text:
\section{Introduction}

It is well-known that Entanglement of quantum state is a valuable physical
resource in quantum information theories\cite{nielsen00}. It makes the quantum
teleportation\cite{bennett93} and superdense coding\cite{bennett92} possible within the quantum
mechanical law. Furthermore, it is the physical resource which is responsible for the
speed-up of the quantum computer\cite{vidal03-1}. In this reason there has been a
flurry of activity recently in the research of entanglement.

Many new properties for the entanglement of three- or higher-qubit pure states have been
reported in the recent papers\cite{pure}. However, it is in general much more difficult
to understand the properties of entanglement for the mixed states except bipartite case.
These difficulties are mainly originated from the fact that the mixed state entanglement is
defined by a convex-roof extension\cite{benn96,uhlmann99-1} of the pure state entanglement.
In order to compute the entanglement defined by convex-roof method one should derive the
optimal decomposition for the given mixed state. Generally, however, it is highly non-trivial
task to derive the optimal decomposition for the arbitrary mixed states.
This computational difficulty makes it difficult to characterize the multipartite entanglement
for the mixed states.

For the bipartite qubit states, fortunately, Wootters found how to derive the optimal
decomposition for the concurrence, entanglement measure for the bipartite states, in
Ref.\cite{form2,form3}. Thus, one can compute the concurrence ${\cal C} (\rho)$ for the
arbitrary mixed states $\rho$ by Wootters formula
\begin{equation}
\label{two4}
{\cal C} (\rho) = \mbox{max} (0, \lambda_1 - \lambda_2 - \lambda_3 - \lambda_4)
\end{equation}
where $\lambda_i$'s are the eigenvalues, in decreasing order, of the Hermitian matrix
$$
\sqrt{\sqrt{\rho} (\sigma_y \otimes \sigma_y) \rho^* (\sigma_y \otimes \sigma_y)
\sqrt{\rho}}. $$

Complete understanding on the bipartite concurrence makes it possible to derive the
purely three-party entanglement, called three-tangle, for the three-qubit pure
states\cite{tangle1}. This arises from the observation that the three-qubit pure state
$|\psi_{ABC}\rangle$ satisfies the following inequality
\begin{equation}
\label{ckw1}
{\cal C}^2_{AB} + {\cal C}^2_{AC} \leq {\cal C}^2_{A (BC)}
\end{equation}
where
${\cal C}_{AB}$ and ${\cal C}_{AC}$ are concurrences for the reduced states
$\rho^{AB} = \mbox{Tr}_C |\psi_{ABC}\rangle \langle \psi_{ABC}|$ and
$\rho^{AC} = \mbox{Tr}_B |\psi_{ABC}\rangle \langle \psi_{ABC}|$, and ${\cal C}_{A (BC)}$ is
a concurrence between a pair $BC$ and $A$. Therefore, ${\cal C}_{A (BC)}$ represents an
total entanglement of the qubit $A$ arising due to the remaining qubits. For pure state
${\cal C}^2_{A (BC)}$ reduces to $4 \mbox{det} \rho^A$, where
$\rho^A = \mbox{Tr}_{BC} |\psi_{ABC}\rangle \langle \psi_{ABC}|$ and is called one-tangle.
In this sense, the inequality (\ref{ckw1}) indicates that the total one-tangle is greater than
sum of two-tangles. In addition, this observation naturally implies that
$\tau_{ABC} \equiv {\cal C}^2_{A (BC)} - ({\cal C}^2_{AB} + {\cal C}^2_{AC})$, which is called
three-tangle, represents the purely three-way entanglement. For three-qubit pure state
$|\psi\rangle = \sum_{i,j,k=0}^1 a_{ijk} |ijk\rangle$, the three-tangle $\tau_{ABC}$
becomes\cite{tangle1}
\begin{equation}
\label{3-tangle-1}
\tau_{ABC} = 4 |d_1 - 2 d_2 + 4 d_3|,
\end{equation}
where
\begin{eqnarray}
\label{3-tangle-2}
& &d_1 = a^2_{000} a^2_{111} + a^2_{001} a^2_{110} + a^2_{010} a^2_{101} +
                                                              a^2_{100} a^2_{011
}
                                                              \\   \nonumber
& &d_2 = a_{000} a_{111} a_{011} a_{100} + a_{000} a_{111} a_{101} a_{010} +
         a_{000} a_{111} a_{110} a_{001}
                                                              \\   \nonumber
& &\hspace{1.0cm} +
         a_{011} a_{100} a_{101} a_{010} + a_{011} a_{100} a_{110} a_{001} +
         a_{101} a_{010} a_{110} a_{001}
                                                              \\   \nonumber
& &d_3 = a_{000} a_{110} a_{101} a_{011} + a_{111} a_{001} a_{010} a_{100}.
\end{eqnarray}
The three-tangle defined by Eq.(\ref{3-tangle-1}) exactly coincides with the modulus of a
Cayley's hyperdeterminant\cite{cay1845,miy03} and is an invariant quantity under the
local $SL(2,\mathbb{C})$ transformation\cite{ver03,lei04}.

The three-tangle (\ref{3-tangle-1}) has following two important properties. Firstly, for a
completely separable ($A-B-C$) and biseparable ($A-BC$, $B-AC$, $AB-C$) states $\tau_{ABC}$
becomes zero. This means that the three-tangle is truly the pure three-party quantity related
to the entanglement. Secondly, the three-tangles for the
Greenberger-Horne-Zeilinger(GHZ)\cite{green89} and W\cite{dur00-1} states defined
\begin{equation}
\label{ghzandw}
|GHZ\rangle = \frac{1}{\sqrt{2}} \left( |000\rangle + |111\rangle \right)
\hspace{1.0cm}
|W\rangle = \frac{1}{\sqrt{3}} \left( |001\rangle + |010\rangle + |100\rangle \right)
\end{equation}
become
\begin{equation}
\label{3-tangle-ghz-w}
\tau_{ABC} (|GHZ\rangle) = 1 \hspace{1.0cm} \tau_{ABC} (|W\rangle) = 0.
\end{equation}
Since the whole three-qubit pure states can be classified by completely separable,
biseparable, GHZ-type, and W-type states through stochastic local operation and classical
communication(SLOCC)\cite{dur00-1}, Eq.(\ref{ghzandw}) indicates that the three-tangle does
not properly reflect the three-party entanglement of the W-type states.

For the mixed states the three-tangle is defined by a convex-roof
method\cite{benn96,uhlmann99-1} as follows:
\begin{equation}
\label{mixed-3-tangle}
\tau_{ABC} (\rho) = \min \sum_i p_i \tau_{ABC} (\rho_i)
\end{equation}
where the minimum is taken over all possible ensembles of pure states. The pure state ensemble
corresponding to the minimum $\tau_{ABC}$ is called optimal decomposition. It is in general
highly difficult to derive the optimal decomposition for the arbitrary mixed states.
Fortunately, Lohmayer et al\cite{tangle2} have derived recently the optimal decomposition
when the mixed state $\rho$ is
\begin{equation}
\label{mixed-ghz-w}
\rho (p) = p |GHZ\rangle \langle GHZ| + (1-p) |W\rangle \langle W|
\end{equation}
and have computed the three-tangle explicitly. They also have found that the
Coffman-Kundu-Wootters(CKW) inequality (\ref{ckw1}) holds for mixed states as well as
pure states.
Subsequently, the three-tangle for the rank-$2$ mixed state composed of the generalized
GHZ and generalized W states has been computed in Ref.\cite{tangle3}. In Ref.\cite{tangle4}
furthermore, the optimal decompositions and the three-tangle for the rank-$3$ mixed state
composed of GHZ, W, and flipped W states are also explicitly derived. Most recently, the
three-tangle for the rank-$4$ mixed states composed of $4$-different GHZ states are explicitly
computed in Ref.\cite{jung09-1}.

On the other hand, in order to reflect the three-party entanglement of the W-type states
properly we need to define new three-party entanglement measure different from the three-tangle.
One of the candidate is a $\pi$-tangle discussed in Ref.\cite{ou07-1}. The $\pi$-tangle is
defined in terms of the global negativities\cite{vidal01-1} defined
\begin{equation}
\label{negativity-1}
{\cal N}^A = || \rho^{T_A} || - 1 \hspace{1.0cm}
{\cal N}^B = || \rho^{T_B} || - 1 \hspace{1.0cm}
{\cal N}^C = || \rho^{T_C} || - 1
\end{equation}
where $||R|| = \mbox{Tr} \sqrt{R R^{\dagger}}$, and the superscripts $T_A$, $T_B$ and $T_C$
represent the partial transpositions for the $A$-qubit, $B$-qubit and $C$-qubit respectively.
Due to the separability criterion via partial transposition\cite{peres96,horod96,horod97}
it is easy to show that the global negativities
vanish for the separable states. It is worthwhile noting that the computation of the global
negativities is relatively simple compared to concurrence or three-tangle for the mixed
states since it does not need the convex-roof extension. In addition, the negativities also
satisfy the monogamy inequality
\begin{equation}
\label{monogamy-2}
{\cal N}_{AB}^2 + {\cal N}_{AC}^2 \leq {\cal N}_{A(BC)}^2
\end{equation}
like concurrence. Then, the $\pi$-tangle is defined as
\begin{equation}
\label{pi-1}
\pi_{ABC} = \frac{1}{3} (\pi_A + \pi_B + \pi_C )
\end{equation}
where
\begin{equation}
\label{pi-2}
\pi_A = {\cal N}_{A(BC)}^2 - ({\cal N}_{AB}^2 + {\cal N}_{AC}^2) \hspace{.5cm}
\pi_B = {\cal N}_{B(AC)}^2 - ({\cal N}_{AB}^2 + {\cal N}_{BC}^2) \hspace{.5cm}
\pi_C = {\cal N}_{(AB)C}^2 - ({\cal N}_{AC}^2 + {\cal N}_{BC}^2).
\end{equation}
It is easy to show that the $\pi$-tangles for $|GHZ\rangle$ and $|W\rangle$ become
\begin{equation}
\label{pi-ghz-w}
\pi_{ABC} (|GHZ\rangle) = 1 \hspace{1.0cm}
\pi_{ABC} (|W\rangle) = \frac{4}{9} (\sqrt{5} - 1) \sim 0.55.
\end{equation}
Thus the $\pi$-tangle reflects the three-party entanglement of the W-type states unlike the
three-tangle.

In this paper we would like to explore the physical role of the three-party entanglement in the
real quantum information process. In order to discuss this issue we adopt the
tripartite teleportation scheme discussed in Ref.\cite{karl98}. Similar issue was discussed
in Ref.\cite{08-mixed}, where the physical role of the concurrence is discussed in the
bipartite teleportation through noisy channels. Ref.\cite{08-mixed} has shown that the
concurrences of the mixed state quantum channels arising due to some noises vanish in the
region of $\bar{F} \leq 2 / 3$, where $\bar{F}$ is an average fidelity between initial
Alice's unknown state and final Bob's state. Since $\bar{F} = 2/3$ corresponds to the best
possible score when Alice and Bob communicate with each other through the classical
channel\cite{popescu94-1}, this result indicates that the entanglement of the quantum
channel is a genuine physical resource for the teleportation process.

This paper is organized as follows. In section II we re-formulate the tripartite teleportation
process\cite{karl98} in terms of the density matrix. This re-description allows us to
formulate the tripartite teleportation process when quantum channel is mixed state. The
several basic quantities are calculated in this section, which are essential for the
calculation of various fidelities in next sections. In section III we compute the accomplice's
fidelities and receiver's fidelities when the tripartite teleportation process is performed
through noisy channels.
In section IV we compute the $\pi$-tangles for the various noisy channels.
It is shown that the $\pi$-tangles for all noise channels decrease with increasing
the decoherence parameter $\kappa t$. This is in fact expected due to the fact that
the decoherence in general disentangles the entanglement of quantum states like
``sudden death''. The $\pi$-tangle for X- and Z-noise channels vanish at the
$\kappa t \rightarrow \infty$ limit. However, the $\pi$-tangles for Y- and isotropy-noise
channels are found to be non-zero at the finite range of $\kappa t$.
In section V we compute the three-tangles for the X- and Z-noise channels. It is shown that
the three-tangle for the Z-noise channel is exactly same with the corresponding $\pi$-tangle.
The three-tangle for the X-noise channel is shown to have three different expressions depending
on the range of $\kappa t$. Since the channels for the Y- and isotropy-noises are
rank-$8$ mixed states, there is no general method to compute the three-tangles. However, we
derived the upper bound of these three-tangles.
In section VI we analyze the $\pi$-tangle and three-tangle by making use of the receiver's
fidelities. The $\pi$-tangle seems to be too large to have a nice physical interpretation.
The three-tangle also seems to be too small by similar manner. This analysis strongly
suggests that we may need different three-party entanglement measure whose value is between
three-tangle and $\pi$-tangle.

\section{Basic Quantities}

In this section we want to re-formulate the tripartite teleportation scheme in terms of the
density matrices\cite{yeo03-1}. It involves sender (Alice), accomplice (Bob) and
receiver (Charlie). Initially they share each single qubit of the GHZ state, {\it i.e.}
$\rho_{GHZ} = |GHZ\rangle_{234} \langle GHZ|$. The purpose of the tripartite teleportation is
as follows. Firstly, Alice at location $2$ should transport a single qubit state
\begin{equation}
\label{unknown-1}
\rho_{in} = |\psi_{in}\rangle \langle \psi_{in}| \hspace{1.0cm}
|\psi_{in}\rangle = \cos \left(\frac{\theta}{2}\right) e^{i \phi/2} |0\rangle +
                    \sin \left(\frac{\theta}{2}\right) e^{-i \phi/2} |1\rangle
\end{equation}
to the receiver, Charlie, at location $4$ with fidelity $\bar{F}_C$ as high as possible
with the help of the accomplice, Bob, at location $3$. At the second time Alice should
transport $\rho_{in}$ to the accomplice, Bob, with fidelity $\bar{F}_B$ as high as
possible. Of course, we cannot make $\bar{F}_B = \bar{F}_C = 1$ due to
no-cloning/broadcast theorems\cite{wootters82,barnum96}. The task is accomplished if one
can make $\bar{F}_B$ and $\bar{F}_C$ as high as possible. In this sense the tripartite
teleportation scheme is similar to a quantum copier
(cloning device)\cite{buzek96-1,buzek97-1,buzek97-2,gisin97-1}.

From the postulate of quantum mechanics on composite systems the initial state of the
tripartite teleportation process should be
\begin{equation}
\label{initial-1}
\rho_{in} \otimes \rho_{GHZ}.
\end{equation}
As will be discussed below $\rho_{GHZ}$ will be changed into $\varepsilon (\rho_{GHZ})$
when noise is introduced when Alice, Bob and Charlie prepare the GHZ state initially,
where $\varepsilon$ is a quantum operation\cite{nielsen00}.

At the next stage Alice performs a projective measurement by preparing a set of the
measurement operators $\{M_1, M_2, M_3, M_4 \}$ with
\begin{equation}
\label{projective-1}
M_1 = |\Phi^+\rangle \langle \Phi^+|    \hspace{.5cm}
M_2 = |\Phi^-\rangle \langle \Phi^-|    \hspace{.5cm}
M_3 = |\Psi^+\rangle \langle \Psi^+|    \hspace{.5cm}
M_4 = |\Psi^-\rangle \langle \Psi^-|,
\end{equation}
where
\begin{equation}
\label{bell-basis}
|\Phi^{\pm}\rangle = \frac{1}{\sqrt{2}} \left( |00\rangle \pm |11\rangle \right)_{12}
\hspace{1.0cm}
|\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}} \left( |01\rangle \pm |10\rangle \right)_{12}.
\end{equation}
Since $|\Phi^{\pm}\rangle$ and $|\Psi^{\pm}\rangle$ form a Bell basis, the operators satisfy
the completeness constraint
\begin{equation}
\label{complete-1}
\sum_m M_m^{\dagger} M_m = I.
\end{equation}
From the quantum mechanical postulates the probability $P_{m}$, probability that the result
of the Alice's measurement is $m$, is given by
\begin{equation}
\label{probability-1}
P_m = \mbox{Tr} \left[ \left(M_m^{\dagger} M_m \otimes I_{34} \right)
                       \left( \rho_{in} \otimes \rho_{GHZ} \right) \right]
\end{equation}
and the state of the system after the Alice's measurement reduces to
\begin{equation}
\label{post-1}
\tilde{\rho}_m = \frac{1}{P_m} (M_m \otimes I_{34}) \left( \rho_{in} \otimes \rho_{GHZ} \right)
                    (M_m \otimes I_{34})^{\dagger}.
\end{equation}
For our case we have $P_1 = P_2 = P_3 = P_4 = 1/4$. After measurement, Alice broadcasts her
measurement outcome to Bob and Charlie via a classical channel.

Next, our concern is moved to the subsystem of Bob and Charlie. This process can be performed
by tracing out the Alice's subsystem, {\it i.e}
\begin{equation}
\label{partial-1}
\pi_{3,4}^m = \mbox{Tr}_{1,2} \left(\tilde{\rho}_m \right).
\end{equation}
After then, the accomplice, Bob, performs a projective measurement again by preparing a set of
measurement operators $\{N_1, N_2 \}$ with
\begin{equation}
\label{projective-2}
N_1 = |\mu^+\rangle \langle \mu^+| \hspace{1.0cm}
N_2 = |\mu^-\rangle \langle \mu^-|,
\end{equation}
where
\begin{equation}
\label{basis-2}
|\mu^+\rangle = \sin \nu |0\rangle + \cos \nu |1\rangle     \hspace{1.0cm}
|\mu^-\rangle = \cos \nu |0\rangle - \sin \nu |1\rangle.
\end{equation}
Since $|\mu^+\rangle$ and $|\mu^-\rangle$ form a basis for the Bob's qubit, the completeness
condition
\begin{equation}
\label{complete-2}
N_1^{\dagger} N_1 + N_2^{\dagger} N_2 = I
\end{equation}
is naturally satisfied. From the quantum mechanical postulates again the probability $q_{mn}$,
probability that the result of the Bob's measurement is $n$ on condition that the outcome of
Alice's measurement is $m$, reduces to
\begin{equation}
\label{probability-2}
q_{mn} = \mbox{Tr} \left[ (N_n \otimes I_4) \pi_{3,4}^m \right]
\end{equation}
and the state of the system after the Bob's measurement becomes
\begin{equation}
\label{post-2}
\tilde{\pi}_{mn} = \frac{1}{q_{mn}} (N_n \otimes I_4) \pi_{3,4}^m (N_n \otimes I_4)^{\dagger}.
\end{equation}
For our case $q_{mn}$ becomes
\begin{equation}
\label{probability-3}
q_{11} = q_{21} = q_{32} = q_{42} = \frac{1}{2} \left(1 - \cos 2 \nu \cos \theta \right)
\hspace{.5cm}
q_{12} = q_{22} = q_{31} = q_{41} = \frac{1}{2} \left(1 + \cos 2 \nu \cos \theta \right).
\end{equation}

After then, our concern is moved to the subsystem of Charlie by tracing out the Bob's
subsystem, {\it i.e.}
\begin{equation}
\label{partial-2}
\chi_4^{mn} = \mbox{Tr}_3 \left( \tilde{\pi}_{mn} \right).
\end{equation}
Finally, Charlie takes an appropriate unitary transformation to his own qubit
\begin{equation}
\label{unit-1}
\tau_{mn} = u_4^{mn} \chi_4^{mn} \left(u_4^{mn} \right)^{\dagger}.
\end{equation}
The unitary operator $u_4^{mn}$ becomes
\begin{equation}
\label{unit-2}
u_4^{11} = u_4^{22} = I  \hspace{.5cm}
u_4^{12} = u_4^{21} = \sigma_z  \hspace{.5cm}
u_4^{31} = u_4^{42} = \sigma_x  \hspace{.5cm}
u_4^{32} = u_4^{41} = \sigma_y
\end{equation}
where $\sigma_i$ is usual Pauli matrices. At this stage the tripartite teleportation process
is terminated.

Now, we want to discuss the tripartite teleportation process through a noisy channel.
If noise is introduced at the initial stage when Alice, Bob, and Charlie share their each
single qubit of $|GHZ\rangle$, $\rho_{GHZ}$ is, in general, changed into the mixed state.
The mixed state can be derived by solving a master equation in the Lindblad
form\cite{lindblad76}
\begin{equation}
\label{lindblad}
\frac{\partial \rho}{\partial t} = -i [H_S, \rho] + \sum_{i, \alpha}
\left(L_{i,\alpha} \rho L_{i,\alpha}^{\dagger} - \frac{1}{2}
\left\{ L_{i,\alpha}^{\dagger} L_{i,\alpha}, \rho \right\} \right)
\end{equation}
where the Lindblad operator
$L_{i,\alpha} \equiv \sqrt{\kappa_{i,\alpha}} \sigma^{(i)}_{\alpha}$ acts on the
$i$th qubit and describes decoherence. Of course, the operator
$\sigma^{(i)}_{\alpha}$ denotes the
Pauli matrix of the $i$th qubit with $\alpha = x,y,z$. The constant $\kappa_{i,\alpha}$
is approximately equal to the inverse of decoherence time. In this paper we will assume for
simplicity that the constant $\kappa_{i,\alpha}$ is independent of $i$ and $\alpha$, {\it i.e.}
$\kappa_{i,\alpha} = \kappa$.

Solutions of Eq.(\ref{lindblad}) for the
$(L_{2,x}, L_{3,x}, L_{4,x})$, $(L_{2,y}, L_{3,y}, L_{4,y})$,
$(L_{2,z}, L_{3,z}, L_{4,z})$, and isotropy noises were solved explicitly in Ref.\cite{all1}.
The spectral decompositions of the results are as follows:
\begin{eqnarray}
\label{summary1}
& &\varepsilon_X (\rho_{GHZ}) = x |GHZ, 1\rangle \langle GHZ, 1| +
\frac{1-x}{3} \bigg[|GHZ, 3\rangle \langle GHZ, 3|
                                                           \\   \nonumber
& & \hspace{2.0cm} + |GHZ, 5\rangle \langle GHZ, 5| +
|GHZ, 7\rangle \langle GHZ, 7| \bigg]  \hspace{1.0cm}
\left( x = \frac{1}{4} (1 + 3 e^{-4 \kappa t}) \right)   \\  \nonumber
& &\varepsilon_Y (\rho_{GHZ}) = \frac{y_+^3}{8} |GHZ, 1\rangle \langle GHZ, 1| +
\frac{y_-^3}{8} |GHZ, 2\rangle \langle GHZ, 2|           \\  \nonumber
& & \hspace{2.0cm} + \frac{y_+ y_-^2}{8} \bigg[ |GHZ, 3\rangle \langle GHZ, 3|
+ |GHZ, 5\rangle \langle GHZ, 5| +
|GHZ, 7\rangle \langle GHZ, 7| \bigg]                  \\   \nonumber
& & \hspace{2.0cm} + \frac{y_+^2 y_-}{8} \bigg[ |GHZ, 4\rangle \langle GHZ, 4|
+ |GHZ, 6\rangle \langle GHZ, 6| +
|GHZ, 8\rangle \langle GHZ, 8| \bigg]                  \\   \nonumber
& & \hspace{11.0cm} (y_{\pm} = 1 \pm e^{-2 \kappa t})     \\   \nonumber
& &\varepsilon_Z (\rho_{GHZ}) = z |GHZ, 1\rangle \langle GHZ, 1| +
(1-z) |GHZ, 2\rangle \langle GHZ, 2|  \hspace{1.0cm} \left(z = \frac{1}{2} (1 + e^{-6 \kappa t})
                                                                      \right)
                                                                             \\   \nonumber
& &\varepsilon_I (\rho_{GHZ}) = \frac{1 + 3 p^2 + 4 p^3}{8} |GHZ, 1\rangle \langle GHZ, 1|
+  \frac{1 + 3 p^2 - 4 p^3}{8} |GHZ, 2\rangle \langle GHZ, 2|
                                                                            \\  \nonumber
& & \hspace{3.0cm} + \frac{1 - p^2}{8} \bigg[ I - (|000\rangle \langle000|
                                            + |111\rangle \langle111| ) \bigg]
\hspace{1.0cm} (p = e^{-4 \kappa t})
\end{eqnarray}
where the subscripts $X$, $Y$, $Z$, and $I$ represent the type of noise channels, and
\begin{eqnarray}
\label{summary2}
& &|GHZ, 1\rangle = \frac{1}{\sqrt{2}} \left( |000\rangle + |111\rangle \right) \hspace{1.0cm}
|GHZ, 2\rangle = \frac{1}{\sqrt{2}} \left( |000\rangle - |111\rangle \right)
                                                                           \\   \nonumber
& &|GHZ, 3\rangle = \frac{1}{\sqrt{2}} \left( |001\rangle + |110\rangle \right) \hspace{1.0cm}
|GHZ, 4\rangle = \frac{1}{\sqrt{2}} \left( |001\rangle - |110\rangle \right)
                                                                           \\   \nonumber
& &|GHZ, 5\rangle = \frac{1}{\sqrt{2}} \left( |010\rangle + |101\rangle \right) \hspace{1.0cm}
|GHZ, 6\rangle = \frac{1}{\sqrt{2}} \left( |010\rangle - |101\rangle \right)
                                                                           \\   \nonumber
& &|GHZ, 7\rangle = \frac{1}{\sqrt{2}} \left( |011\rangle + |100\rangle \right) \hspace{1.0cm}
|GHZ, 8\rangle = \frac{1}{\sqrt{2}} \left( |011\rangle - |100\rangle \right).
\end{eqnarray}

\begin{center}
\begin{tabular}{c||c|c|c} \hline
quantities & no noise and $Z$ noise & $X$ and $Y$ noises  & Isotropy noise  \\  \hline \hline
$P_1, P_2, P_3, P_4$ & $\frac{1}{4}$ & $\frac{1}{4}$ & $\frac{1}{4}$     \\    \hline
$q_{11}, q_{21}, q_{32}, q_{42}$ & $\frac{1}{2} (1 - \cos 2 \nu \cos \theta)$ &
$\frac{1}{2} (1 - \cos 2 \nu \cos \theta e^{-4 \kappa t})$ &
$\frac{1}{2} (1 - \cos 2 \nu \cos \theta e^{-8 \kappa t})$   \\   \hline
$q_{31}, q_{41}, q_{12}, q_{22}$ & $\frac{1}{2} (1 + \cos 2 \nu \cos \theta)$ &
$\frac{1}{2} (1 + \cos 2 \nu \cos \theta e^{-4 \kappa t})$ &
$\frac{1}{2} (1 + \cos 2 \nu \cos \theta e^{-8 \kappa t})$   \\   \hline
\end{tabular}

\vspace{0.1cm}
Table I: Basic Quantities in Tripartite Teleportation
\end{center}
\vspace{0.5cm}
The probabilities $P_m$'s and $q_{mn}$'s in the noisy channels can be directly computed
by changing $\rho_{GHZ}$ into the mixed states (\ref{summary1}) in Eq.(\ref{probability-1})
and Eq.(\ref{probability-2}). The results for the $(L_{2,x}, L_{3,x}, L_{4,x})$,
$(L_{2,y}, L_{3,y}, L_{4,y})$, $(L_{2,z}, L_{3,z}, L_{4,z})$\footnote{For simplicity, we will
use the terminology X-, Y-, and Z-noises together for $(L_{2,x}, L_{3,x}, L_{4,x})$,
$(L_{2,y}, L_{3,y}, L_{4,y})$, $(L_{2,z}, L_{3,z}, L_{4,z})$ noises},
and isotropy noise channels
are summarized in Table I. As Table I indicated, $P_m$'s and $q_{mn}$'s in the various
noisy channels reduce to $P_i = 1/4 (i=1, \cdots, 4)$ and Eq.(\ref{probability-3}) when
$\kappa = 0$ limit.

\section{Fidelities}

\begin{center}
\begin{tabular}{c||c|c} \hline
Type of noise &  $F_C (\theta, \phi)$ & $\bar{F}_C$
                                                                      \\   \hline \hline
no noise & $1 - \frac{1}{2} (1 - \sin 2 \nu ) \sin^2 \theta$ & $\frac{1}{3} (2 + \sin 2 \nu)$
                                                            \\ \hline
X noise & $\frac{1}{2} [(1 + \sin^2 \theta \cos^2 \phi \sin 2 \nu)$
& $\frac{1}{6} [ (3 + \sin 2 \nu)$   \\
{}  & $ + e^{-4 \kappa t} (\cos^2 \theta + \sin^2 \theta \sin^2 \phi \sin 2 \nu) ]$
& $+ e^{-4 \kappa t} (1 + \sin 2 \nu) ]$ \\  \hline
Y noise & $\frac{1}{2} [1 + e^{-2 \kappa t} \sin^2 \theta \sin^2 \phi \sin 2 \nu +
e^{-4 \kappa t} \cos^2 \theta$
& $\frac{1}{6} [3 + e^{-2 \kappa t} \sin 2 \nu$   \\
{} & $+ e^{-6 \kappa t} \sin^2 \theta \cos^2 \phi \sin 2 \nu]$
& $ + e^{-4 \kappa t} + e^{-6 \kappa t} \sin 2 \nu]$      \\   \hline
Z noise & $1 - \frac{1}{2} (1 - \sin 2 \nu e^{-6 \kappa t}) \sin^2 \theta$
& $\frac{1}{3} [2 + e^{-6 \kappa t} \sin 2 \nu]$           \\   \hline
Isotropy noise & $\frac{1}{2} [1 + e^{-8 \kappa t} \cos^2 \theta + e^{-12 \kappa t}
\sin^2 \theta \sin 2 \nu]$  &
$\frac{1}{6} [3 + e^{-8 \kappa t} + 2 \sin 2 \nu e^{-12 \kappa t}]$   \\   \hline

\end{tabular}

\vspace{0.1cm}
Table II: Charlie's fidelity
\end{center}
\vspace{0.5cm}

The Charlie's fidelity, which measures how well the initial state $\rho_{in}$ is transported
to the Charlie's final state, can be computed as follows. Since Charlie's final state is
$\tau_{mn}$ provided that Alice and Bob measure $m$ and $n$ respectively, one can define
the fidelity $F_{mn}^C$ in this case as a form
\begin{equation}
\label{fidel-C-1}
F_{mn}^C = \mbox{Tr} \left[ \tau_{mn} \rho_{in} \right].
\end{equation}
Averaging over all possible measurement outcomes, we can define the Charlie's fidelity in a
form
\begin{equation}
\label{fidel-C-2}
F_C (\theta, \phi) = \sum_{m=1}^{4} \sum_{n=1}^2 P_{m} q_{mn} F_{mn}^C.
\end{equation}
Finally, averaging $F_C (\theta, \phi)$ over all possible input states, we can define the
Charlie's average fidelity $\bar{F}_C$ as follows:
\begin{equation}
\label{fidel-C-3}
\bar{F}_C = \frac{1}{4 \pi} \int_0^{\pi} d\theta \int_0^{2 \pi} d\phi \sin \theta
F_C (\theta, \phi).
\end{equation}
When there is no noise, $F_C (\theta, \phi)$ and $\bar{F}_C$ becomes
\begin{equation}
\label{fidel-C-4}
F_C (\theta, \phi) = 1 - \frac{1}{2} (1 - \sin 2 \nu) \sin^2 \theta   \hspace{1.0cm}
\bar{F}_C = \frac{1}{3} (2 + \sin 2 \nu).
\end{equation}
Thus, the Charlie's fidelities depend on the set of Bob's measurement operators. If Bob
chooses $\nu = \pi / 4$, $\bar{F}_C$ reaches to its maximum $\bar{F}_C = 1$, which means the
perfect teleportation from Alice to Charlie.

The Charlie's fidelities $F_C (\theta, \phi)$ and $\bar{F}_C$ are summarized in Table II when
the mixed states changed from $|GHZ\rangle$ by various noises are introduced
as a quantum channel. Comparing Table II with Table I of
Ref.\cite{all1}, one can realize that the Charlie's fidelities with $\nu = \pi/4$ exactly
coincide with fidelities of the bipartite teleportation when same noises are introduced
initially in the quantum channel.

In the tripartite teleportation scheme, however, there are additional fidelities
between Alice's state $\rho_{in}$ and Bob's final state.
Since Bob's final state after his measurement is
$N_1$ or $N_2$ defined in Eq.(\ref{projective-2}) with respective probability
$\sum_{i=1}^4 P_i q_{i1}$ or $\sum_{i=1}^4 P_i q_{i2}$, the Bob's final fidelities
can be defined as
\begin{eqnarray}
\label{fidel-B-1}
& &F_B^T (\theta, \phi) = \mbox{Tr} [N_1 \rho_{in}] \sum_{i=1}^4 P_i q_{i1} +
                        \mbox{Tr} [N_2 \rho_{in}] \sum_{i=1}^4 P_i q_{i2}  \\  \nonumber
& &\bar{F}_B^T = \frac{1}{4 \pi} \int_0^{\pi} d\theta \int_0^{2\pi} d\phi \sin \theta
                                     F_B^T (\theta, \phi).
\end{eqnarray}
If one computes $F_B^T (\theta, \phi)$ and $\bar{F}_B^T$ for X-, Y-, Z-, and
isotropy-noise channels, one can show that they are all same as
\begin{equation}
\label{fidel-B-2}
F_B^T (\theta, \phi) = \bar{F}_B^T = \frac{1}{2}.
\end{equation}
This is too small because the optimal value for a classical teleportation scheme is
$2/3$.

\begin{center}
\begin{tabular}{c||c|c|c} \hline
Type of noise &  Alice's outcome & $F_B^m (\theta, \phi)$ & $\bar{F}_B^m$
                                                                      \\   \hline \hline
X-noise & $m=1, 2$ & $\frac{1}{2} + \frac{1}{2} \cos 2 \nu \cos 2 \theta$ &
$\frac{1}{6} \left(3 + e^{-4 \kappa t} \cos^2 2 \nu \right)$             \\
and & {} & $\times \left(\cos 2 \nu \cos \theta - \sin 2 \nu \sin \theta \right)
                         e^{-4 \kappa t}$ & {}                      \\   \cline{2-4}
Y-noise & $m=3, 4$ & $\frac{1}{2} - \frac{1}{2} \cos 2 \nu \cos 2 \theta$ &
$\frac{1}{6} \left(3 - e^{-4 \kappa t} \cos^2 2 \nu \right)$     \\
{} & {} & $\times \left(\cos 2 \nu \cos \theta - \sin 2 \nu \sin \theta \right)
                         e^{-4 \kappa t}$ & {}                   \\   \hline
no-noise & $m=1, 2$ & $\frac{1}{2} \bigg( 1 + \cos^2 2 \nu \cos^2 \theta $ &
$\frac{2}{3} - \frac{1}{6} \sin^2 2 \nu$                          \\
and & {} & $ - \sin 2 \nu \cos 2 \nu \sin \theta \cos \theta \cos \phi \bigg)$ & {}
                                                                         \\   \cline{2-4}
Z-noise & $m=3, 4$ & $\frac{1}{2} \bigg( 1 - \cos^2 2 \nu \cos^2 \theta$ &
$\frac{1}{3} + \frac{1}{6} \sin^2 2 \nu$     \\
{} & {} & $+ \sin 2 \nu \cos 2 \nu \sin \theta \cos \theta \cos \phi \bigg)$ & {}
                                                                            \\   \hline
{} & $m=1, 2$ & $\frac{1}{2} + \frac{1}{2} \cos 2 \nu \cos 2 \theta$ &
$\frac{1}{6} \left(3 + e^{-8 \kappa t} \cos^2 2 \nu \right)$             \\
Isotropy & {} & $\times \left(\cos 2 \nu \cos \theta - \sin 2 \nu \sin \theta \right)
                         e^{-8 \kappa t}$ & {}                      \\   \cline{2-4}
noise & $m=3, 4$ & $\frac{1}{2} - \frac{1}{2} \cos 2 \nu \cos 2 \theta$ &
$\frac{1}{6} \left(3 - e^{-8 \kappa t} \cos^2 2 \nu \right)$     \\
{} & {} & $\times \left(\cos 2 \nu \cos \theta - \sin 2 \nu \sin \theta \right)
                         e^{-8 \kappa t}$ & {}                   \\   \hline
\end{tabular}

\vspace{0.1cm}
Table III: Bob's fidelities just after Alice broadcasts her outcome.
\end{center}
\vspace{0.5cm}

However, one can define the Bob's fidelities at the stage just after Alice broadcasts her
measurement outcome to Bob and Charlie via classical channel. If Alice's outcome is $m$,
then the
Bob's fidelities can be defined as
\begin{eqnarray}
\label{fidel-B-3}
& &F_{B}^m (\theta, \phi) = q_{m1} \mbox{Tr} [N_1 \rho_{in}] +
                            q_{m2} \mbox{Tr} [N_2 \rho_{in}]
                                                                   \\   \nonumber
& &\bar{F}_B^m = \frac{1}{4\pi} \int_0^{\pi} d\theta \int_0^{2\pi} d \phi \sin \theta
                             F_{B}^m (\theta, \phi).
\end{eqnarray}
When there is no noise, it is straightforward to show that $F_{B}^m (\theta, \phi)$ and
$\bar{F}_B^m$ become
\begin{eqnarray}
\label{fidel-B-4}
& &F_{B}^{m=1} (\theta, \phi) = F_{B}^{m=2} (\theta, \phi) =
\frac{1}{2} \left[1 + \cos^2 2 \nu \cos^2 \theta - \sin 2 \nu \cos 2 \nu \sin \theta
                    \cos \theta \cos \phi \right]           \\   \nonumber
& &F_{B}^{m=3} (\theta, \phi) = F_{B}^{m=4} (\theta, \phi) =
\frac{1}{2} \left[1 - \cos^2 2 \nu \cos^2 \theta + \sin 2 \nu \cos 2 \nu \sin \theta
                    \cos \theta \cos \phi \right]           \\   \nonumber
& & \bar{F}_B^{m=1} = \bar{F}_B^{m=2} = \frac{2}{3} - \frac{1}{6} \sin^2 2 \nu
                                                           \\   \nonumber
& & \bar{F}_B^{m=3} = \bar{F}_B^{m=4} = \frac{1}{3} + \frac{1}{6} \sin^2 2 \nu.
\end{eqnarray}
When $m=1$ or $2$, $\bar{F}_B^m$ reaches to its maximum value $2/3$ if $\nu=0$ and $\nu=\pi/2$.
At the same time the Charlie's fidelity $\bar{F}_C$ becomes to its minimum value $2/3$.
When $\bar{F}_B^m$ reaches to its minimum value $1/2$ at $\nu=\pi/4$, $\bar{F}_C$ becomes
to its maximum value $1$. Thus, one can increase/decrease $\bar{F}_B^m$ at the cost of
decreasing/increasing $\bar{F}_C$.

\begin{figure}[ht!]
\begin{center}
\includegraphics[height=10cm]{fig1.eps}
\caption[fig1]{The $\nu$- and $\kappa t$-dependence  of $\bar{F}_C$ and
$\bar{F}_B^m$ with $m=1, 2$ when the type of noise is $(L_{2,z}, L_{3,z}, L_{4,z})$. The upper
and lower surfaces correspond to $\bar{F}_C$ and $\bar{F}_B^m$ respectively. The difference
between $\bar{F}_C$ and $\bar{F}_B^m$ is maximized in the $\kappa t \rightarrow 0$ limit.
However, this difference becomes negligible with increasing $\kappa t$. This is due to
the fact that decoherence is a major dominant effect in the region of large $\kappa t$.}
\end{center}
\end{figure}

The Bob's fidelities $F_B^m (\theta, \phi)$ and $\bar{F}_B^m$ are summarized in Table III when
the various noisy channels are introduced. One of the interesting points of Table III is that
the Bob's fidelities for the Z-noisy channel is independent of the noise parameter $\kappa$
while the Charlie's fidelities is dependent on $\kappa$ as Table II indicated. The $\nu$- and
$\kappa t$-dependence of $\bar{F}_C$ and $\bar{F}_B^m (m=1,2)$ in $(L_{2,z}, L_{3,z}, L_{4,z})$
noisy channel is plotted together in Fig. 1.
The upper surface in the figure corresponds to $\bar{F}_C$
and the lower one to $\bar{F}_B^m$. The difference between $\bar{F}_C$ and $\bar{F}_B^m$ is
averagely maximized when $\kappa = 0$, which means there is no noise. If, however, $\kappa t$
becomes larger and larger, the difference between two fidelities becomes negligible. This is
due to the fact that the effect of noise is significant compared to the choice of $\nu$ in the
Bob's measurement. One can find a similar behaviors in the other noisy channels although we
have not presented the $\nu$- and $\kappa t$-dependence of the fidelities explicitly in this
paper.

\section{$\pi$-tangle}

In this section we will compute the $\pi$-tangle of the various noisy channels
defined in Eq.(\ref{pi-1}). When there is no noise, it is easy to show that
\begin{equation}
\label{pi-11}
|| \rho_{GHZ}^{T_A} || = || \rho_{GHZ}^{T_B} || = || \rho_{GHZ}^{T_C} || = 2,
\end{equation}
which results in
\begin{equation}
\label{pi-12}
{\cal N}_{A(BC)} = {\cal N}_{B(AC)} = {\cal N}_{C(AB)} = 1.
\end{equation}
In addition, one can show that there is no contribution to the entanglement from the
two-tangles in GHZ state:
\begin{equation}
\label{pi-13}
{\cal N}_{AB} = {\cal N}_{AC} = {\cal N}_{BC} = 0.
\end{equation}
Thus, $\pi$-tangle for the GHZ state is simply
\begin{equation}
\label{pi-14}
\pi_{ABC}^{GHZ} = 1,
\end{equation}
which indicates that the GHZ state is a maximally entangled state.

\begin{center}
\begin{tabular}{c||c} \hline
Type of noise &   $\pi$-tangle
                                                                      \\   \hline \hline
no noise & $1$                                                 \\   \hline
X noise &
$ e^{-8 \kappa t}$   \\    \hline
Y noise &
$\frac{1}{64} \bigg[ |1 - 3 e^{-2 \kappa t} - e^{-4 \kappa t} - e^{-6 \kappa t}| $  \\
{} &
 $- (1 - 3 e^{-2 \kappa t} - e^{-4 \kappa t} - e^{-6 \kappa t}) \bigg]^2$  \\ \hline
Z noise & $e^{-12 \kappa t}$
                                     \\   \hline
Isotropy &  $\frac{1}{64} \bigg[ |1 - e^{-8 \kappa t} - 4 e^{-12 \kappa t}|$  \\
noise &
$ - (1 - e^{-8 \kappa t} - 4 e^{-12 \kappa t}) \bigg]^2$   \\ \hline

\end{tabular}

\vspace{0.1cm}
Table IV: The $\pi$-tangles for the various noisy channels
\end{center}
\vspace{0.5cm}

The $\pi$-tangles for $(L_{2,x}, L_{3,x}, L_{4,x})$, $(L_{2,y}, L_{3,y}, L_{4,y})$,
$(L_{2,z}, L_{3,z}, L_{4,z})$, and isotropy channels can be computed straightforwardly.
For all noisy channels ${\cal N}_{A(BC)} = {\cal N}_{B(AC)} = {\cal N}_{C(AB)}$ and
${\cal N}_{AB} = {\cal N}_{AC} = {\cal N}_{BC} = 0$ hold. This seems to be due to the fact that
we considered only same-axis noisy channels. The $\pi$-tangles for the various noisy channels
are summarized at Table IV.  The interesting fact Table IV indicates is that while
the $\pi$-tangles for the X- and Z-noise channels vanish at $\kappa t \rightarrow \infty$
limit, those for the Y- and isotropy-noise channels goes to zero at
$y_* \leq \kappa t \leq \infty$ and $i_* \leq \kappa t \leq \infty$ respectively, where
\begin{eqnarray}
\label{pi-15}
& &y_* = \ln \frac{1 + (19 + 3 \sqrt{33})^{1/3} + (19 - 3 \sqrt{33})^{1/3}}{3}
       \sim 0.609378                                        \\   \nonumber
& &i_* = \frac{1}{4} \ln \frac{(54 + 3 \sqrt{321})^{1/3} + (54 - 3 \sqrt{321})^{1/3}}{3}
       \sim 0.146435.
\end{eqnarray}

\begin{figure}[ht!]
\begin{center}
\includegraphics[height=6cm]{fig2a.eps}
\includegraphics[height=6cm]{fig2b.eps}
\includegraphics[height=6cm]{fig2c.eps}
\includegraphics[height=6cm]{fig2d.eps}
\caption[fig2]{ The $\kappa t$ dependence of $\pi$-tangles and Charlie's average
fidelities $\bar{F}_C$ in $(L_{2,x}, L_{3,x}, L_{4,x})$ (Fig. 2a),
$(L_{2,y}, L_{3,y}, L_{4,y})$ (Fig. 2b), $(L_{2,z}, L_{3,z}, L_{4,z})$ (Fig. 2c), and
isotropy (Fig. 2d) noisy channels. }
\end{center}
\end{figure}

The $\kappa t$-dependence of $\pi$-tangles together with Charlie's average fidelity $\bar{F}_C$
for the various noisy channels are plotted in Fig. 2. In the Z-noisy channel the $\pi$-tangle
vanishes at $\kappa t = \infty$ and at this limit $\bar{F}_C$ goes to $2/3$ regardless of
$\nu$, which is a classical fidelity limit. In the X-noise channel $\bar{F}_C$ goes to
$(3 + \sin 2 \nu) / 6$ at $\kappa t = \infty$. When $\nu = \pi / 4$, this also goes to
$2 /3$. Therefore, the $\pi$-tangles for X- and Z-noise channels seem to show a nice
connection between the Charlie's average fidelity and three-way entanglement of the
given channel.

However, this nice property is not maintained in the Y- and isotropy-noise channel. In the
Y-noise channel the $\pi$-tangle vanishes at $y_* \leq \kappa t$. At $\kappa t= y_*$
the Charlie fidelity reduces to
$ 0.166667 (3.08738 + 0.321426 \sin 2 \nu) $,
whose maximum is $f_Y = 0.568314$. Thus $f_Y$ is much less than the classical fidelity
limit $2/3$. Similar behavior can be found in the isotropy channel. In this channel the
$\pi$-tangle vanishes at $i_* \leq \kappa t$. At $\kappa t = i_*$ the maximum Charlie's
fidelity becomes $f_I = 0.609159$, which is also less than the classical limit $2/3$.

\section{three-tangle}
In this section we would like to discuss the three-tangles for the various noisy channels
expressed in Eq.(\ref{summary1}).

\subsection{ $(L_{2,z}, L_{3,z}, L_{4,z})$ noisy channel}
Let us consider the pure state
\begin{equation}
\label{tangle-z-1}
|Z(z, \varphi)\rangle = \sqrt{z} |GHZ,1\rangle - e^{i \varphi} \sqrt{1-z} |GHZ,2\rangle
\end{equation}
where $z = (1 + e^{-6 \kappa t}) / 2$. It is easy to show that the three-tangle of
$|Z(z, \varphi)\rangle$ is
\begin{equation}
\label{tangle-z-2}
\tau_3 \left(|Z(z, \varphi)\rangle \right) = (1 - 2 z + 2 z^2) - 2 z (1-z) \cos 2\varphi.
\end{equation}
Thus, $\tau_3 \left(|Z(z, \varphi)\rangle \right)$ has a minimum at $\varphi=0$ and
$\varphi=\pi$, i.e.
\begin{equation}
\label{tangle-z-3}
\tau_3 \left(|Z(z, 0)\rangle \right) = \tau_3 \left(|Z(z, \pi)\rangle \right) = (1 - 2 z)^2.
\end{equation}
In terms of the terminologies of Ref.\cite{oster07} $(1-2z)^2$ forms a convex characteristic
curve in $(z, \tau_3 \left(|Z(z, \varphi)\rangle \right))$ plane. In addition, one can show
straightforwardly that $\varepsilon(\rho_{GHZ})$ defined in Eq.(\ref{summary1}) can be
decomposed into
\begin{equation}
\label{tangle-z-4}
\varepsilon(\rho_{GHZ}) = \frac{1}{2} |Z(z, 0)\rangle \langle Z(z,0)| + \frac{1}{2}
|Z(z, \pi)\rangle \langle Z(z,\pi)|.
\end{equation}
If Eq.(\ref{tangle-z-4}) is optimal, then the three-tangle for $\varepsilon(\rho_{GHZ})$ is
$(2z - 1)^2$. Since this coincides with the convex characteristic curve, Eq.(\ref{tangle-z-4})
should be the optimal decomposition. Thus, the three-tangle for the $\varepsilon(\rho_{GHZ})$
is
\begin{equation}
\label{tangle-z-5}
\tau_{ABC}^z = (1 - 2 z)^2 = e^{-12 \kappa t}.
\end{equation}
It is interesting to note that the three-tangle and $\pi$-tangle are same with each other
in this channel.

\subsection{ $(L_{2,x}, L_{3,x}, L_{4,x})$ noisy channel}
Before we start computation, it is worthwhile noting that
as shown in Ref.\cite{jung09-1} the state
\begin{equation}
\label{tangle-x-1}
\Pi_{GHZ} = \frac{1}{3} \bigg[ |GHZ,3\rangle \langle GHZ,3| + |GHZ,5\rangle \langle GHZ,5|
           + |GHZ,7\rangle \langle GHZ,7| \bigg]
\end{equation}
has vanishing three-tangle. This fact is shown in appendix A.

Now, let us consider a pure state
\begin{eqnarray}
\label{tangle-x-2}
& & |X(x, \varphi_1, \varphi_2, \varphi_3)\rangle = \sqrt{x} |GHZ,1\rangle -
e^{i \varphi_1} \sqrt{\frac{1-x}{3}} |GHZ,3\rangle \\   \nonumber
& & \hspace{3.0cm} -
e^{i \varphi_2} \sqrt{\frac{1-x}{3}} |GHZ,5\rangle  -
e^{i \varphi_3} \sqrt{\frac{1-x}{3}} |GHZ,7\rangle
\end{eqnarray}
where $x = (1 + 3 e^{-4\kappa t})/4$. Then it is easy to show that the three-tangle of
$|X(x, \varphi_1, \varphi_2, \varphi_3)\rangle$ becomes
\begin{eqnarray}
\label{tangle-x-3}
& &\tau_3 \left(|X(x, \varphi_1, \varphi_2, \varphi_3)\rangle \right)   \\   \nonumber
& & = \Bigg| x^2 + \frac{(1-x)^2}{9} \left( e^{4i\varphi_1} + e^{4i\varphi_2} +
                                           e^{4i\varphi_3} \right) -
\frac{2}{3} x (1-x) \left( e^{2i\varphi_1} + e^{2i\varphi_2} + e^{2i\varphi_3} \right)
                                                                        \\   \nonumber
& & \hspace{.5cm} - \frac{2}{9} (1-x)^2 \left( e^{2i (\varphi_1 + \varphi_2)} +
e^{2i (\varphi_1 + \varphi_3)} + e^{2i (\varphi_2 + \varphi_3)}   \right) -
\frac{8 \sqrt{3}}{9} \sqrt{x (1 - x)^3} e^{i (\varphi_1 + \varphi_2 + \varphi_3)}  \Bigg|.
\end{eqnarray}

The vectors $ |X(x, \varphi_1, \varphi_2, \varphi_3)\rangle$ has following properties.
The three-tangle of it has the largest zero at $x=x_0 \equiv 3/4$ and
$\varphi_1 = \varphi_2 = \varphi_3 = 0$. The vectors $|X(x, 0, 0, 0)\rangle$,
$|X(x, 0, \pi, \pi)\rangle$, $|X(x, \pi, 0, \pi)\rangle$ and $|X(x, \pi, \pi, 0)\rangle$ have
same three-tangles. Finally, $\varepsilon_X (\rho_{GHZ})$ can be decomposed into
\begin{eqnarray}
\label{tangle-x-4}
& &\varepsilon_X (\rho_{GHZ}) = \frac{1}{4} \Bigg[
|X(x, 0, 0, 0)\rangle \langle X(x, 0, 0, 0)| +
|X(x, 0, \pi, \pi)\rangle \langle X(x, 0, \pi, \pi)|    \\  \nonumber
& & \hspace{2.0cm}
+ |X(x, \pi, 0, \pi)\rangle \langle X(x, \pi, 0, \pi)| +
|X(x, \pi, \pi, 0)\rangle \langle X(x, \pi, \pi, 0)|   \Bigg].
\end{eqnarray}

When $x \leq x_0$, one can construct the optimal decomposition in the following form:
\begin{eqnarray}
\label{tangle-x-5}
& &\varepsilon_X (\rho_{GHZ}) = \frac{x}{4 x_0} \Bigg[
|X(x_0, 0, 0, 0)\rangle \langle X(x_0, 0, 0, 0)| +
|X(x_0, 0, \pi, \pi)\rangle \langle X(x_0, 0, \pi, \pi)|    \\  \nonumber
& & \hspace{2.0cm}
+ |X(x_0, \pi, 0, \pi)\rangle \langle X(x_0, \pi, 0, \pi)| +
|X(x_0, \pi, \pi, 0)\rangle \langle X(x_0, \pi, \pi, 0)|   \Bigg]  \\  \nonumber
& & \hspace{6.0cm}
+ \frac{x_0 - x}{x_0} \Pi_{GHZ}.
\end{eqnarray}
Since $\Pi_{GHZ}$ has the vanishing three-tangle, one can show easily
\begin{equation}
\label{tangle-x-6}
\tau^X_{ABC} = 0  \hspace{1.0cm} \mbox{when} \hspace{,3cm} x \leq x_0 = 3/4.
\end{equation}

Now, let us consider the three-tangle of $\varepsilon_X (\rho_{GHZ})$ in the region
$x_0 \leq x \leq 1$. Since Eq.(\ref{tangle-x-4}) is an optimal decomposition at $x=x_0$,
one can conjecture that it is also optimal in the region $x_0 \leq x$. As will be shown
shortly, however, this is not true at the large-$x$ region. If we compute the three-tangle
under the condition that Eq.(\ref{tangle-x-4}) is optimal at $x_0 \leq x$, its expression
becomes
\begin{equation}
\label{tangle-x-7}
\alpha_I^X (x) = x^2 - \frac{1}{3} (1 - x)^2 - 2 x (1 - x) - \frac{8 \sqrt{3}}{9}
\sqrt{x (1 - x)^3}.
\end{equation}
However, one can show straightforwardly that $\alpha_I^X (x)$ is not a convex function
in the region $x \geq x_*$, where
\begin{equation}
\label{tangle-x-8}
x_* = \frac{1}{4} \left( 1 + 2^{1/3} + 4^{1/3} \right) \approx 0.961831.
\end{equation}
Therefore, we need to convexify $\alpha_I^X (x)$ in the region $x_1 \leq x \leq 1$ to make
the three-tangle to be convex function, where $x_1$ is some number between
$x_0$ and $x_*$. The number $x_1$ will be determined shortly.

In the large $x$-region one can derive the optimal decomposition in a form:
\begin{eqnarray}
\label{tangle-x-9}
& &\varepsilon_X (\rho_{GHZ})    \\   \nonumber
& & = \frac{1-x}{4 (1-x_1)} \Bigg[
|X(x_1, 0, 0, 0)\rangle \langle X(x_1, 0, 0, 0)| +
|X(x_1, 0, \pi, \pi)\rangle \langle X(x_1, 0, \pi, \pi)|    \\  \nonumber
& & \hspace{2.5cm}
+ |X(x_1, \pi, 0, \pi)\rangle \langle X(x_1, \pi, 0, \pi)| +
|X(x_1, \pi, \pi, 0)\rangle \langle X(x_1, \pi, \pi, 0)|   \Bigg]  \\  \nonumber
& & \hspace{4.0cm}
+ \frac{x - x_1}{1 - x_1} |GHZ,1\rangle \langle GHZ,1|
\end{eqnarray}
which gives a three-tangle as
\begin{equation}
\label{tangle-x-10}
\alpha_{II}^X (x, x_1) = \frac{1 - x}{1 - x_1} \alpha_I^X (x_1) + \frac{x - x_1}{1 - x_1}.
\end{equation}
Since $d^2 \alpha_{II}^X / dx^2 = 0$, there is no convex problem if $\alpha_{II}^X (x, x_1)$
is a three-tangle in the large-$x$ region.
The constant $x_1$ can be fixed from the condition of minimum $\alpha_{II}^X$,
i.e. $\partial \alpha_{II}^X (x, x_1) / \partial x_1 = 0$, which gives
\begin{equation}
\label{tangle-x-11}
x_1 = \frac{1}{4} (2 + \sqrt{3}) \approx 0.933013.
\end{equation}
As expected $x_1$ is between $x_0$ and $x_*$. Thus, finally the three-tangle for
$\varepsilon_X (\rho_{GHZ})$ becomes
\begin{eqnarray}
\label{tangle-x-12}
\tau_{ABC}^X = \left\{     \begin{array}{cc}
                        0   & \hspace{1.0cm}  x \leq x_0    \\
                        \alpha_I^X (x)    & \hspace{1.0cm}   x_0 \leq x \leq x_1    \\
                        \alpha_{II}^X (x, x_1)    & \hspace{1.0cm}    x_1 \leq x \leq 1
                           \end{array}                                    \right.
\end{eqnarray}
and the corresponding optimal decompositions are Eq.(\ref{tangle-x-5}), Eq.(\ref{tangle-x-4})
and Eq.(\ref{tangle-x-9}) respectively.
In terms of $\kappa t$ $\tau_{ABC}^X$ reduces to
\begin{eqnarray}
\label{tangle-x-13}
\tau_{ABC}^X = \left\{    \begin{array}{cc}
                 \alpha_{II}^X (x, x_1)    & \hspace{1.0cm} 0 \leq \kappa t \leq \mu_1^X  \\
                 \alpha_I^X (x)    & \hspace{1.0cm} \mu_1^X \leq \kappa t \leq \mu_2^X \\
                 0                &  \hspace{1.0cm}  \mu_2^X \leq \kappa t \leq \infty
                          \end{array}                                                 \right.
\end{eqnarray}
where  $x = (1 + 3 e^{-4 \kappa t}) / 3$ and
\begin{equation}
\label{tangle-x-14}
\mu_1^X = -\frac{1}{4} \ln \frac{4 x_1 - 1}{3} \approx 0.0233899
\hspace{1.0cm}
\mu_2^X = - \frac{1}{4} \ln \frac{2}{3} \approx 0.101366.
\end{equation}

\subsection{$(L_{2,y}, L_{3,y}, L_{4,y}$) noisy channel}

The mixed state $\varepsilon_Y (\rho_{GHZ})$ given in Eq.(\ref{summary1}) can be re-written as
\begin{equation}
\label{tangle-y-1}
\varepsilon_Y (\rho_{GHZ}) = \xi \Pi_1^{GHZ} (Y_1) + (1 - \xi) \Pi_2^{GHZ} (Y_2)
\end{equation}
where
\begin{eqnarray}
\label{tangle-y-2}
& &\Pi_1^{GHZ} (Y_1) = Y_1 |GHZ,1\rangle \langle GHZ,1| \\  \nonumber
& &  \hspace{2.0cm}
+ \frac{1 - Y_1}{3} \bigg[ |GHZ,3\rangle \langle GHZ,3| + |GHZ,5\rangle \langle GHZ,5| +
  |GHZ,7\rangle \langle GHZ,7| \bigg]
                                                                     \\   \nonumber
& &\Pi_2^{GHZ} (Y_2) = Y_2 |GHZ,2\rangle \langle GHZ,2|   \\  \nonumber
& &  \hspace{2.0cm}
+ \frac{1 - Y_2}{3} \bigg[ |GHZ,4\rangle \langle GHZ,4| + |GHZ,6\rangle \langle GHZ,6| +
  |GHZ,8\rangle \langle GHZ,8| \bigg].
\end{eqnarray}
In Eq.(\ref{tangle-y-2}) the constants are given by
\begin{equation}
\label{tangle-y-3}
\xi = \frac{y_+ (y_+^2 + 3 y_-^2)}{8}     \hspace{1.0cm}
Y_1 = \frac{y_+^2}{y_+^2 + 3 y_-^2}       \hspace{1.0cm}
Y_2 = \frac{y_-^2}{3 y_+^2 + y_-^2}
\end{equation}
where $y_{\pm} = 1 \pm e^{-2 \kappa t}$. It is worthwhile noting that $\Pi_2^{GHZ} (Y_2)$ is
local-unitary (LU) equivalent to $\Pi_1^{GHZ} (Y_2)$, i.e.
$$ \Pi_1^{GHZ} (Y_2) = (\sigma_z \otimes \openone \otimes \openone) \Pi_2^{GHZ} (Y_2)
(\sigma_z \otimes \openone \otimes \openone)^{\dagger}. $$
Since the three-tangle is LU-invariant quantity, the three-tangle for $\Pi_2^{GHZ} (Y_2)$ should
be equal to that for $\Pi_1^{GHZ} (Y_2)$. Since $\Pi_1^{GHZ} (Y_2)$ can be obtained from
$\varepsilon_X (\rho_{GHZ})$ by replacing $x$ by $Y_2$, one can compute the three-tangle for
$\Pi_2^{GHZ} (Y_2)$ directly from Eq.(\ref{tangle-x-12}). Since, furthermore, $Y_2 \leq 1/4$
in the entire range of $\kappa t$, the three-tangle for $\Pi_2^{GHZ} (Y_2)$ should be
zero.

Since $\varepsilon_Y(\rho_{GHZ})$ is rank-$8$ mixed state, it seems to be highly difficult to
compute the three-tangle. Still we do not know how to compute it analytically. However, one can
compute its upper bound as following. Since the three-tangle for $\Pi_2^{GHZ} (Y_2)$ is zero
and the three-tangle for the mixed state is obtained by the convex-roof method,
Eq.(\ref{tangle-y-1}) implies that the three-tangle for $\varepsilon_Y(\rho_{GHZ})$ should be less
than $\xi$ times three-tangle for $\Pi_1^{GHZ} (Y_1)$. Since $\Pi_1^{GHZ} (Y_1)$ is same with
$\varepsilon_X(\rho_{GHZ})$ if $x$ is replaced by $Y_1$, one can compute the upper bound of
the three-tangle for $\varepsilon_Y(\rho_{GHZ})$, $\tau_{ABC}^{Y:UB}$ directly from
Eq.(\ref{tangle-x-12}). The superscript UB stands for upper bound. The final result of this
upper bound can be summarized as
\begin{eqnarray}
\label{tangle-y-4}
\tau_{ABC}^{Y:UB} = \left\{     \begin{array}{cc}
                \xi \alpha_{II}^X (Y_1, x_1)  & \hspace{1.0cm} 0 \leq \kappa t \leq \nu_1^*  \\
                \xi \alpha_I^X (Y_1)    & \hspace{1.0cm} \nu_1^* \leq \kappa t \leq \nu_2^*  \\
                0                       & \hspace{1.0cm} \nu_2^* \leq \kappa t \leq \infty
                                 \end{array}                   \right.
\end{eqnarray}
where
\begin{equation}
\label{tangle-y-5}
\nu_1^* = -\frac{1}{2} \ln (\sqrt{3} - 1) \sim 0.155953   \hspace{1.0cm}
\nu_2^* = \frac{1}{2} \ln 2 \sim 0.346574.
\end{equation}
Of course, $x_1$ is given in Eq.(\ref{tangle-x-11}).

\subsection{isotropy noisy channel}

The mixed state $\varepsilon_I (\rho_{GHZ})$ given in Eq.(\ref{summary1}) can be re-written as
\begin{equation}
\label{tangle-i-1}
\varepsilon_I (\rho_{GHZ}) = \zeta \Sigma_1^{GHZ} + (1 - \zeta) \Sigma_2^{GHZ}
\end{equation}
where
\begin{eqnarray}
\label{tangle-i-2}
& &\Sigma_1^{GHZ} = \left(\frac{1}{2} + \frac{2 p^3}{1 + 3 p^2} \right)
                                 |GHZ,1\rangle \langle GHZ,1|           \\   \nonumber
& &\hspace{3.0cm} +
\left(\frac{1}{2} - \frac{2 p^3}{1 + 3 p^2} \right) |GHZ,2\rangle \langle GHZ,2|
                                                            \\   \nonumber
& &\Sigma_2^{GHZ} = \frac{1}{6} \left\{ I - \left(|000\rangle \langle 000| + |111\rangle
                                                  \langle 111| \right) \right\}
\end{eqnarray}
and
\begin{equation}
\label{tangle-i-3}
\zeta = \frac{1 + 3 p^2}{4}
\end{equation}
with $p = e^{-4 \kappa t}$.

The state $\varepsilon_I (\rho_{GHZ})$ is rank-8 mixed state and we do not know how to
compute the three-tangle of it exactly. Since, however, the three-tangle of $\Sigma_2^{GHZ}$
is zero, one can compute at least the upper bound as $\zeta$ times three-tangle of
$\Sigma_1^{GHZ}$. This upper bound can be easily computed by making use of the analytical
result of the three-tangle for the Z-noise channel.
The final result of this upper bound is
\begin{equation}
\label{tangle-i-4}
\tau_{ABC}^{I:UB} = \frac{4 p^6}{1 + 3 p^2} = \frac{4 e^{-24 \kappa t}}{1 + 3 e^{-8 \kappa t}}
\end{equation}
where the superscript UB stands for the upper bound.

\section{Conclusion}

\begin{figure}[ht!]
\begin{center}
\includegraphics[height=6cm]{fig3a.eps}
\includegraphics[height=6cm]{fig3b.eps}
\includegraphics[height=6cm]{fig3c.eps}
\caption[fig3]{ The $\kappa t$ dependence of three-tangle and $\pi$-tangle for
$(L_{2,x}, L_{3,x}, L_{4,x})$ (Fig. 3a),
$(L_{2,y}, L_{3,y}, L_{4,y})$ (Fig. 3b), and
isotropy (Fig. 3c) noisy channels. }
\end{center}
\end{figure}

In this paper we computed the $\pi$-tangles explicitly for the mixed states summarized in
Eq.(\ref{summary1}). It is shown that the $\pi$-tangles for the X- and Z-noisy channels
vanish at $\kappa t \rightarrow \infty$, where the maximum Charlie's fidelities reduce to
the classical limit $2/3$. However, this nice property is not maintained for Y- and
isotropy-noise channels. For Y-noise the $\pi$-tangle vanishes at $y_* \leq \kappa t$, where
$ y_*$ is given at Eq.(\ref{pi-15}). At $\kappa t = y_*$ the maximum Charlie's fidelity becomes
$0.57$, which is much less than the classical limit. For isotropy noise the $\pi$-tangle
vanishes at $i_* \leq \kappa t$. At $\kappa t = i_*$ the maximum Charlie's fidelity
becomes $0.61$, which is also less than the classical limit. Although the $\pi$-tangle was
constructed in Ref.\cite{ou07-1} to reflect the three-party entanglement of the W-type
states, it does not seem to give a meaningful interpretation in the real quantum information
process.

We also computed the three-tangles for the X- and Z-noise channels. The remarkable fact is that
the three-tangle for the Z-noise channel is exactly same with the corresponding
$\pi$-tangle. Therefore, the three-tangle for the Z-noise channel vanishes at
$\kappa t \rightarrow \infty$, where all Charlie's fidelities reduce to the classical limit
regardless of Bob's measurement outcome. For X-noise the $\kappa t$-dependence of the
three tangle is plotted in Fig. 3(a). For comparison we plotted the corresponding $\pi$-tangle
together. As Fig. 3(a) has shown, the three-tangle is much less than the corresponding
$\pi$-tangle. In this channel the three-tangle vanishes at $\mu_2^X \leq \kappa t$, where
$\mu_2^X = -(1/4) \ln (2/3)$. At $\kappa t = \mu_2^X$ the Charlie's fidelity becomes
$(11 + 5 \sin 2 \nu) / 18$. When, therefore, $\nu = (1/2) \sin^{-1} (1/5) \sim 0.100679$,
Charlie's fidelity reduces to the classical limit $2/3$. However, the maximum Charlie's fidelity
goes to $8/9$, which is much larger that the classical limit.

The $\kappa t$-dependence of
$\tau^{Y:UB}_{ABC}$ and $\tau^{I:UB}_{ABC}$ are plotted in Fig. 3(b) and Fig. 3(c) respectively.
For comparison we plotted the corresponding $\pi$-tangle together. Fig. 3(b) shows that
$\tau^{Y:UB}_{ABC}$ is larger than the corresponding $\pi$-tangle at
$0 \leq \kappa t \leq 0.315$. Fig. 3(c) shows that $\tau^{I:UB}_{ABC}$ is larger than the
corresponding $\pi$-tangle in the entire range of $\kappa t$. This is due to the fact that
$\tau^{Y:UB}_{ABC}$ and $\tau^{I:UB}_{ABC}$ are merely the upper bounds of the real
three-tangles for Y- and isotropy-noise channels. If the calculational tool for the
three-tangle of the arbitrary three-party mixed states are developed someday,
the real three-tangles computed via this tool should be less than the corresponding
$\pi$-tangles.

In this paper we examined the physical meaning of the three-tangle and $\pi$-tangle
in the real quantum information process. We adopted the three-party teleportation via
various noisy channels as a model of quantum process.
It is shown that the $\pi$-tangle seems to be too large to have a meaningful interpretation.
Although we cannot compute the three-tangles for Y- and isotropy-noise channels due to their
high rank, the results for X- and Z-noise seems to imply the fact that the three-tangle
is too small to have meaningful interpretation. Probably we need a different three-party
entanglement measure whose value is between three-tangle and $\pi$-tangle.

{\bf Acknowledgement}:
This work was supported by the Kyungnam University
Foundation Grant, 2008.

redpajama_set_name: RedPajamaArXiv


text:
The Hindu Newspaper
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: VF page is not transferring to standard record detail page I am trying to display contact list in the visualforce page and when user click on the contact or on the detail button VF page should transfer to the standard record detail page.
I am trying both ways using outputlink and button but either ways is not working. please help.
Visualforce
   <apex:page standardController="Contact" extensions="ContacSearch" showHeader="false" standardStylesheets="false" sidebar="false" applyHtmlTag="false" applyBodyTag="false" docType="html-5.0">
<apex:form >
<apex:pageBlock >

             <apex:pageblocksection >
                 <apex:pageblocksectionitem >
                    <apex:outputText style="font-weight:600" value="Search Contact"/>
                 </apex:pageblocksectionitem>
             </apex:pageblocksection>

                   First Name
                    <apex:inputText value="{!firstname}" />

                   Last Name
                   <apex:inputText value="{!lastname}" />

                <apex:commandButton value="Search"  action="{!searchContacts}" reRender="contact-table" />

             <apex:pageblocksection >
                 <apex:pageblocksectionitem >
                    <apex:outputText style="font-weight:900" value="Contact Details"/>
                 </apex:pageblocksectionitem>
             </apex:pageblocksection>

                <apex:pageBlock >
                <apex:pageBlockTable style="width:50%" id="contact-table" value="{!contacts}" var="c">

                    <apex:column style="width:100px" >
                       <apex:facet name="header">First Name</apex:facet>
                         {!c.FirstName}
                    </apex:column>
                    <apex:column style="width:100px" >

                    <apex:outputLink value="{!URLFOR($Action.Contact.View, c.Id)}">
                    {!c.LastName}
                    </apex:outputLink>

                      </apex:column>
                  <apex:column style="width:100px" >
                  <apex:facet name="header">Detail</apex:facet>
                     <apex:commandButton value="Goto DetailPage" action="{!gotoContactDetailPage}" reRender="form">
                     <apex:param name="contactid" value="{!c.Id}" assignTo="{!contactId}"/>
                     </apex:commandButton>

                 </apex:column>
                </apex:pageBlockTable>

          </apex:pageBlock>

</apex:pageBlock>
</apex:form>
</apex:page>

Controller
public with sharing class ContacSearch{

    public ContacSearch(ApexPages.StandardController controller) {
    }

    public List<Contact> contacts { get; set; }
    public String firstname { get; set; }
    public String lastname { get; set; }
    public string contactId {get; set;}

    public ContacSearch()
    {
        contacts = new List<Contact>();
    }

    public PageReference searchContacts()
    {
        contacts = [select Id, FirstName, LastName from Contact
                    where FirstName = :firstname
                    and LastName = :lastname];
        return null;
    }
    public PageReference gotoContactDetailPage()
    {
        return new PageReference('/' + contactId);
    }
}

A:

*

*<apex:outputLink/> to navigate to contact details

<apex:outputLink value="{!URLFOR($Action.Contact.View, c.Id)}">
        {!c.LastName}
 </apex:outputLink>

*<apex:commandLink/> as button to navigate to Contact Detail page without the use of controller method:

<apex:commandLink action="{!URLFOR($Action.Contact.View, c.Id)}"
value="Goto DetailPage"
target="_blank" styleClass="btn" style="padding: 4px; text-decoration: none;"/>

*Navigation with the use of <apex:commandButton/> through Controller method:

<apex:commandButton value="Goto DetailPage" action="{!gotoContactDetailPage}" reRender="form">
    <apex:param name="contactid" value="{!c.Id}" assignTo="{!contactId}"/>
</apex:commandButton>

public PageReference gotoContactDetailPage()
{
    return (new ApexPages.StandardController (new Contact(Id=contactId))).view();
}

redpajama_set_name: RedPajamaStackExchange


text:
RatProofChickenFeeder.net
Are You Ready to Save Money Feeding Your Chickens? How much Money are you wasting by feeding rats, mice, and wild birds?
Rat Proof Feeders
Parts and Upgrade Kits
Vermin dangers
Comparison to Other Feeders
MA 2017 Feeder
Stop the waste now
The Dangers of Allowing Wild
Birds and Rats in Your
Chicken's Feed
Allowing rats, squirrels, and wild birds to eat out of your poultry feeders is never a good idea. Look at how seriously the commercial farms take bio security, they know that a single vermin can introduce disease and kill of tens of thousands of birds. Many of the diseases carried by rats, squirrels, and wild birds are not lethal to the wild pests but chickens will have no natural immunity to the disease. But having chickens and rats isn't inevitable.
Bird Flu or Avian Flu
Are you worried about bird flu or Avian flu is spread by wild birds? With the outbreak of Bird Flu in early 2015 it is very important to prevent wild birds from sharing feed and water with your flock. Authorities in several states have already begun testing and euthanizing backyard chicken flocks and it is reported that they will test flocks up to seven miles from any outbreak. A wild bird proof chicken feeder is cheap insurance that your flock will not be exposed. If you are having problems with wild birds like starlings, house sparrows, English sparrows, pigeons, or grackles this is the solution.
Avian Influenza is also called Bird Flu but that is a bit misleading because there are different strains of the virus and mutations of the virus so in reality there are many forms of Bird Flu. The term Avian Influenza also refers to those forms of the virus that are adapted to birds, forms that are carried by birds but are seldom lethal to the wild birds, and those forms of the virus that can cross over into humans. The outcome can range from a simple case of flu, respiratory failure, pneumonia, or conjunctivitis. Wild birds and chickens aren't a good mix.
The two common bird flu viruses are H5N1 and H7N9 and most of the time the birds are carriers but generally will have immunity but domestic poultry will get sick and even the they don't die they will have to be destroyed to stop the spread of the virus. People seldom get these two types of bird flu virus and when they do it is usually people that live in close contact with the birds, sometimes living and sleeping with infected domestic poultry.
Even rarer are cases where bird flu has passed from human to human and while the numbers of people that have developed the disease is small the mortality rate is appalling. What scientists really fear is that the virus mutations are so very different from regular flu viruses that we won't have immunity against the virus. Consider the American Indian population that was wiped out after European visitors showed up in the New World.
But is the danger that real? Looking back in history shows that there have been many viruses that mutated from birds but only three grew into pandemics. First was the 1957 H2N2 virus that mutated from avian and human viruses. Then in 1968 N3N2 erupted into a pandemic. But the scariest form of avian flu was the infamous "Spanish Flu" of 1918 that killed a half million Americans and 50 million across the world.
The most recent epidemic in recent times was in 1997 when H5N1 swept through Hong Kong where 18 people were hospitalized and one third of them died. Hong Kong killed a million and a half chickens in order to eradicate the disease. In addition- and perhaps most frightening is the fact that there were rare situations where this virus spread from person to person in 1997. H5N1 didn't just cause flu symptoms to appear, it was very aggressive to the point where the patient's own immune system turned against their own body, reminiscent of the Spanish Flue at the turn of the century. H5N1 has only killed 167 people since 2003 but only about twice that many came down with it.
Thankfully the virus needed an avian host to transfer to humans so infected humans didn't pass along the infections. Scientists learned that H5N1 laced a protein called hemagglutinin, which is common in common human flu viruses. The protein has a spike that allows it to grab hold of cells in the human nose and throat.
This protein allows a virus to be spread airborne with a sneeze or cough. H5N1 prefers to live in the intestinal tract of birds and that is all that has saved millions if not billions of people throughout the world. But viruses mutate easily and should H5N1 mutate and acquire a hemagglutinin protein coating we would see unstoppable pandemics across the world with as high as 43% mortality rates.
Unfortunately, viruses change all the time. Thus, the virus could mutate to a form that prefers to attach itself to the sinus passages at any time. In fact, the likelihood of a pandemic outbreak from a mutation of H5N1 in the future, according to some scientists, is as high 43%. Outbreaks in Vietnam and Thailand showed that the two most common anti viral medications were useless against H5N1. More worrisome is that the virus has spread to ducks, usually in a low level mutation but there have been dead wild ducks found with a high mortality mutation of the virus.
What States Have Experienced Avian Flu?
As of mid March 2015 here is a partial list of states that have either had an outbreak recently or since 2000 or have made plans to
deal with an outbreak. We will post the missing states as we compile the information.
Officials are downplaying any concern but they do warn people to get a flu shot just in case.
Sent out advisory to hunters after the reports from Washington and Oregon bird flu cases surfaced.
Continues to test humans for potential bird flu cases.
The discovery of a highly pathogenic H5N2 avian flu in turkeys in Arkansas has caused Mexico, Canada, and the European Union to
join China and Korea's ban on American poultry products. Over 40,000 turkeys in Boon County Arkansas were destroyed in an
attempt to contain the contagion. Arkansas is one of the major poultry producing states and stock prices of the major poultry producers plummeted which in turn is going to result in lobbyists pressuring legislators to increase regulations on backyard chickens and small free range flocks.
In January a California turkey flock was quarantined and all of the birds were destroyed after routine testing turned up bird flu. Then in February a second commercial flock was found with infected birds.
Hasn't reported on any out breaks but one Colorado blogger had it right when she pointed out that a single backyard flock turning up infected would create immediate trade sanctions against all poultry product coming from that state.
Is well prepared for an outbreak as one case had been discovered some years ago. Connecticut has a large egg production industry and has programs in place to monitor flocks and diagnose any outbreaks.
Had two flocks testing positive back in 2004 which led to testing seventy five flocks around the two commercial operations and poultry
import bans from eleven nations. Delaware immediately destroyed tens of thousands of birds in the two flocks, canceled all meetings of farmers, canceled sales of farming equipment and auctions, and also stopped the sale of live poultry in the state. With a multi billion dollar brooder industry Delaware would take draconian measures to eradicate any outbreaks. One of the farms that had an outbreak, Purdue Farms, blamed a local backyard flock that was raising birds for the New York market One local news story pointed out that a Virginia outbreak led to four months of exterminating flocks, over five million birds slaughtered, before the outbreak was contained.
Has a large wild bird population but so far there is limited information on any outbreaks or concern.
Has been studying bird flu since the early 200's and has a strong research organization based at the colleges. Like most states it is
monitoring for bird flu but no new cases have been reported.
Has hundreds of thousands of Chinese visitors each year and is very close to Asia so they have been studying the bird flu but there are no reports of any poultry outbreaks.
In Mid January Idaho confirmed a case of backyard chickens infected with bird flu after some falcons died of the disease. The farm was about twenty miles from the falconry and the farm was quarantined and the flock euthanized immediately.
Has yet to suffer any outbreak of bird flu but health officials and universities are monitoring any suspicious poultry deaths.
Has not reported any outbreaks but has contingency plans in place should an outbreak occur.
One local story pointed out the connection between backyard flocks and disease: "Also, and Thompson says this is critical, all his turkeys live indoors. When he started in the business 35 years ago, the turkeys ranged outside. We raised them out in the fresh air and it was just really, really wonderful" he said, "and you are going to hear some sarcasm right now: What was so nice about it was we had all kinds of wild birds swooping in and eating out of the same feeders that the turkeys were eating out of. The wild birds would repay their inadvertent hosts by sharing whatever diseases they were carrying. Say what you will about animal welfare, but moving the birds inside did protect them from those threats. Many small backyard flocks still have contact with wild birds. In fact, that's what officials think happened in Washington, Oregon and Idaho where they've found the highly contagious form of bird flu. Iowa State University professor K. J. Yoon says these cases in the Pacific Northwest were handled exactly as planned.
Bird flu has been detected in a backyard flock of chickens and ducks in northeast Kansas. The Kansas Department of Agriculture said in a news release late Friday that the discovery was made in Leavenworth County. Backyard poultry owners in the county are urged to contact the department. Additionally, parts of Cherokee and Crawford counties in the southeast part of the state are under surveillance after a confirmed case of the H5N2 strain of bird flu was found just across the state line in Jasper County, Missouri, this month.
In 2009 bird flu was found on a Kentucky farm, resulting in 20,000 birds being euthanized. The Kentucky Department of Agriculture is
conducted surveillance on backyard flocks within a two-mile radius of the farm after the outbreak.
;Health officials are predicting that the bird flu will make its way to Louisiana. The state immediately ordered a ban on incoming poultry unless they were from tested and verified disease free flocks. The state tests commercial farms and has partnered with the Game and Fish Department to test backyard flocks if needed.
Bird flu hit Maryland in 2004 leading officials to kill 325,000 infected birds and another 40,000 that tested negative but were killed because they belonged to one of the farmers than had an infected flock. They tested over 1,000 flocks and some insiders said that they destroyed flocks that tested negative just to reassure foreign countries that they had contained the outbreak.
Had an outbreak in 2012 that killed some swans but no poultry farms were impacted.
Minnesota is the nation's top turkey producing state and reported on March 4th that rapid response protocols from the federal government and poultry industry were used. Once H5N2 was confirmed at the farms the surviving turkeys were killed . Officials quarantined the facilities an declared six mile radius restricted zones. They tested nearby farms but found no signs of other infected flocks
Has reported no outbreaks but officials are pointing to the Mississippi valley as an avenue that would spread the disease.
Officials quarantined two Missouri farms, one in the Southwest, the other in the center of the state. All the turkeys were killed.
Confirmed a case of bird flu in late January. However it was a wild duck, a species that usually carries the virus without harm to the host.
Had a low level bird flu outbreak in 2014, leading Hong Kong to outlaw the import of poultry products from one New Jersey county. Four of the 35 live bird markets in New Jersey had birds that tested positive for the disease.
Diseases spread by Rats, Squirrels, and other Rodents
Hantavirus Pulmonary Syndrome, HPS
Caused by a virus spread by mice and rats. Breathing in dust contaminated with rodent urine or droppings, by direct contact with rodents, or from bite wounds.
Hemorrhagic Fever with Renal Syndrome
Spreads in the same manner as above but can be spread from person to person in rare cases. Typically a European and Asian
Typically in West Africa and caused by the same manners as above as well as food contaminated with rodent urine or droppings. May spread from human to human by contact.
Is a world wide problem caused by a bacteria found in food or water contaminated with rodent urine. Dust or water can carry the bacteria through the skin or mucous membranes on contact.
Lymphocytic Chorio-meningitis
Caused by a virus carried by the common house mouse and is found world wide. Same vector as above, contaminated dust, direct contact with rodents or their urine and droppings, or bite wounds.
Plague…. And yes, that is Bubonic Plague
Caused by a rodent carried flea that harbors bacteria and is found in Africa, Asia, South America, and Western United States. You have to be bitten by a flea or in direct contact with an infected animal. Carried by most wild rodents like squirrels, rats, prairie dogs, and chipmunks.
Rat bite fever
Caused by bites or scratches from rats and mice and is world wide. Contaminated food or water carries the bacterial or you can catch it by handling a dead rodent.
Carried worldwide by rats and mice through a bacteria that can be ingested with contaminated food or water.
South American Arenaviruses
(Argentine hemorrhagic fever, Bolivian hemorrhagic fever, Sabiá-associated hemorrhagic fever, Venezuelan hemorrhagic fever) Mostly dangerous in South America and is caused by the same vectors as above.
Carried world wide by rodents of all types including squirrels. Can be spread by contact with dead rodents, infected ticks or deer flies, contaminated food or water, or breathing in the bacteria
Diseases carried wild birds
There are said to be as many as sixty diseases carried by wild birds.
Here are a few:
Colibacillosis
Yersiniosis
Pasteurellosis
Chlamydiosis
Warts (viral papillomas)
Avian pox
Histoplasmosis is a respiratory disease that may be fatal. It results from a fungus growing in dried bird droppings.
◾Candidiasis is a yeast or fungus infection spread by pigeons. The disease affects the skin, the mouth, the respiratory system, the intestines and the urogenital tract, especially the vagina. It is a growing problem for women, causing itching, pain and discharge.
◾Cryptococcosis is caused by yeast found in the intestinal tract of pigeons and starlings. The illness often begins as a pulmonary disease and may later affect the central nervous system. Since attics, cupolas, ledges, schools, offices, warehouses, mills, barns, park buildings, signs, etc. are typical roosting and nesting sites, the fungus is apt to found in these areas.
◾St. Louis Encephalitis, an inflammation of the nervous system, usually causes drowsiness, headache and fever. It may
even result in paralysis, coma or death. St. Louis encephalitis occurs in all age groups, but is especially fatal to persons over age 60. The disease is spread by mosquitoes which have fed on infected house sparrow, pigeons and house finches carrying the Group B virus responsible for St. Louis encephalitis.
◾Salmonellosis often occurs as "food poisoning" and can be traced to pigeons, starlings and sparrows. The disease bacteria are found in bird droppings; dust from droppings can be sucked through ventilators and air conditioners, contaminating food and cooking surfaces in restaurants, homes and food processing plants.
◾E.coli. Cattle carry E. coli 0157:H7. When birds peck on cow manure, the E. coli go right through the birds and the bird droppings can land on or in a food or water supply.
How about parasites carried by birds?
◾Bed bugs (Cimex lectularius) may consume up to five times their own weight in blood drawn from hosts which include humans and some domestic animals. In any extreme condition, victims may become weak and anemic. Pigeons, starlings and house sparrows are known to carry bed bugs.
◾Chicken mites (Dermanyssus gallinae) are known carriers of encephalitis and may also cause fowl mite dermatitis and acariasis. While they subsist on blood drawn from a variety of birds, they may also attack humans. They have been found on pigeons, starlings and house sparrows.
◾Yellow mealworms (Tenebrio molitor), perhaps the most common beetle parasites of people in the United States, live in pigeon nests. It is found in grain or grain products, often winding up in breakfast cereals, and may cause intestinal canthariasis and hymenolespiasis.
◾West Nile Virus while West Nile is technically not transmitted to humans from birds, humans can get infected by the bite of a
mosquito who has bitten an infected bird. The obvious lesson is that the fewer birds there are in any given area, the better. This translates into a smaller chance of an infected bird in that area, a smaller chance of a mosquito biting an infected bird and then biting a human.
Stop the wild birds like starlings, house sparrows ,English sparrows, pigeons, or grackles.
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Interested in having Joshua Coburn speak at your event, school, business, or one on one?
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Home / Gift Air Fresheners / Gift Souvenir Las Vegas / GFT-Las Vegas-reg / 500 units / SHIPPING INCLUDED!
Gift Souvenir Las Vegas / GFT-Las Vegas-reg / 500 units / SHIPPING INCLUDED!
500 UNITS / SOUVENIR / GIFT Air Fresheners. Custom "Las Vegas" Registration Plate Air fresheners. Packed in master box of 500 units. Production time 10-14 business days.


OPEN: 4/25 1-4PM!! Cozy meets Contemporary in this luxurious home!Be a culinary genius in the gourmet kitchen, complete with SS appliances, pantry,& quartz counters. Host movie night in the Theater room, with screen and projector. End the night with a glass of wine from the 400 bottle rack (removable). Cozy up by one of the 3 fireplaces, either in the Great Room, lower living area, or outdoors on the gorgeous patio.Peaceful green views.
This content last updated on 4/23/19 6:56 PM PDT. Some properties which appear for sale on this web site may subsequently have sold or may no longer be available.


Occupying a sun splashed parcel, this modern home is part of a community estate with a pool, clubhouse and tennis court for neighbourhood use. Wonderfully located, The Ponds School is 600m away while John Palmer Public School and The Ponds Shopping Centre are within walking distance. This residence has a light-filled and airy position on a corner block. The master bedroom opens to a covered balcony capturing summer breezes and streetscape views. A walk-through robe links to a dual-basin ensuite and there is an extra built-in robe for extensive storage. A fourth bedroom downstairs has versatility as a study, with built-in storage.
The galley-style kitchen looks out to the alfresco and has a gas cooktop, stainless steel oven and Caesarstone benchtops. Accessed through the lounge room, the alfresco offers an intimate space for outdoor entertaining. It merges with a pristine lawn that wraps around the house. It's a short walk to buses and Ironbark Lake. Stanhope Village shops are close and Rouse Hill Town Centre is a short drive, providing extra retail options and access to the new train station.


text:
Try to Remember (The Way We Were) by Gladys Knight & the Pips
"They Just Can't Stop It" The (Games People Play) - Spinners
#6 in 1975
Just a Little Bit of You - Michael Jackson
That's the Way of the World - Earth, Wind & Fire
Lovin' You - Minnie Riperton
Do It In The Name Of Love - Ben E. King
Do It Any Way You Wanna - The People's Choice
Sooner Or Later - The Impressions
Silver Convention - Fly, Robin, Fly
Major Harris - Love Won't Let Me Wait
Gladys Knight & The Pips Top Songs in the Charts Top One Hit Wonders of the 1970s Top R&B One Hit Wonders
Try to Remember (The Way We Were) was the #17 song in 1975 in the R&B charts. The song was performed by Gladys Knight & The Pips. The song also included appearances from The Pips. Comment below with facts and trivia about the song and we may include it in our song facts!
Sound Reaction Test - How Fast Can You React?
What Was the Most Popular Christmas Movie the Year You Were Born?


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Stunt Coordinator Accused by Eliza Dushku of Sexual Assault Dropped by Agent
Joel Kramer has denied the allegations.
The stunt coordinator accused by Eliza Dushku of sexually assaulting her when she was a minor while working on 1994's True Lies has been dropped by his agent.
Worldwide Production Agency (WPA) no longer represents Joel Kramer, president Richard Caleel confirmed Sunday to The Hollywood Reporter.
"WPA has elected to part ways with Joel Kramer based on the allegations of misconduct now being reported," Caleel said. "Such behavior is unacceptable and entirely at odds with the standards of conduct we demand of ourselves, and expect from our clients."
Dushku accused Kramer of assault in a lengthy Facebook post on Saturday, in which she claimed she was "sexually molested" by the then-36-year-old stunt coordinator after Kramer "lured" her to his hotel room during production.
Kramer denied the allegations to THR on Saturday.
Dushku wrote in part, "I remember vividly how he methodically drew the shades and turned down the lights; how he cranked up the air-conditioning to what felt like freezing levels, where exactly he placed me on one of the two hotel room beds, what movie he put on the television (Coneheads); how he disappeared in the bathroom and emerged, naked, bearing nothing but a small hand towel held flimsy at his mid-section."
On Sunday, Sue Booth Forbes, Dushku's legal guardian on the set, said in a statement that the actress was telling the truth.
"I was on the True Lies set for 3 weeks and reported Joel Kramer's inappropriate sexual behavior towards 12-year-old Eliza to a person in authority," Forbes said. "I was met with blank stares and had the sense that I wasn't telling that person anything they didn't already know."


We repair your equipment quickly without hassle and in the interim can also provide you with a loan replacement.
We offer comprehensive, expert assistance for the rapid elimination of malfunctions or machine downtime. HBS supports you with preventive maintenance and regular safety checks as well as DGUV inspections.
We calibrate your stud welding equipment in accordance with DIN EN ISO 14555 and applicable standards.
You will receive technical application support from our field services team and our extensive international dealer network. We will be pleased to advise you on-site at any time.


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RE CONSULTING, INC in Hobbs is a company that specializes in Business Consulting Services, Nec. Our records show it was established in New Mexico.


Founded in March 2012, SCOEP's spring and fall programs offer students from sixth grade through high school aerobic exercise and environmental education through kayaking sessions in Charleston Harbor and on the Black River, near Andrews, SC. The SCOEP program is operated by an independently contracted outfitter with many years of experience serving the greater Charleston community.
SCOEP's programs are based on intensive hands-on learning, teaching conservation and environmental stewardship through immersion in the ecology, history, and natural beauty of South Carolina's Lowcountry. SCOEP guides, who are trained naturalists with certifications in kayaking, first aid, and lifeguarding, take students on a four-hour kayaking adventure that combines local history, environmental education, and team-building.
Since its inception, more than 13,000 students have taken part in SCOEP field trips.