is called a threshold scheme. A $(k,n)$ classical threshold scheme has $n$ share holders, of which any $k$ are sufficient to reconstruct the secret, while any set of $k-1$ or fewer shares has  no  information  about  the  secret

Quantum secret sharing (QSS) \cite{hillery, cleve99} deals with the problem of sharing of both classical as well as quantum secrets, however the resources used for secret sharing are purely quantum mechanical in nature. A typical protocol for quantum secret sharing, like many other tasks in quantum cryptography, uses entanglement  as a cardinal resource, mostly pure entangled states. Karlsson et al.\cite{karlsson} studied quantum secret sharing protocols using bipartite pure entangled states as resources. Many authors investigated the concept of quantum secret sharing  using tripartite pure entangled states and  multi-partite states like graph  states \cite{bandyopadhyay,bagherinezhad,lance,gordon,zheng, markham,markham08}. Q. Li et al. \cite{li} proposed semi-quantum secret sharing protocols taking maximally entangled GHZ state as resource.
In a realistic situation however, the secret sharing of classical or quantum information involves transmission of qubits through noisy channels that give rise to mixed states. Several works were done in this context where mixed and noisy entangled states were taken as a resource for the purpose of secret sharing\cite{indranil, satya, indranil1}
Quantum secret sharing has also been realized in experiments \cite{tittel, schmid, schmid1, bogdanski}

In most of the quantum secret sharing scenarios, the protocols ensure security only if some of the share holders are dishonest instead of all. However these protocols fail to address the security, when the dealer after sharing finds all the share holders to be dishonest. In that situation they can collaborate with each other and reveal the secret without the dealer's consent. Hence, once the secret is shared, the dealer has no way to bring the secret back to herself/himself.
In this work we address this problem of quantum secrets by taking a simple scenario when there is one dealer Alice and the secret is shared between Bob and Charlie. Here in our protocol we empower the dealer to keep a part of the  quantum secret (which we refer to as \textbf{quantum share}) with herself which enables her to stop the reconstruction of the secret by Bob and Charlie and consequently retrieve the secret when she finds both of them to be dishonest.
This empowerment comes at the cost of one extra qubit with the dealer. So for a three party scenario (one dealer and two additional share holders) we need a four qubit entangled state as a resource for the execution of the protocol. However if the dealer (Alice) is not revoking the secret, then all the shareholders can reconstruct the secret using our protocol at their location. In addition of introducing our protocol, we
explicitly work out with an class four qubit pure entangled states as a resource. We explicitly find out the ranges of the input state parameters under which both revocation and reconstruction of the secrets will be possible.  \\

In section II we introduce our revocation and reconstruction protocol for one dealer and two share holders and  introduce the concept of \textbf{quantum share} of the secret. In section III we execute our protocol with a four qubit pure entangled state as a resource. In particular we find out the range based on the input state parameters under which both revocation at the dealer's location and the reconstruction at the share holder's location will be possible.

\section{THE PARTIAL SECRET SHARING PROTOCOL: QUANTUM SHARE OF THE SECRET  }

In this section we first discuss what is the difficulty in carrying out this protocol in 3-party, 3-qubit scenario. Further we give an outline of our protocol in general and introduce the concept of \textbf{quantum share} of secret in that process.

\subsection{Going Beyond Three Qubit Secret Sharing }
In most of the existing Quantum Secret Sharing (QSS) protocols, the dealer (Alice), who is having the secret will first split the quantum information by combining her secret with the resource state and then does a two qubit measurement of the secret qubit and the resource qubit shared with her. In this case Alice, Bob and Charlie (Bob and Charlie are the two parties with whom Alice wants to share the secret) are sharing a three qubit resource among them with each possessing one qubit.
The measurement will result in the secret being shared into two quantum qubits with the other two parties, Bob and Charlie. Alice will only have classical outcome of the measurement but not an quantum part of the secret that she has shared. The strength of 3-party, 3-qubit protocol is that if one of the share holders (Bob or Charlie) is dishonest, he can not reveal the secret. The existing protocol gives full security to it. However if both of them are dishonest they can collaborate with each other to get hold of the secret, and Alice can not do anything about it.  This can allow Bob and Charlie, if dishonest
to collude and reconstruct the exact secret qubit with a probability of $1/4$.

In the QSS protocol which uses a 3-qubit GHZ resource, Alice, the dealer
begins by taking her secret, combines it with her 3-qubit GHZ resource

\\

Alice now conveys the result of her measurement to Bob or Charlie. Assuming Bob now measures his state in the Hadamard basis, Charlie is able to reconstruct the secret. However he needs two bits of classical information in order to reconstruct the secret.

Hence, after Alice now conveys the result of her measurement in Bell basis to Bob, Charlie can reconstruct Alice's state but only with Bob's assistance. Bob must measure his state and send the result to Charlie. Without Bob's information, Charlie can not reconstruct Alice's state. If Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest, she does not share her results of measurement in the Bell basis with Bob or Charlie. Even then, Bob and Charlie can reconstruct the exact secret without the help of Alice with a probability of $1/4$. However, if Alice comes to know that either Bob is dishonest or Charlie is dishonest or both are dishonest after conveying the results of her measurement in the Bell basis, there is nothing she can do to prevent Bob and Charlie from reconstructing the secret. Even Charlie and Bob guess Alice's outcome, the state they reconstructed will be known as the actual secret only after application for which the protocol is being used succeeds.

To nullify this probability, we introduce the need of one more qubit with Alice, so that she can also hold a share in the quantum form. This we refer as \textbf{quantum share} of the secret. Thus, Bob and Charlie will be unable to reconstruct the secret without the involvement of Alice. More interestingly if Alice finds both the party to be dishonest, then she can always bring back the secret to her.

\subsection{Our Protocol: A New Quantum Secret Sharing Protocol}
 In this subsection we give an outline of our protocol which requires a 4-qubit resource, with two of the qubits with Alice, the third qubit with Bob and the fourth qubit with Charlie.
 \\

\\

\noindent \textbf{a}. \emph{\textbf{Step 1: Splitting of Quantum Information or the sharing  phase.}}

\noindent Alice combines her secret, which is in the state $|\psi\rangle_{s}=\alpha|0\rangle_s + \beta|1\rangle_s$, (where $|\alpha|^2+|\beta|^2=1$) with the 4-Qubit entangled resources $|X\rangle_{abcd}$, where the qubits $a$ and $b$ are with Alice, while $c$ and $d$ are with Bob and Charlie respectively. She measures the secret $s$ and the qubit $a$  in the Bell basis, and she gets one of the four possible outcomes $\{\psi^{\pm}, \phi^{\pm}\}$ (Bell states) and corresponding to each measurement outcomes, Alice, Bob and Charlie share three qubit entanglement among each other. The secret is shared in form of three qubit entanglement among these three parties (see Fig).\\

\noindent \textbf{b}. \emph{\textbf{Step 2: Reconstruction of the secret.}}
Alice then uses her second qubit (\textbf{quantum share of the resource}) to participate in the reconstruction of the secret along with Bob or Charlie. If Charlie needs to reconstruct the secret, Alice and Bob measure their respective particles in the Hadamard basis. At this point, if Alice comes to know that either Bob or Charlie is dishonest or both are dishonest, she will not measure her state and the chance that Charlie retrieves the secret is zero. In addition to that Alice can bring back her secret by asking Bob and Charlie to do their measurements in the Hadamard basis. The importance of the protocol is that the dealer Alice holds a quantum share of the secret with herself, which empowers her to stop the revealing of the secret  and consequently bring the entire secret back to herself without revealing it even after sharing  (see Fig).
}

\begin{figure}[h]

    \includegraphics[width=.575\textwidth,left]{Revoc-Comb.pdf}

\end{figure}

\section{ CHARACTERIZATION OF FOUR QUBIT RESOURCE FOR OUR PROTOCOL}
\noindent In this section we consider a particular class of four qubit state and investigate whether they will be useful to be used as resource states for our protocol \cite{verstraete}.

\\

\noindent We begin with the first $4-$qubit normalized $G_{abcd}$ state,
\begin{eqnarray}
&& |G_{abcd}\rangle={}\nonumber\\&&\(\frac{a+d}{2}(|0000\rangle+|1111\rangle)+ \(\frac{a-d}{2}(|1100\rangle+|0011\rangle){}\nonumber\\&& + \(\frac{b+c}{2}}(|0101\rangle+|1010\rangle) + \(\frac{b-c}{2}(|0110\rangle+|1001\rangle)
\end{eqnarray}
with $a$, $b$, $c$ and $d$ being complex numbers and the normalization condition being $|a|^2+|b|^2+|c|^2+|d|^2 = 1$.
Here, the first two qubits are with Alice, the third one with Bob and the last one with Charlie. Here Alice is the dealer who shares the secret with both Bob and Charlie. The secret to be shared by Alice is given by the one qubit state $|\psi_{S}\rangle = \alpha|0\rangle+\beta|1\rangle$ where $|\alpha|^2+|\beta|^2=1$. \\
The combined state of the system with the message and the resource is given by,
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=\{\alpha|0\rangle+\beta|1\rangle\}\otimes {}\nonumber\\&&
\frac{a+d}{2}\((|0000\rangle+|1111\rangle) + \frac{a-d}{2}\((|1100\rangle+|0011\rangle) {}\nonumber\\&& +  \frac{b+c}{2}\((|0101\rangle+|1010\rangle) + \frac{b-c}{2} \((|0110\rangle+|1001\rangle).
\end{eqnarray}
\\
\noindent Our next target is to see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie with the state $G_{abcd}$ as a resource.  Revocation is important when Alice decides to change the secret or
Alice guesses that recipients are no longer trustworthy, or there is an update of secret in higher level application using secret sharing as a subroutine.
Now if we rewrite the combined system with the secret qubit $S$ and the first qubit of Alice in the Bell basis we have
\begin{eqnarray}
&&|\psi_{S}\rangle\otimes |G_{abcd}\rangle
=(\(|\phi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle+\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle+\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle+\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle+\beta|001\rangle)]+{}\nonumber\\&&(\(|\phi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|000\rangle-\beta|111\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|011\rangle-\beta|100\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|101\rangle-\beta|010\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|110\rangle-\beta|001\rangle)]+{}\nonumber\\&&(\(|\psi^+\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle+\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle+\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle+\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle+\beta|110\rangle)]+{}\nonumber\\&&(\(|\psi^-\rangle)\[[(\frac{a+d}{2\sqrt{2}})(\alpha|111\rangle-\beta|000\rangle)+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})(\alpha|100\rangle-\beta|011\rangle)+(\frac{b+c}{2\sqrt{2}})(\alpha|010\rangle-\beta|101\rangle)+{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})(\alpha|001\rangle-\beta|110\rangle)].\\{}\nonumber
\end{eqnarray}

\noindent Now Alice measures both the secret qubit and the first qubit of the resource state in the Bell basis. There will be four possible measurement outcomes $\{\phi^\pm, \psi^\pm\}$ as a result of Alice's measurement. In correspondence to various measurement outcomes obtained by Alice in the Bell basis, Alice, Bob and Charlie's qubits collapse into the states as given in TABLE I. It is important to note that the information about the secret is shared among three parties Alice, Bob and Charlie and Alice being the dealer also has a hold on the part of the secret. The Alice's qubit that is shared in form of a three qubit state with Bob and Charlie is the \textbf{quantum share} of the secret. Next we see how the dealer Alice is revoking back the secret which is already shared with Bob and Charlie.

\begin{table}[h]
\caption{\bf  Alice, Bob and Charlie's Combined  State}
\begin{center}
\begin{tabular}{|c|c|}
\hline \parbox{0.9in}{\ \\\begin{center}Alice's Measurement Outcomes\end{center}\\ \ \\} & \parbox{1.75in}{\ \\\begin{center}Alice, Bob and Charlie's Combined State\end{center}\ \\ \ \\}\\
\hline $|\phi^\pm \rangle$  & \{\(\frac{a+d}{\sqrt{2}}[\alpha|000\rangle\pm\beta|111\rangle]+ \\ & \(\frac{a-d}{\sqrt{2}}[\alpha|011\rangle\pm\beta|100\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|101\rangle\pm\beta|010\rangle]+ \\ & \(\frac{b-c}{\sqrt{2}}[\alpha|110\rangle\pm\beta|001\rangle]\}  \\
\hline $|\psi^\pm\rangle$ & \{\(\frac{a+d}{\sqrt{2}}[\alpha|111\rangle\pm\beta|000\rangle]+\\ & \(\frac{a-d}{\sqrt{2}}[\alpha|100\rangle\pm\beta|011\rangle]+ \\ & \(\frac{b+c}{\sqrt{2}}[\alpha|010\rangle\pm\beta|101\rangle]+ \\ &
\(\frac{b-c}{\sqrt{2}}[\alpha|001\rangle\pm\beta|110\rangle]\}   \\
\hline
\end{tabular}
\end{center}
\end{table}
\\

\noindent \textbf{Revocation of Secret on Alice's location} Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie (with Bob's and Charlie's qubit are written in $|+\rangle$ and $|-\rangle$ basis), when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$, can be rewritten as,
\begin{eqnarray}
|Y_{234}^{\phi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\}.
\end{eqnarray}
\\

\noindent Now if we write Bob and Charlie's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$ can be rewritten as,

\begin{eqnarray}
|Y_{234}^{\psi^\pm}\rangle&=&(\frac{a+d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B}(|+\rangle-|-\rangle)_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}]\pm{}\nonumber\\&&\beta[|1\rangle_{A}(|+\rangle-|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle+|-\rangle)_{C}|0\rangle_{C}] +{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[|1\rangle_{A}(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[|0\rangle_{A}(|+\rangle+|-\rangle)_{B} (|+\rangle-|-\rangle)_{C}|1\rangle_{C}]\} .
\end{eqnarray}
\\

\noindent Now if Bob and Charlie carry out the measurement in Hadamard basis, the resultant state on Alice's side is given by the following:\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\phi^{\pm} \rangle$  :}\\

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha\pm b\beta)|0\rangle+\sqrt{2}(b\alpha\pm a\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \pm c\beta)|0\rangle-\sqrt{2}(c\alpha \pm d\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(d\alpha \mp c\beta)|0\rangle+\sqrt{2}(c\alpha \mp d\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(a\alpha \mp b\beta)|0\rangle-\sqrt{2}(b\alpha \mp a\beta)|1\rangle. $\\

\noindent \textbf{Bob and Charlie's joint measurement outcome when Alice's initial measurement outcome is $|\psi^{\pm} \rangle$  :}

\noindent When Bob and Charlie measure in the Hadamard Basis $|+\rangle|+\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(b\alpha\pm a\beta)|0\rangle+\sqrt{2}(a\alpha\pm b\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis: $|+\rangle|-\rangle$, Alice's resultant state is as follows:\\
$\sqrt{2}(c\alpha \pm d\beta)|0\rangle-\sqrt{2}(d\alpha \pm c\beta)|1\rangle.$

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|+\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(c\alpha \mp d\beta)|0\rangle-\sqrt{2}(d\alpha \mp c\beta)|1\rangle. $

\noindent When Bob and Charlie measure in the Hadamard Basis $|-\rangle|-\rangle$, Alice's resultant state is as follows:\\
$-\sqrt{2}(b\alpha \mp b\beta)|0\rangle-\sqrt{2}(a\alpha \mp b\beta)|1\rangle $.\\

\noindent Now if Bob and Charlie sends their measurement outcomes inform of two cbits with the encoding $+ \rightarrow 0$, $- \rightarrow 1$, Alice will be able to revive the secret with the applications of appropriate Pauli operator. However this will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which revocation will be possible. Next in the form of following theorem, we give the range of input parameters for which Alice will be able to revive the secret.

\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Alice to retrieve the secret are as follows:
\begin{enumerate}

     \item If $a$ and $c$ are chosen as real numbers, then we have $b$ and $d$ as purely imaginary numbers such that: \\
     (a) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$ (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$

     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $a$ and $d$ are chosen to be real numbers, then we have $b$ and $c$ as purely imaginary numbers such that:
     (a) $a^2 \leq \frac{1}{2}$ or  $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c)\ $|b|^2 = \frac{1}{2} - a^2$
     (d)\ $|c|^2 = \frac{1}{2} - d^2$

     \item If $b$ and $c$ are chosen to be real numbers, then we have $a$ and $d$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b)\ $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d)\ $|d|^2 = \frac{1}{2} - c^2$

     \item If $b$ and $d$ are chosen to be real numbers, then we have $a$ and $c$ as purely imaginary numbers such that:
     (a) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (b) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (c) $|a|^2 = \frac{1}{2} - b^2$
     (d) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
The union of conditions 1, 2, 3 and 4 given above gives all the necessary values of a, b, c and d for Alice to retrieve the secret.
\end{theorem}
\\

\noindent For the protocol to be successful, the reconstruction at the shareholder's place should be possible in the normal scenario when all the share holder's are not dishonest. Next we consider the reconstruction of the secret at the Charlie's place (say). The same procedure can be repeated for Bob also.

\noindent \textbf{Reconstruction of Secret on Charlie's location:} Now if we write Alice and Bob's qubit in the Hadamard basis, then the combined shared system of Alice, Bob and Charlie, when Alice's initial measurement is $|\phi^{\pm}\rangle$ can be rewritten as:
\begin{eqnarray}
|Y_{234}^{\phi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}]|1\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]\}.
\end{eqnarray}

The combined shared system, when Alice's initial measurement is $|\psi^{\pm}\rangle$ is given by
\begin{eqnarray}
|Y_{234}^{\psi^{\pm}}\rangle&=
(\frac{a+d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle0|-\rangle)_{B}|0\rangle_{C}]\}+{}\nonumber\\&&(\frac{a-d}{2\sqrt{2}})\{\alpha[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|0\rangle_{C}]\pm{}\nonumber\\&&\beta[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|1\rangle_{C}]\}+{}\nonumber\\&&(\frac{b+c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}]|0\rangle_{C}\pm{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}] -{}\nonumber\\&&(\frac{b-c}{2\sqrt{2}})\{[\alpha[(|+\rangle+|-\rangle)_{A} (|+\rangle+|-\rangle)_{B}|1\rangle_{C}]+{}\nonumber\\&&\beta[(|+\rangle-|-\rangle)_{A} (|+\rangle-|-\rangle)_{B}|0\rangle_{C}]\}.
\end{eqnarray}

\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\phi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle]$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle] $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle] $\\

\noindent \textbf{Alice and Bob's joint measurement outcomes when Alice's initial measurement outcome is $|\psi^{\pm}\rangle$ :}

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|+\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(a+b+c-d)\alpha-(a+b-c+d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|+\rangle|-\rangle$, Charlie's resultant state is as follows: $\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(a-b-c-d)\beta)]|0\rangle+[(-a+b+c+d)\alpha-(-a+b-c-d)\beta)]|1\rangle].$

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|+\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a-b+c+d)\alpha-(-a+b+c+d)\beta)]|0\rangle+[(a-b-c-d)\alpha-(-a+b-c-d)\beta)]|1\rangle]. $

\noindent When Alice and Bob measure in the Hadamard Basis $|-\rangle|-\rangle$, Charlie's resultant state is as follows:
$\frac{1}{\sqrt{2}}[[(a+b-c+d)\alpha-(a+b+c-d)\beta)]|0\rangle+[(-a-b-c+d)\alpha-(a+b-c+d)\beta)]|1\rangle]. $\\

\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{alice-recon-final.pdf}
\end{figure}

\noindent Now if Alice and Bob send their measurement outcomes inform of two cbits with the same encoding $+ \rightarrow 0$, $- \rightarrow 1$, Charlie can reconstruct the initial secret by applying appropriate Pauli operators on his obtained state. Just like in the case of revocation, here reconstruction of the secret will not be possible for all values of $a,b,c,d$. There will be certain range of the input parameters for which the reconstruction will take place. Next in the form of following lemma, we give the range of input parameters for which Charlie will be able to revive the secret.
\begin{theorem}
For a Quantum Secret Sharing protocol to be implemented using the ${G}_{abcd}$ state, the necessary conditions for Charlie to reconstruct the secret are as follows:
\begin{enumerate}

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $a$ and $c$ are chosen as real numbers, and $b$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = -d$ and $a = -c$}
     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$
     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $a$ and $d$ are chosen as real numbers, and $b$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$b = c$ and $a = d$}

     (b) $a^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|b|^2 = \frac{1}{2} - a^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

     \item
     If $b$ and $c$ are chosen as real numbers, and $a$ and $d$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = -d$ and $b = -c$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $c^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq c \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|d|^2 = \frac{1}{2} - c^2$

     \item
     If $b$ and $d$ are chosen as real numbers, and $a$ and $c$ are chosen as imaginary numbers, then we have:\\
     (a) {$a = c$ and $b = d$}

     (b) $b^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq b \leq \frac{1}{\sqrt{2}}$
     (c) $d^2 \leq \frac{1}{2}$ or $-\frac{1}{\sqrt{2}} \leq d \leq \frac{1}{\sqrt{2}}$

     (d) $|a|^2 = \frac{1}{2} - b^2$
     (e) $|c|^2 = \frac{1}{2} - d^2$

\end{enumerate}
\end{theorem}
\\
\begin{figure}[h]

  \includegraphics[width=.75\textwidth,left]{charlie-recon-final.pdf}
\end{figure}

\noindent For the proposed protocol to be successful there will be common range of the values of $a,b,c,d$ for which both revocation and reconstruction will be possible which we get by the union of conditions 1, 2, 3, 4, 5, 6, 7 and 8 in theorem 2 given above.

The intersection of the plane and the cylinder in the figure above gives all possible values of the tuple $(a, b, d)$ for case 2 in theorem 2, and where $c$ will have a unique value.

\\

\\

\section{Conclusion:} In this article we  address the revocation and reconstruction of quantum secret for  1-dealer and 2-share holders scenario  using 4-qubit entangled states. Here Alice, (the dealer) also possesses a \textbf{quantum share}  of the  secret and can always bring back her secret with the help of that quantum share when she finds other shareholders to be dishonest. The protocol ensures the normal reconstruction of the secret at shareholder's location otherwise. The actual advantage of the protocol is, it prevents the  shareholders from any possibility of
reconstructing the secret if they are dishonest and collude.  We not only give the basic idea of the protocol but also exemplify with the help of four qubit state. Our protocol will ensure an additional level of security to the normal quantum secret sharing protocols when all the share holders are dishonest.\\

\\
\noindent \textit{\bf Note:} For proof of \textbf{Theorem 1} and \textbf{Theorem 2}, please see the Appendix.\\