text:
\section{Introduction} \label{sec_Intro}

Entanglement swapping \cite{ZZHE93} is a phenomenon where 2 or more qubits which didn't interact in
the past are brought into an entangled state. For example, two parties Alice and Bob share the
Bell states $\ket{\Phi^+}_{12}$ and $\ket{\Phi^+}_{34}$ such that Alice is in possession of qubits 1
and 3 and Bob of qubits 2 and 4. If Alice performs a measurement in the Bell basis on qubits 1 and 3
she will end up with one of the four Bell states. Simultaneously Bob's qubits 2 and 4 are also
brought into a Bell state. This measurement can be described by the equation
\begin{eqnarray}
\ket{\Phi^+}_{12} \otimes \ket{\Phi^+}_{34}
= &\dfrac{1}{2} \Bigl(&
\ket{\Phi^+}_{13} \ket{\Phi^+}_{24} + \ket{\Phi^-}_{13} \ket{\Phi^-}_{24} \nonumber \\
&+&\ket{\Psi^+}_{13} \ket{\Psi^+}_{24} + \ket{\Psi^-}_{13} \ket{\Psi^-}_{24} \Bigr)
\label{eq_EntSwap}
\end{eqnarray}
This equation shows that Alice's result is completely random, i.e. she will obtain any of the four
Bell states with equal probability. But the state of the qubits 2 and 4 is completely determined by
the result of Alice's measurement. This correlation is used in a number of key distribution
protocols \cite{WZT05,Cab01,LSZW03,GGWZ05,Cab00,Son04,Cab00_2} to share a secret key between Alice
and Bob. Since the practical realization of entanglement swapping is rather complex to achieve with
today's technology none of these QKD protocols has been implemented yet.
\par
Also Li et al. presented a QKD protocol \cite{LWWSZ06} based on entanglement swapping which not only
produces random key bits but also certain key bits. This is achieved by introducing a Pauli
operation in eq. (\ref{eq_EntSwap}) as it will be described in detail in section
\ref{sec_OrigProt}. As it is shown in \refcite{LWWSZ06} this protocol is secure against an
intercept/resend attack as well as a collective attack. But we will present an attack strategy in
section \ref{sec_AttStrat} which is based on a 6-qubit state. This state is more complex than the
system of the sender and the receiver, but it provides an adversary with full information about the
key shared between the two parties.
\par
The main idea of this attack is that the adversary Eve entangles herself with Alice and Bob using a
state which preserves the correlations of Alice's Bell state measurement. Further this state
contains additional information which allows Eve to eavesdrop Bob's secret result. An attack where
Eve is entangled with one party was presented by Zhang, Li and Guo \cite{ZLG00_1} on the protocol
of Cabello \cite{Cab00_2} but it wouldn't work for the protocol in \refcite{LWWSZ06} since Eve
wouldn't stay undetected.
\par
We will also make a proposal how to secure the protocol against this new attack strategy. This can
be achieved using a Hadamard operation and we will describe the new version in detail in section
\ref{sec_ModProt}. In section \ref{sec_SecAnal} we will discuss the security of our new version.

\section{The Original Protocol} \label{sec_OrigProt}

Alice creates $2n$ EPR pairs, each in a Bell state, e.g. $\ket{\Phi^+}$. In the following we will
focus on the simple case $n=1$, where Alice and Bob share the states $\ket{\Phi^+}_{12}$ and
$\ket{\Phi^+}_{34}$, which are publicly known. In \refcite{LWWSZ06} Li et al. start the protocol with
preshared entangled states which is a very strong assumption because the qubits have to be shared
somehow between the two parties. But in the security analysis in \refcite{LWWSZ06} an adversary Eve is
granted the possibility to interact with the Bell states. Thus we can conclude that there have to be
some qubits in transit between Alice and Bob. There are two possible ways to share the Bell states
between the legitimate parties: either one party, e.g. Alice, prepares both Bell states and sends
two qubits to the other party or both parties prepare one Bell state and they exchange two qubits.
\par
The first scenario is insecure against a simple intercept-resend attack: If Alice prepares the Bell
states $\ket{\Phi^+}_{12}$ and $\ket{\Phi^+}_{34}$ and sends qubits 2 and 4 to Bob Eve can intercept
these qubits and perform a Bell state measurement on them. This will bring qubits 1 and 3 also into
a Bell state which is known to Eve. Then she forwards qubits 2 and 4 to Bob. Thus Eve has full
information about Alice's and Bob's secret measurement result and is able to eavesdrop the key
perfectly. Therefore we will assume that Alice prepares $\ket{\Phi^+}_{12}$ and Bob prepares
$\ket{\Phi^+}_{34}$ and that they exchange the qubits 2 and 3 (cf. picture (2) in figure
\ref{pic_OrigProt}).
\begin{figure}[pb]
\centerline{\psfig{file=fig_orig_protocol.eps,width=0.85\textwidth}}
\vspace*{8pt}
\caption{An illustration of the original protocol by Li at al. $^9$. Here $\sigma_x$ is
chosen for Alice's secret operation and $\ket{\Psi^-}_{13}$ as Alice's result of the Bell state
measurement.}
\label{pic_OrigProt}
\end{figure}
\par
Alice randomly chooses one of the four Pauli operations, $I$, $\sigma_x$, $\sigma_y$ or $\sigma_z$
and applies it to qubit 1 (cf. picture (3) in figure \ref{pic_OrigProt}). Alice's operation will
generally be described as $\sigma_A^{(i)}$, where the superscript $(i)$ denotes that it is applied
on qubit $i$. As a next step Alice performs a Bell state measurement on qubits 1 and 3 in her
possession (cf. picture (4) in figure \ref{pic_OrigProt}).
\par
With this measurement the effect of entanglement swapping arises and entangles Bob's qubits 2
and 4. Following the equation
\begin{eqnarray}
\sigma_A^{(1)} \ket{\Phi^+}_{12} \otimes \ket{\Phi^+}_{34} =
&\dfrac{1}{2} \Bigl(&
\ket{\Phi^+}_{13} \sigma_A^{(2)} \ket{\Phi^+}_{24} + \ket{\Phi^-}_{13} \sigma_A^{(2)} \ket{\Phi^-}_{24} \nonumber \\
&+&\ket{\Psi^+}_{13} \sigma_A^{(2)} \ket{\Psi^+}_{24} + \ket{\Psi^-}_{13} \sigma_A^{(2)} \ket{\Psi^-}_{24} \Bigr)
\label{eq_CESP_ESPO_PD_ESwOp}
\end{eqnarray}
Alice can determine in which state Bob's qubits are after her Bell state measurement if she applied
the operation $\sigma_A$ on qubit 1. Additionally, she can compute also what the state of qubits 2
and 4 would be, if she didn't apply any operation on qubit 1 using eq. (\ref{eq_EntSwap}) from
above. Now Alice announces to Bob that she made a Bell state measurement but keeps her result secret.
\par
To infere Alice's result Bob performs a Bell state measurement on qubit 2 and 4 in his possession.
Since he does not know yet which operation Alice applied and in which state qubits 1 and 2 have been
before the measurement he does not know the exact state of qubits 1 and 3. With the help of eq.
(\ref{eq_EntSwap}) he can at least deduce in which state qubits 1 and 3 should be if Alice
didn't apply anything to qubit 1. This is called Alice's imaginary result.
\par
To get the correct state of qubits 1 and 3 Bob asks Alice about her result but keeps his own result
secret. Using the information about Alice's and his own result Bob can deduce which Pauli operation
Alice applied. Alice and Bob now share information about Alice's randomly chosen operation and the
result Bob obtained, which is also random. They use these two pieces of information to extract a
classical raw key. Therefore they agree beforehand on a mapping of Bell states onto classical 2-bit
strings, e.g.
\begin{equation}
\ket{\Phi^+} \longrightarrow 00 \quad \ket{\Phi^-} \longrightarrow 01 \quad
\ket{\Psi^+} \longrightarrow 10 \quad \ket{\Psi^-} \longrightarrow 11
\label{eq_CESP_ESPO_PD_Map}
\end{equation}
as well as of Pauli operations onto classical 2-bit strings, i.e.
\begin{equation}
I \longrightarrow 00 \quad \sigma_x \longrightarrow 01 \quad
\sigma_y \longrightarrow 10 \quad \sigma_z \longrightarrow 11
\end{equation}
\par
If we take the general case with $2n$ EPR pairs Alice and Bob repeat these steps for all the
pairs. In the end they publicly compare a certain number of bits of the generated raw key to detect
an eavesdropper.

\section{A Perfect Attack Strategy} \label{sec_AttStrat}

In this section we discuss an attack strategy where Eve is able to perfectly eavesdrop the secret
key of Alice and Bob without being detected. Therefore she uses the six qubit state (cf. picture (1)
in figure \ref{pic_Attack})
\begin{eqnarray}
\ket{\delta}_{PQRSTU} = &\dfrac{1}{2\sqrt{2}} \Bigl(&
\ket{000000}_{PQRSTU} + \ket{001101}_{PQRSTU} \nonumber \\
&+&\ket{010111}_{PQRSTU} + \ket{011010}_{PQRSTU} \nonumber \\
&+&\ket{100110}_{PQRSTU} + \ket{101011}_{PQRSTU} \nonumber \\
&+&\ket{110001}_{PQRSTU} + \ket{111100}_{PQRSTU}
\Bigr)
\label{eq_EveInitState}
\end{eqnarray}
This state is rather complex and not easy to generate (cf. appendix) but it has
the special property that it can be written as
\begin{eqnarray}
\ket{\delta}_{PQRSTU} = &\dfrac{1}{2} \Bigl(&
\ket{\Phi^+}_{PR} \otimes \ket{\Phi^+}_{QS} \otimes \ket{\Phi^+}_{TU} \nonumber \\
&+&\ket{\Phi^-}_{PR} \otimes \ket{\Phi^-}_{QS} \otimes \ket{\Phi^-}_{TU} \nonumber \\
&+&\ket{\Psi^+}_{PR} \otimes \ket{\Psi^+}_{QS} \otimes \ket{\Psi^+}_{TU} \nonumber \\
&+&\ket{\Psi^-}_{PR} \otimes \ket{\Psi^-}_{QS} \otimes \ket{\Psi^-}_{TU}
\Bigr)
\label{eq_EveStateProp}
\end{eqnarray}
That means, a measurement on qubits $P$ and $R$ as well as $Q$ and $S$ yields in the same result as
it would be expected from entanglement swapping in eq. (\ref{eq_EntSwap}). Further the remaining two
qubits $T$ and $U$ are in the same state as the qubits $Q$ and $S$. Eve can use this fact to obtain
enough information to deduce the secret measurement results of Alice and Bob if she somehow manages
that the two parties share the state $\ket{\delta}$ instead of two Bell states. That can be achieved
easily using entanglement swapping on qubits 2 and 3 in transit together with her six qubits.
\par
When Alice sends out qubit 2 and Bob qubit 3 Eve preforms Bell state measurements on both
of them with respective qubits from $\ket{\delta}_{PQRSTU}$ (cf. picture (2) in figure
\ref{pic_Attack}). A problem is that Eve, at first, doesn't know which initial state Alice prepared.
But, as we will show in the next paragraphs, she can overcome this problem fairly easy. Eve's first
Bell state measurement on qubits 2 and $P$ is described in eq. (\ref{eq_CESP_ESPO_AS_ES_ES1}). (As
already pointed out above, we will use $\sigma_A$ to describe Alice's secret Pauli operation.)
\begin{eqnarray}
\lefteqn{\sigma_A^{(1)} \ket{\Phi^+}_{12} \otimes \ket{\delta}_{PQRSTU} =} \hspace{30pt} \nonumber \\
& &\sigma_A^{(1)} \frac{1}{\sqrt{2}} \Bigl(\ket{00}_{12} + \ket{11}_{12}\Bigr) \otimes
\frac{1}{2\sqrt{2}} \Bigl( \ket{000000}_{PQRSTU} + \ket{001101}_{PQRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{010111}_{PQRSTU} + \ket{011010}_{PQRSTU} + \ket{100110}_{PQRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{101011}_{PQRSTU} + \ket{110001}_{PQRSTU} + \ket{111100}_{PQRSTU}\Bigr) \nonumber \\
=&\dfrac{1}{2} \Biggl( &\ket{\Phi^+}_{2P} \otimes \sigma_A^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{1QRSTU} + \ket{001101}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{010111}_{1QRSTU} + \ket{011010}_{1QRSTU} + \ket{100110}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{101011}_{1QRSTU} + \ket{110001}_{1QRSTU} + \ket{111100}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Phi^-}_{2P} \otimes \sigma_A^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{1QRSTU} + \ket{001101}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{010111}_{1QRSTU} + \ket{011010}_{1QRSTU} - \ket{100110}_{1QRSTU} \nonumber \\
& &\qquad\quad{}- \ket{101011}_{1QRSTU} - \ket{110001}_{1QRSTU} - \ket{111100}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Psi^+}_{2P} \otimes \sigma_A^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000110}_{1QRSTU} + \ket{001011}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{010001}_{1QRSTU} + \ket{011100}_{1QRSTU} + \ket{100000}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{101101}_{1QRSTU} + \ket{110111}_{1QRSTU} + \ket{111010}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Psi^-}_{2P} \otimes \sigma_A^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000110}_{1QRSTU} + \ket{001011}_{1QRSTU} \nonumber \\
& &\qquad\quad{}+ \ket{010001}_{1QRSTU} + \ket{011100}_{1QRSTU} - \ket{100000}_{1QRSTU} \nonumber \\
& &\qquad\quad{}- \ket{101101}_{1QRSTU} - \ket{110111}_{1QRSTU} - \ket{111010}_{1QRSTU} \Bigr)
\Biggr)
\label{eq_CESP_ESPO_AS_ES_ES1}
\end{eqnarray}
At this point Eve can not clearly say which state the qubits 1, $Q, R, S, T$ and $U$ are in. All
four results of her Bell state measurement are equally likely and she has absolutely no information
about the operation $\sigma_A$. Nevertheless, Eve is able to transform the 6-qubit state in a way
that it is always in the state $\sigma_A^{(1)} \ket{\delta}_{1QRSTU}$. Therefore she performs a
$\sigma_x$ operation on qubits $S$ and $T$ if she gets $\ket{\Psi^+}_{2P}$ or does nothing if her
result is $\ket{\Phi^+}_{2P}$. To correct the negative signs in case Eve obtains a
$\ket{\Phi^-}_{2P}$ she performs a $\sigma_z$ on qubits $S$ and $U$. In case of $\ket{\Psi^-}_{2P}$
she first has to apply the $\sigma_x$ operator on qubits $S$ and $T$ and then the $\sigma_z$
operator on qubits $S$ and $U$.
\par
The second Bell state measurement is performed on qubits 3 and $S$. Again, the resulting 6-qubit
state is not entirely known but it can be one out of four possible states. We won't describe this
Bell state measurements explicitly but we want to stress that Eve is able to bring the 6 qubits in
the state $\sigma_A^{(1)} \ket{\delta}_{1QR4TU}$. Therefore she uses a similar method as described
above applying the Pauli operations $\sigma_x$ and $\sigma_z$ on qubits $R$, $T$ and $U$.
\par
Now Eve sends qubit $R$ to Alice and qubit $Q$ to Bob and keeps qubits $T$ and $U$ by herself (cf.
picture (4) in figure \ref{pic_Attack}). Alice and Bob, who are not aware of Eve's intervention
follow the protocol and perform Bell state measurements on their respective particles (cf. pictures
(5) and (6) in figure \ref{pic_Attack}). If we look at the 6-qubit state
$\sigma_A^{(1)} \ket{\delta}_{1QR4TU}$, Alice's Bell state measurement on qubits 1 and $R$ changes
the state in a way that it leaves the qubit pairs $Q$, 4 and $T$, $U$ in an entangled state. This
can be seen from the alternative description of the state $\sigma_A^{(1)} \ket{\delta}_{1QR4TU}$ in
eq. (\ref{eq_CESP_ESPO_AS_ES_ESAlice}) and picture (7) in figure \ref{pic_Attack}).
\begin{eqnarray}
\lefteqn{\sigma_A^{(1)} \ket{\delta}_{1QR4TU} =} \hspace{30pt} \nonumber \\
& &\sigma_A^{(1)} \frac{1}{2\sqrt{2}} \Bigl( \ket{000000}_{1QR4TU} + \ket{001101}_{1QR4TU} \nonumber \\
& &\qquad\quad\;{}+ \ket{010111}_{1QR4TU} + \ket{011010}_{1QR4TU} + \ket{100110}_{1QR4TU} \nonumber \\
& &\qquad\quad\;{}+ \ket{101011}_{1QR4TU} + \ket{110001}_{1QR4TU} + \ket{111100}_{1QR4TU}\Bigr) \nonumber \\
=& \dfrac{1}{2} \Biggl(&\sigma_A^{(1)} \ket{\Phi^+}_{1R} \otimes \frac{1}{2} \Bigl(
\ket{0000}_{Q4TU} + \ket{0011}_{Q4TU} \nonumber \\
& &\qquad\qquad\qquad{}+ \ket{1100}_{Q4TU} + \ket{1111}_{Q4TU} \Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Phi^-}_{1R} \otimes \frac{1}{2} \Bigl(
\ket{0000}_{Q4TU} - \ket{0011}_{Q4TU} \nonumber \\
& &\qquad\qquad\qquad{}- \ket{1100}_{Q4TU} + \ket{1111}_{Q4TU} \Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Psi^+}_{1R} \otimes \frac{1}{2} \Bigl(
\ket{0101}_{Q4TU} + \ket{0110}_{Q4TU} \nonumber \\
& &\qquad\qquad\qquad{}+ \ket{1001}_{Q4TU} + \ket{1010}_{Q4TU} \Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Psi^-}_{1R} \otimes \frac{1}{2} \Bigl(
\ket{0101}_{Q4TU} + \ket{0110}_{Q4TU} \nonumber \\
& &\qquad\qquad\qquad{}- \ket{1001}_{Q4TU} - \ket{1010}_{Q4TU} \Bigr)
\Biggr)
\label{eq_CESP_ESPO_AS_ES_ESAlice}
\end{eqnarray}
Alice gets each possible result with equal probability of $\frac{1}{4}$ as she would expect it.
Further, Bob's Bell state measurement will end up with certainty in a specific Bell state, which is
fully dependent on Alice's result. Due to the special choice of the state $\ket{\delta}$ the
correlation between his and Alice's result is preserved as they would expect it if no eavesdropper
was present. Thus the presence of Eve won't be detected when Alice and Bob compare some bits of the
raw key. Additionally, it leaves the qubits $T$ and $U$, which are in Eve's possession, in the very
same state Bob obtains as result (cf. also picture (7) in figure \ref{pic_Attack}). This is easy to
see if we look at the alternative description in the Bell basis of the remaining four qubits $Q$, 4,
$T$ and $U$ after Alice's measurement
\begin{eqnarray}
\frac{1}{2} \Bigl(
 \sigma_A^{(1)} \ket{\Phi^+}_{1R} \otimes \ket{\Phi^+}_{Q4} \otimes \ket{\Phi^+}_{TU}
&+&\sigma_A^{(1)} \ket{\Phi^-}_{1R} \otimes \ket{\Phi^-}_{Q4} \otimes \ket{\Phi^-}_{TU} \nonumber \\
+\sigma_A^{(1)} \ket{\Psi^+}_{1R} \otimes \ket{\Psi^+}_{Q4} \otimes \ket{\Psi^+}_{TU}
&+&\sigma_A^{(1)} \ket{\Psi^-}_{1R} \otimes \ket{\Psi^-}_{Q4} \otimes \ket{\Psi^-}_{TU}
\Bigr)
\label{eq_CESP_ESPO_AS_ES_ESBob}
\end{eqnarray}
\par
At this time, Eve has full information about Bob's secret measurement. That means, Eve has as much
information as Bob and thus can obtain the classical secret raw key in the same way Bob does: From
her result of the Bell state measurement Eve can compute Alice's imaginary result. Moreover, when
Alice publicly announces her result, Eve is able to infere which Pauli operation Alice has chosen.
Thus Eve knows both parts of the shared secret information. Therefore she easily can infere the
classical bit string which is used as a raw key by Alice and Bob and so she is able to obtain the
secret key, too.
\par
Instead of performing the Bell state measurements on the qubit pairs 2, $P$ and 3, $S$ as soon as
she receives them, Eve has the opportunity to delay her measurement and immediately send qubit $R$
to Alice (cf. picture (2) in figure \ref{pic_ModProtAttack2}). With this strategy Eve is able to
overcome operations Alice applies on qubit 1 but she introduces a much higher error rate as we will
describe in section \ref{sec_SecAnal}.
Another, but rather strong, assumption is that Eve is in control of Alice's and Bob's EPR source.
That means she is able to distribute the state $\ket{\delta}_{PQRSTU}$ between her and the two
parties without further Bell state measurements. In this case Alice's measurement on qubits $P$ and
$R$ leaves Bob's qubits in a correlated state and Eve with the full information about the results,
as described above. Since Alice and Bob can prepare their EPR source themself this assumption is not
very practical.

\section{The Modified Protocol} \label{sec_ModProt}

The protocol can be improved by performing a Hadamard operation before Alice applies her secret
Pauli operation on qubit 1 (cf. picture (3) in figure \ref{pic_ModProt}). The Hadamard operation
alters the initial state in a way that Eve can't eavesdrop the secret key without introducing a
certain error rate. Thus Alice and Bob can detect her easily as it will be shown in section
\ref{sec_SecAnal}.
\par
Due to the use of the additional Hadamard operation it is not important any more wether Alice
prepares both Bell states or each party prepares a Bell state by its own. To stay consistent with
the above descriptions of the protocol and the attack we will discuss the scenario where Alice
prepares the state $\ket{\Phi^+}_{12}$ and Bob prepares $\ket{\Phi^+}_{34}$ (for the simple case
$n=1$). Alice sends out qubit 2 to Bob and he sends qubit 3 to Alice. When she receives Bob's qubit
she randomly applies either the identity operator $I$ or the Hadamard operator
\begin{equation}
H = \frac{1}{\sqrt{2}} \left( \begin{array}{cc}1 &1 \\ 1 &-1\end{array} \right)
\end{equation}
on qubit 1. This alters the initial state to
\begin{equation}
H^{(1)} \ket{\Phi^+}_{12} \otimes \ket{\Phi^+}_{34} =
\ket{\omega^+}_{12} \otimes \ket{\Phi^+}_{34}
\label{eq_AltInitState}
\end{equation}
where $\ket{\omega^+}_{12}$ is one of the four possible superpositions of Bell states, i.e.
\begin{eqnarray}
H^{(1)} \ket{\Phi^\pm}_{12} = \ket{\omega^\pm}_{12} &=&
\frac{1}{\sqrt{2}} \Bigl( \ket{\Phi^\mp}_{12} \pm \ket{\Psi^\pm}_{12} \Bigr) \nonumber \\
H^{(1)} \ket{\Psi^\pm}_{12} = \ket{\chi^\pm}_{12} &=&
\frac{1}{\sqrt{2}} \Bigl( \ket{\Psi^\mp}_{12} \pm \ket{\Phi^\pm}_{12} \Bigr)
\label{eq_SupBell}
\end{eqnarray}
Then Alice applies her secret Pauli operation $\sigma_A$ on qubit 1 and performs a Bell state
measurement on qubits 1 and 3 (cf. picture (5) in figure \ref{pic_ModProt}). This Bell state
measurement can be described as
\begin{eqnarray}
\ket{\omega^+}_{12} \otimes \ket{\Phi^+}_{34} =
&\dfrac{1}{2} \Bigl(& \ket{\Phi^+}_{13} \otimes \ket{\omega^+}_{24}
+\ket{\Phi^-}_{13} \otimes \ket{\omega^-}_{24} \nonumber \\
&+&\ket{\Psi^+}_{13} \otimes \ket{\chi^+}_{24}
+\ket{\Psi^-}_{13} \otimes \ket{\chi^-}_{24}
\Bigr)
\end{eqnarray}
\begin{figure}[pb]
\centerline{\psfig{file=fig_orig_protocol_attack.eps,width=0.85\textwidth}}
\caption{An illustration of the new attack strategy described in section \ref{sec_AttStrat}. Again
$\sigma_x$ is chosen as Alice's secret operation and $\ket{\Psi^-}_{1R}$ as Alice's result of the
Bell state measurement.}
\label{pic_Attack}
\end{figure}
Further she publicly tells Bob wether she has applied the $H$ operator or not. If she
did so, Bob applies the $H$ operator on qubit 2 and otherwise he does nothing (cf. picture (6) in
figure \ref{pic_ModProt}). As we can see from eq. (\ref{eq_SupBell}) a repeated application of the
Hadamard operator eliminates the superposition and the correlation from the original Bell state
measurement is preserved.
In the end Bob performs a Bell measurement on qubits 2 and 4 and computes Alice's actual and
imaginary result as it is described in the original protocol.

\section{Security Analysis} \label{sec_SecAnal}

In the following we will provide a security analysis of the modified protocol. Since the application
of the Hadamard operation is just a minor change of the original protocol, the modified version is
also secure against an intercept/resend attack as well as a collective attack, as described in
\refcite{LWWSZ06}. Therefore we will first inspect Eve's probability to stay undetected if she
follows the same attack strategy just described in section \ref{sec_AttStrat}. If Alice does not
apply the Hadamard operator Eve's attack will be successfull as we already pointed out above. Thus
we will discuss the case where Alice performs the Hadamard operation on qubit 1 (cf. figure
\ref{pic_ModProtAttack}). Due to the fact that the Hadamard operation alters the initial state as we
have seen in eq. (\ref{eq_AltInitState}) Eve's first Bell state measurement on qubits 2 and $P$ will
also change as described in eq.
(\ref{eq_ModProt_EveBSM}).
\begin{eqnarray}
\lefteqn{\sigma_A^{(1)} H^{(1)} \ket{\Phi^+}_{12} \otimes \ket{\delta}_{PQRSTU} =} \hspace{30pt} \nonumber \\
&\dfrac{1}{2} \Biggl(&
\ket{\Phi^+}_{2P} \otimes \sigma_A^{(1)} H^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{1QRSTU} + \ket{001101}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{010111}_{1QRSTU} + \ket{011010}_{1QRSTU} + \ket{100110}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{101011}_{1QRSTU} + \ket{110001}_{1QRSTU} + \ket{111100}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Phi^-}_{2P} \otimes \sigma_A^{(1)} H^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{1QRSTU} + \ket{001101}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{010111}_{1QRSTU} + \ket{011010}_{1QRSTU} - \ket{100110}_{1QRSTU} \nonumber \\
& &\quad{}- \ket{101011}_{1QRSTU} - \ket{110001}_{1QRSTU} - \ket{111100}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Psi^+}_{2P} \otimes \sigma_A^{(1)} H^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000110}_{1QRSTU} + \ket{001011}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{010001}_{1QRSTU} + \ket{011100}_{1QRSTU} + \ket{100000}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{101101}_{1QRSTU} + \ket{110111}_{1QRSTU} + \ket{111010}_{1QRSTU} \Bigr) \nonumber \\
&+&\ket{\Psi^-}_{2P} \otimes \sigma_A^{(1)} H^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000110}_{1QRSTU} + \ket{001011}_{1QRSTU} \nonumber \\
& &\quad{}+ \ket{010001}_{1QRSTU} + \ket{011100}_{1QRSTU} - \ket{100000}_{1QRSTU} \nonumber \\
& &\quad{}- \ket{101101}_{1QRSTU} - \ket{110111}_{1QRSTU} - \ket{111010}_{1QRSTU} \Bigr)
\Biggr)
\label{eq_ModProt_EveBSM}
\end{eqnarray}
As it has been shown in section \ref{sec_AttStrat} Eve is able to change the resulting states of her
measurements such that she will always end up with a slight variation of her initial state. A
similar argument holds for Eve's second measurement on qubits 3 and $S$. Thus the state after these
two measurements is (cf. picture (6) in figure \ref{pic_ModProtAttack})
\begin{equation}
\ket{\delta^\prime}_{1QR4TU} = \sigma_A^{(1)} H^{(1)} \ket{\delta}_{1QR4TU}
\label{eq_EveAltState}
\end{equation}
Then Eve sends out qubit $R$ to Alice and qubit $Q$ to Bob. When Alice performs her Bell state
measurement the state $\ket{\delta^\prime}_{1QR4TU}$ can be described as in eq.
(\ref{eq_ModProt_AliceBSM}) (cf. also picture (7) in figure \ref{pic_ModProtAttack}).
\begin{eqnarray}
\lefteqn{\sigma_A^{(1)} H^{(1)} \ket{\delta}_{1QR4TU} =} \hspace{30pt} \nonumber \\
& &\sigma_A^{(1)} H^{(1)} \frac{1}{2\sqrt{2}} \Bigl(
  \ket{000000}_{1QR4TU} + \ket{001101}_{1QR4TU} \nonumber \\
& &\qquad\quad{}+ \ket{010111}_{1QR4TU} + \ket{011010}_{1QR4TU} + \ket{100110}_{1QR4TU} \nonumber \\
& &\qquad\quad{}+ \ket{101011}_{1QR4TU} + \ket{110001}_{1QR4TU} + \ket{111100}_{1QR4TU}
\Bigr) \nonumber \\
=& \dfrac{1}{2} \Biggl(&
\sigma_A^{(1)} \ket{\Phi^+}_{1R} \otimes \frac{1}{2\sqrt{2}} \Bigl(
\ket{0000}_{Q4TU} - \ket{0011}_{Q4TU} \nonumber \\
& &\qquad\quad{}- \ket{1100}_{Q4TU} + \ket{1111}_{Q4TU} + \ket{0101}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{0110}_{Q4TU} + \ket{1001}_{Q4TU} + \ket{1010}_{Q4TU}
\Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Phi^-}_{1R} \otimes \frac{1}{2\sqrt{2}} \Bigl(
\ket{0000}_{Q4TU} + \ket{0011}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{1100}_{Q4TU} + \ket{1111}_{Q4TU} - \ket{0101}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{0110}_{Q4TU} + \ket{1001}_{Q4TU} - \ket{1010}_{Q4TU}
\Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Psi^+}_{1R} \otimes \frac{1}{2\sqrt{2}} \Bigl(
\ket{0101}_{Q4TU} - \ket{0110}_{Q4TU} \nonumber \\
& &\qquad\quad{}- \ket{1001}_{Q4TU} + \ket{1010}_{Q4TU} + \ket{0000}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{0011}_{Q4TU} + \ket{1100}_{Q4TU} + \ket{1111}_{Q4TU}
\Bigr) \nonumber \\
&+&\sigma_A^{(1)} \ket{\Psi^-}_{1R} \otimes \frac{1}{2\sqrt{2}} \Bigl(
\ket{0101}_{Q4TU} + \ket{0110}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{1001}_{Q4TU} + \ket{1010}_{Q4TU} - \ket{0000}_{Q4TU} \nonumber \\
& &\qquad\quad{}+ \ket{0011}_{Q4TU} + \ket{1100}_{Q4TU} - \ket{1111}_{Q4TU}
\Bigr)
\Biggr)
\label{eq_ModProt_AliceBSM}
\end{eqnarray}

\begin{figure}[pb]
\centerline{\psfig{file=fig_mod_protocol.eps,width=0.85\textwidth}}
\caption{An illustration of the modified protocol presented in section \ref{sec_ModProt}. Alice
performs the Hadamard operation before she applies her secret operation $\sigma_x$. Again,
$\ket{\Psi^-}_{13}$ is assumed to be Alice's result of the Bell state measurement.}
\label{pic_ModProt}
\end{figure}

For reasons of simplicity we will assume that Alice's result is $\sigma_A^{(1)} \ket{\Phi^-}_{1R}$
as it is illustrated in picture (8) in figure \ref{pic_ModProtAttack} (the computations for any
other result are analog). Bob also applies the Hadamard operator on qubit $Q$ in his possession and
thus alters the state of qubits $Q,4,T$ and $U$ to
\begin{eqnarray}
&\dfrac{1}{4} \Bigl(&
   \ket{0000}_{Q4TU} + \ket{1000}_{Q4TU} + \ket{0011}_{Q4TU} + \ket{1011}_{Q4TU} \nonumber \\
&+&\ket{0100}_{Q4TU} - \ket{1100}_{Q4TU} + \ket{0111}_{Q4TU} - \ket{1111}_{Q4TU} \nonumber \\
&-&\ket{0101}_{Q4TU} - \ket{1101}_{Q4TU} + \ket{0110}_{Q4TU} + \ket{1110}_{Q4TU} \nonumber \\
&+&\ket{0001}_{Q4TU} - \ket{1001}_{Q4TU} - \ket{0010}_{Q4TU} + \ket{1010}_{Q4TU}
\Bigr)
\end{eqnarray}
which can be alternatively written as
\begin{equation}
\frac{1}{\sqrt{2}} \biggl(
\ket{\Phi^-}_{Q4} \otimes \ket{\omega^-}_{TU} + \ket{\Psi^+}_{Q4} \otimes \ket{\chi^+}_{TU}
\biggr)
\end{equation}
From this it is easy to see that Bob will get the expected result, $\ket{\Phi^-}_{Q4}$ only with
50\% probability. In the other half of the cases he will obtain $\ket{\Psi^+}_{Q4}$. Alice and Bob
will detect this error when they check the correlation between their results.
\par
That means in $\frac{1}{2}$ of the cases, when Alice doesn't apply the Hadamard operation, Eve will
stay undetected with certainty. In the other half of the cases when Alice performs the Hadamard
transformation on qubit 1 Eve will be detected with probability $\frac{1}{2}$. These probabilities
are valid for the simple case ($n=1$) where Alice and Bob share only 2 Bell states. In general,
Alice and Bob share $2n$ entangled qubit pairs and then they will detect Eve with the probability
\begin{equation}
p = 1 - \biggl( \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} \biggr)^n
  = 1 - \biggl(\frac{3}{4}\biggr)^n
\label{eq_ModProt_Error}
\end{equation}
which can be brought close to 1 for large $n$.
\par
A possibility for Eve to overcome the effect of the Hadamard operation is to apply a $H$ operation
herself on the qubits $P$ and $Q$ before the protocol starts. This changes the initial state
$\ket{\delta}_{PQRSTU}$ to $H^{(P)}H^{(Q)}\ket{\delta}_{PQRSTU}$ and in the course of the
protocol Alice and Bob's application of the Hadamard operator will invert Eve's operations. In
detail, eq. \ref{eq_EveAltState} changes to
\begin{equation}
\ket{\delta^\prime}_{1QR4TU} = \sigma_A^{(1)} H^{(1)} H^{(1)} H^{(Q)} \ket{\delta}_{1QR4TU}
= \sigma_A^{(1)} H^{(Q)} \ket{\delta}_{1QR4TU}
\end{equation}
and further, after Bob's application of the Hadamard operation
\begin{equation}
H^{(Q)} \ket{\delta^\prime}_{1QR4TU} = \sigma_A^{(1)} H^{(Q)} H^{(Q)} \ket{\delta}_{1QR4TU}
= \sigma_A^{(1)} \ket{\delta}_{1QR4TU}
\end{equation}
which is equal to the state from eq. \ref{eq_CESP_ESPO_AS_ES_ESAlice}. Thus, the attack reduces to
the version described in section \ref{sec_AttStrat} and Eve will obtain full information about the
key whenever Alice decides to perform the $H$ operation. In all the other cases Eve will introduce
an error equal to eq. \ref{eq_ModProt_Error} due to her Hadamard operations, which is easy to see
following the arguments in the paragraphs above. At this point we want to stress that Alice and Bob
will become suspicious if they find errors only in those cases when Alice does (or does not,
respectively) perform the Hadamard operation. Thus, a better strategy for Eve is to randomly apply
the two Hadamard operations on qubits $P$ and $Q$. She will introduce the same error rate but the
errors will be equally distributed.
\par
Another option for Eve to invert Alice's Hadamard operation is to delay her measurement on the qubit
pairs 2, $P$ and 3, $S$ until Alice publicly announces whether she applied the Hadamard operation or
not. That means, Eve intercepts qubits 2 and 3 in transit between Alice and Bob and immediately
forwards qubit $R$ to Alice (cf. picture (2) in figure \ref{pic_ModProtAttack2}). Alice, convinced
to have received Bob's qubit, applies the Hadamard operation and performs the Bell state measurement
on qubits 1 and $R$ (cf. pictures (3)-(5) in figure \ref{pic_ModProtAttack2}). Due to entanglement
swapping qubit 2 is now entangled with qubits $P$, $Q$, $S$, $T$ and $U$, as
described in eq. \ref{eq_SecAnal_AliceBSM}.
\begin{eqnarray}
\lefteqn{\sigma_A^{(1)} H^{(1)} \ket{\Phi^+}_{12} \otimes \ket{\delta}_{PQRSTU} =} \hspace{30pt} \nonumber \\
&\dfrac{1}{2} \Biggl(&
\ket{\Phi^+}_{1R} \otimes \sigma_A^{(P)} H^{(2)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{PQ2STU} + \ket{001101}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{010111}_{PQ2STU} + \ket{011010}_{PQ2STU} + \ket{100110}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{101011}_{PQ2STU} + \ket{110001}_{PQ2STU} + \ket{111100}_{PQ2STU} \Bigr) \nonumber \\
&+&\ket{\Phi^-}_{1R} \otimes \sigma_A^{(P)} H^{(2)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000000}_{PQ2STU} - \ket{001101}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{010111}_{PQ2STU} - \ket{011010}_{PQ2STU} + \ket{100110}_{PQ2STU} \nonumber \\
& &\quad{}- \ket{101011}_{PQ2STU} + \ket{110001}_{PQ2STU} - \ket{111100}_{PQ2STU} \Bigr) \nonumber \\
&+&\ket{\Psi^+}_{1R} \otimes \sigma_A^{(P)} H^{(2)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000101}_{PQ2STU} + \ket{001000}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{010010}_{PQ2STU} + \ket{011111}_{PQ2STU} + \ket{100011}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{101110}_{PQ2STU} + \ket{110100}_{PQ2STU} + \ket{111001}_{PQ2STU} \Bigr) \nonumber \\
&+&\ket{\Psi^-}_{1R} \otimes \sigma_A^{(P)} H^{(2)} \frac{1}{2\sqrt{2}} \Bigl(
\ket{000101}_{PQ2STU} - \ket{001000}_{PQ2STU} \nonumber \\
& &\quad{}+ \ket{010010}_{PQ2STU} - \ket{011111}_{PQ2STU} + \ket{100011}_{PQ2STU} \nonumber \\
& &\quad{}- \ket{101110}_{PQ2STU} + \ket{110100}_{PQ2STU} - \ket{111001}_{PQ2STU} \Bigr)
\Biggr)
\label{eq_SecAnal_AliceBSM}
\end{eqnarray}

\begin{figure}[pb]
\centerline{\psfig{file=fig_mod_protocol_attack_1.eps,width=0.85\textwidth}}
\caption{An illustration of the attack on the modified protocol as described in section
\ref{sec_SecAnal}. Eve follows the strategy presented in section \ref{sec_AttStrat} and Alice has
chosen to apply the Hadamard operation. Again, $\sigma_x$ is Alice's secret operation and
$\ket{\Psi^-}_{1R}$ is Alice's result of the Bell state measurement.}
\label{pic_ModProtAttack}
\end{figure}

Here, Eve does not know in which state her qubits 2, $P$, $Q$, $S$, $T$ and $U$ are because she does
not know Alice's result. But for our further considerations we assume that Alice's result is
$\ket{\Psi^-}_{1R}$, i.e. Eve is in possession of some state $\sigma_A^{(P)}
\ket{\delta^\prime}_{PQ2STU}$ similar to her initial state $\ket{\delta}_{PQ2STU}$. Next, Alice
publicly announces that she applied the Hadamard operation. Having that knowledge Eve applies a $H$
operation on qubits 2 and $Q$ in her possession inverting the effect of Alice $H$ operation and
preparing for Bob's $H$ operation (cf. picture (7) in figure \ref{pic_ModProtAttack2}). Then she
sends qubit $Q$ to Bob and performs a Bell state measurement on qubits 3 and $S$ which changes the
state to $\sigma_A^{(P)} \ket{\delta^\prime}_{PQ24TU}$. Finally, Bob inverts the Hadamard operation
due to Alice's public announcement (cf. picture (8) in figure \ref{pic_ModProtAttack2}). Similar to
eq. \ref{eq_CESP_ESPO_AS_ES_ESAlice} Bob's Bell state measurement on qubits $Q$ and 4 can be described
as
\begin{eqnarray}
\sigma_A^{(P)}\ket{\delta^\prime}_{PQ24TU} = \dfrac{1}{2} \Bigl(
&-&\ket{\Phi^+}_{Q4} \otimes \ket{\Phi^-}_{P2} \otimes \ket{\Phi^+}_{TU} \nonumber \\
&-&\ket{\Phi^-}_{Q4} \otimes \ket{\Phi^+}_{P2} \otimes \ket{\Phi^-}_{TU} \nonumber \\
&+&\ket{\Psi^+}_{Q4} \otimes \ket{\Psi^-}_{P2} \otimes \ket{\Psi^+}_{TU} \nonumber \\
&+&\ket{\Psi^-}_{Q4} \otimes \ket{\Psi^+}_{P2} \otimes \ket{\Psi^-}_{TU}
\Bigr)
\label{eq_SecAnal_BobBSM}
\end{eqnarray}
It is easy to see that Bob's obtains every Bell state with equal probability of 25\%, i.e. his and
Alice's results are completely uncorrelated. In case Alice does not apply the Hadamard operation Eve
will introduce the same error rate. Therefore, the probability Eve will be detected is
\begin{equation}
p = 1 - \biggl( \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \biggr)^n
  = 1 - \biggl(\frac{1}{4}\biggr)^n
\end{equation}
which converges much faster to 1 compared to the first strategy described above (cf. eq.
\ref{eq_ModProt_Error}).

\begin{figure}[pb]
\centerline{\psfig{file=fig_mod_protocol_attack_2.eps,width=0.85\textwidth}}
\caption{An illustration of the attack strategy with delayed measurement on the modified protocol as
described in section \ref{sec_SecAnal}. Eve waits with her measurement on qubits 2 and $P$
until Alice announces whether she applied the Hadamard operation or not.}
\label{pic_ModProtAttack2}
\end{figure}

\section{Conclusion} \label{sec_Concl}

We showed that the protocol presented by Li et al. \cite{LWWSZ06} is open to an attack strategy
where Eve entangles herself with both legitimate communication parties. Using a special 6-qubit
state which preserves the correlated results of entanglement swapping an adversary is able to obtain
full information about the key Alice and Bob share in the end. We also discussed an improvement
where Alice uses a Hadamard transformation on one qubit to secure the protocol. We showed that due
to the change of the initial state caused by the Hadamard operation an adversary will be detected
with a high probability.

\section*{Acknowledgements}

We would like to thank Christian Kollmitzer and Oliver Maurhart for fruitful discussions and
comments. Further we want to thank our reviewer for showing us an interesting new aspect of this
topic. This work was supported by the Integrated Project SECOQC (Contract No. IST-2003-506813) in
the Sixth Framework Program of the European Union.

\section*{Appendix. Generation of Eve's Initial State} \label{sec_appendix}

Eve's initial state $\ket{\delta}_{PQRSTU}$ from eq. (\ref{eq_EveInitState}) is rather complex and
it might not be producible due to today's physical limitations. Nevertheless, we can show that it
is, in principle, possible for Eve to generate this state using a sequence of Hadamard operations
and controlled Pauli operations.
\par
First we assume that Eve is in possession of 3 sources emitting Bell states. The state of these 6
qubits is
\begin{equation}
\ket{\Phi^-}_{PR} \otimes \ket{\Phi^+}_{QS} \otimes \ket{\Phi^+}_{TU}
\end{equation}
Eve performs a Hadamard operation $H$ on qubits $P$ and $Q$ (indicated by the superscripts $(P)$ and
$(Q)$), which alters the initial state first to
\begin{equation}
H^{(P)} \ket{\Phi^-}_{PR} \otimes \ket{\Phi^+}_{QS} \otimes \ket{\Phi^+}_{TU}
\end{equation}
and further to
\begin{eqnarray}
\lefteqn{H^{(P)} \ket{\Phi^-}_{PR} \otimes H^{(Q)} \ket{\Phi^+}_{QS} \otimes \ket{\Phi^+}_{TU}=} \hspace{30pt} \nonumber \\
&\frac{1}{2} \Bigl(&
  \ket{\Phi^+}_{PR} \otimes \ket{\Phi^-}_{QS} \otimes \ket{\Phi^+}_{TU}
+ \ket{\Phi^+}_{PR} \otimes \ket{\Psi^+}_{QS} \otimes \ket{\Phi^+}_{TU} \nonumber \\
&-&\ket{\Psi^-}_{PR} \otimes \ket{\Phi^-}_{QS} \otimes \ket{\Phi^+}_{TU}
-\ket{\Psi^-}_{PR} \otimes \ket{\Psi^+}_{QS} \otimes \ket{\Phi^+}_{TU}
\Bigr)
\end{eqnarray}
Then Eve applies a $\sigma_z$ operation on qubit $Q$ if qubits $P$ and $R$ are in the Bell state
$\ket{\Phi^+}_{PR}$. Otherwise the identity operator is applied. This action can be described as
\begin{eqnarray}
\oP{\Phi^+}{\Phi^+} \otimes \sigma_z \otimes I \otimes I \otimes I
&+&\oP{\Phi^-}{\Phi^-} \otimes I \otimes I \otimes I \otimes I \nonumber \\
{}+\oP{\Psi^+}{\Psi^+} \otimes I \otimes I \otimes I \otimes I
&-&\oP{\Psi^-}{\Psi^-} \otimes I \otimes I \otimes I \otimes I
\end{eqnarray}
and it changes the state to
\begin{eqnarray}
&\dfrac{1}{2} \Bigl(&
  \ket{\Phi^+}_{PR} \otimes \ket{\Phi^+}_{QS} \otimes \ket{\Phi^+}_{TU}
+ \ket{\Phi^+}_{PR} \otimes \ket{\Psi^-}_{QS} \otimes \ket{\Phi^+}_{TU} \nonumber \\
&+&\ket{\Psi^-}_{PR} \otimes \ket{\Phi^-}_{QS} \otimes \ket{\Phi^+}_{TU}
+ \ket{\Psi^-}_{PR} \otimes \ket{\Psi^+}_{QS} \otimes \ket{\Phi^+}_{TU}
\Bigr)
\end{eqnarray}
Next Eve performs a $\sigma_x$ operation on qubit $P$ and a $\sigma_z$ on qubit $T$ if qubits $Q$
and $S$ are in the state $\ket{\Phi^-}_{QS}$. Alternatively, she performs a $\sigma_z$ operation on
qubit $P$ and a $\sigma_x$ operation on qubit $T$ if qubits qubits $Q$ and $S$ are in the state
$\ket{\Psi^+}_{QS}$. If qubits $Q$ and $S$ are in the state $\ket{\Psi^-}_{QS}$ Eve applies a
$i\sigma_y$ operation on both qubits $P$ and $T$. These three controlled operation can be stated as
\begin{eqnarray}
           I \otimes I \otimes \oP{\Phi^+}{\Phi^+} \otimes I \otimes I
&+&\sigma_x  \otimes I \otimes \oP{\Phi^-}{\Phi^-} \otimes \sigma_z \otimes I \nonumber \\
{}+\sigma_z  \otimes I \otimes \oP{\Psi^+}{\Psi^+} \otimes \sigma_x \otimes I
&+&i\sigma_y \otimes I \otimes \oP{\Psi^-}{\Psi^-} \otimes i\sigma_y \otimes I
\end{eqnarray}
Using this operator Eve is able to bring the 6 qubits into the desired state from eq.
(\ref{eq_EveInitState}).

redpajama_set_name: RedPajamaArXiv


text:
ROCK 96.7ROCK 96.7
Def Leppard's Vivian Campbell 'Dodged Bullet of Radiation'
Def Leppard's Vivian Campbell revealed he will not have to undergo radiation treatments for cancer and will be part of the band's touring this summer.
Def Leppard's Vivian Campbell: 'My Cancer Has Returned'
Discouraging news to report, as Def Leppard guitarist Vivian Campbell has revealed that his cancer has returned.
Spencer Kaufman
Vivian Campbell's Cancer Is in Remission Once Again
There's now some good news to share as Campbell has announced for the second time that his cancer is in remission.
Def Leppard's Vivian Campbell Planning Stem Cell Transplant
Def Leppard guitarist Vivian Campbell discusses his current chemo treatment and talks about plans to receive a stem-cell transplant later this year.
Vivian Campbell Says Cancer 'Seems to Be Coming Back Again'
Legendary Def Leppard / Dio guitarist Vivian Campbell announced in June of last year that he was battling Hodgkin's lymphoma, but had updated fans in November with the good news that the cancer had gone into remission. Unfortunately, in a new interview, the shredder has just revealed that his c…
Vivian Campbell Announces Remission From Lymphoma
Fantastic news! Def Leppard / Dio guitarist Vivian Campbell is in remission from Hodgkin's lymphoma. Campbell was diagnosed with the form of cancer earlier this year, but in a Facebook post from the guitarist on his official page, Vivian Campbell announced that the recovery process has begun.
Def Leppard / Dio Guitarist Vivian Campbell Fighting Cancer
Def Leppard guitarist Vivian Campbell has revealed that he was recently diagnosed with Hodgkin's Lymphoma and has already gone through two of six months of chemotherapy to battle the cancer.
Mary Ouellette
Guitarist Vivian Campbell Remorseful Over Past Comments Made About Ronnie James Dio
Current Def Leppard guitarist Vivian Campbell spent most of his career in the '80s playing in the band Dio with Ronnie James Dio. When Campbell parted ways with Dio in 1986, it wasn't on the best terms and the war of words between the two continued throughout their careers. Now…

redpajama_set_name: RedPajamaCommonCrawl


White Plains, NY The Hudson Gateway Realtor Foundation, the charitable arm of the Hudson Gateway Association of Realtors, presented a check for $3,000 to Drug Crisis in Our Backyard, based in Carmel but serving the Hudson Valley. Drug Crisis in Our Backyard was founded in 2012 by Susan and Steve Salomone, and Carol and Lou Christiansen after the loss of their sons to a battle against heroin. The organization promotes awareness about drug use, assists addicted and at-risk individuals and their families, and implements measures, including legislation that holds accountable organizations and medical institutions that perpetuate drug use through over-prescription of opiates and other drugs. It serves the Hudson Valley region with support groups and various events.
Christiansen is a broker/owner with Café Realty in Mount Kisco, and Steve Salomone is an associate broker with Houlihan Lawrence Commercial Real Estate in Rye Brook.


I hiked Inwangsan Mountain Trail in early spring last year with my colleagues with much excitements. It was part of our outing activity to see historical sites especially those related to people who struggle for human freedom.
It was a refreshing spring day. The sun was bright, and a clear day.
We took off from Sodaemun area and walked all the way to the foot of Inwangsan, which is located in Seoul area. It was a pleasant walk and not far from our office.
We past by the walls on our way up and could see a few traditional Korean houses. More modern apartment and higher buildings are becoming more obviously invading the spacious areas of the old Seoul.
As you start climbing up the lower part of the mountain, you'll notice the lower height walls. These are the old Seoul City walls (formerly called 'Hanyang') which the ministry of tourism is promoting by renovating them.
Reconstructing the old city walls would not only attract hikers and tourist but also relived the memories of the past struggles of the people.
Of course, I can imagine how people in the past crisscrossed that same road and mountain paths. I can imagine ordinary people walking (or riding) with their cows carrying heavy loads on their way either to the markets or back home.
Close to Inwangsan (Inwang Mountain), you can find ritual sites of King Sejo and some castle walls that stretch from Sajik Tunnel to Jahamun.
Inwangsan Mountain is more than 300 meters high and tops over Hongdae District, Sodaemun-gu, Muak-dong, Nusang-dong, Ogin-dong, and Buamdong in Jongno-gu.
The mountain's name literally means "mountain of a generous king" in the Korean language. This mountain approximately covers about 1 million square meters. Just like most mountains in Korea, this mountain is mostly composed of granite elements.
You can obviously observe the granite boulders showing off at the sides and piling up on top of this mountain.
These granite boulders have various and fascinating shapes and forms. Typically, Koreans, just like most humans throughout the world would do, associate some shapes to images and create meanings in them.
For example, there are granite rocks that they call train rock (kicha bawi), skirt rock (Chima bawi), hat rock (moha bawi), among hundreds of shapes and forms that one can associate and create meaning with.
That makes hiking this mountain an interesting experience aside from the nice view of Seoul and the surrounding districts.
Inwangsan is famous not only among its hikers who can have a great view of Seoul and its surrounding districts but since time immemorial.
Famous painters depict this mountain in their paintings where you can find in many museums established in Seoul.
For example, the famous painter Jeong Seon who painted "Inwang jesaekdo" (translated as "After the rain at Mount Inwang").
The painting was painted during the reign of King Yeongjo of Joseon Dynasty in 1751. You can find this great painting at the Ho-Am Art Museum in Yongin, Gyeonggi Province.
In addition, the Fortress Walls of Seoul envelops Inwang Mountain where you can also find the tourist sites--Inwangsa (Inwang Temple) and Guksadang Shrine.
Most temples attract foreign visitors, like me, due to various reasons such as its sense of tranquility, divine presence... you can add more!
Guksadang is a shamanist shrine. This shrine houses the spirits of Joseon Dynasty's King Taejo and other generals. Guksadang was first built in Namsan Mountain (in 1395) but got destroyed during the war in 1925 and restored where it is located now.
Take Gyeongbukgung Station-Seoul Subway Line 3, Exit 3. Then go straight ahead for about 90 meters to find Gyeongbukgung Bus Stop. Take Bus #09 and then get off at Ogin Apartment complex. From here, it will be about 600 meters to Inwangsan Mountain.
With confidence, there are many points of origin and directions depending on your choices and point of origin or where you live or start from.
So, there is so much to see in this favorite mountain, although honestly, I've haven't yet explored all of them. That would be part of my update then for this page!
Thanks for reading through and hope you'll visit this blog more often to discover more about this mountain and anything interesting about Seoul and South Korea in general.


The singer, who now co-hosts ITV's Loose Women, told fellow panellists on the show she agreed with former Spice Girl Mel C, who said she did not let her daughter watch raunchy videos.
She said: "I think I went through the same process myself. Having two daughters myself. When you're an artist and you don't have responsibilities within your household, you don't think about it, and when you see your own child being affected or influenced that's when it does become a concern.
"For me when I was younger, when I was doing the Superstar video - was that a skirt?"
She went on: " Now I feel responsible. Not only for my own daughter but other people's daughters and someone like J-Lo at 42 years old, I'm sorry she should be thinking about other people's kids as well and that's my problem."
Mel C said her daughter was a "big fan of Rihanna" but her videos were "too sexual " for her to be allowed to watch them.


Been busy the last week or so on the System.Messaging/AMQP integration on Mono, I updated the RabbitMQ client library to version 1.5.3, only to have them release a new version of the library (1.5.5) literally the day after. Mirco Bauer, who handles the packaging for Debian, ran into a bit of a problem with the way we included the RabbitMQ library. Unfortunately the AMQP spec files are not considered Free under the terms of the DFSG, which is not much of a problem as we do not use them during the build. However, the .cs files that are generated from the specs contained comments that were extracted verbatim from said specs, which make the .cs files non-free too. Anyway, to cut a long story short, I've patched the RabbitMQ code generator and pushed the patch back upstream so that we can generate the .cs files without comments and I'm in the process of fixing the generated files in Mono's trunk and 2.4 branch.
As far as actual new functionality goes, I've updated the MessageEnumerator support, so that MoveNext(TimeSpan timeout) and RemoveCurrent(TimeSpan timeout) are both supported. I've also added calls for all of the other methods through to Mono.Messaging provider layer, unfortunately the way transactions work in AMQP make it impossible use the transactional methods of MessageEnumerator (personally I blame the System.Messaging API design, but that's a rant for a different blog post). If you need transactions, then you stick to the methods in MessageQueue, which are supported.
P.S. Hello Monologue, thanks to Miguel for kindly adding me to the group of Mono bloggers.


text:
Google In Talks With Jio, Airtel Over Project Taara For Internet Through Light Beams
Kritti Bhalla
Google's Project Taara has been piloted in Andhra Pradesh in India and Kenya in Africa
The tech can bring connectivity to remote areas and resolve hurdles such as laying optic fibers
Google India decided to kill its "Station" project in 2020 citing 4G connectivity and cheap data tarrifs
Tech giant Google is reportedly in talks with Reliance Jio and Bharti Airtel to pilot high-speed internet and telecom connectivity using light beams. The initiative is a part of Google's 'Project Taara' which aims to provide internet connectivity through light beams.
The initiative was started by Google X, a company within Google that is known to work on so-called 'moonshots'. Taara builds on the company's earlier efforts for internet connectivity solutions such as Project Loon and others. .
Project Taara, which has been piloted in Andhra Pradesh in India and Kenya in Africa, uses free-space optical communications like light beams to provide internet connectivity to far-flung regions at a speed of 20 gigabytes per second. This is a significant increase from the current one gigabyte per second that most customers have access to currently.
The project can bring connectivity to remote areas and resolve hurdles such as optic fiber installation and other infrastructure, which is said to have intrigued both Airtel and Reliance Jio to be a part of the project.
The technology sends light beams instead of radio waves between transmitters and receivers attached to servers placed on buildings and structures. "This will be a game-changer since it can connect one to remote regions, have internet connectivity across buildings without cables and reduce the burden on towers," a person aware of the development told ET.
"Taara links offer a cost-effective and quickly deployable way to bring high-speed connectivity to remote areas. Taara links help plug critical gaps to major access points, like cell towers and WiFi hotspots, and have the potential to help thousands of people access the educational, business, and communication benefits of the web." Google X's website says.
Between 2015-2020, Google India had provided internet to 415 railway stations in India as a part of its global "Station" project to improve connectivity. However, it decided to kill the project in India earlier this year citing increased 4G connectivity and cheap data tarrifs.
"India, specifically now has among the cheapest mobile data per GB in the world, with mobile data prices having reduced by 95% in the last 5 years, as per TRAI in 2019. Today, Indian users consume close to 10 GB of data, each month, on average," it said.
The company added that after the Indian government, several governments and local entities have kicked off their own initiatives to provide easier, cost-effective access to the internet for everyone, beyond railway stations.
Google Technology


In this new video, young scientists talk about the challenges they faced doing participatory research and how they worked creatively to overcome them.
Maria Fernandez, Honorary Research Fellow, Bioversity International, introduces this new video, in which young scientists talk about the challenges they faced doing participatory research and how they worked creatively to overcome them.
Rural communities manage natural resources under diverse and changing landscape and market conditions throughout the developing world. Over the past 20 years scientists, concerned with the sustainability of smallholder farm and forest resilience have become increasingly aware of the need to link research for development with the needs and experiences of these communities. Our scientists are challenging themselves to find, develop and use gender-responsive participatory research methods and tools to listen to the voices of different groups of actors and generate high quality and useful science.
Participatory research is about providing local people with opportunities to share their knowledge and experience with each other while also sharing it with researchers. It is about tackling problems that are relevant to communities in ways that draw on their own successful experiences and knowledge and integrating them with a conventional scientific perspective. Doing successful participatory research entails acquiring a set of skills.
Bioversity International's Gender Research Fellowship Programme which ran from 2013-14, provided an opportunity for five Research Fellows affiliated with national partner institutes to take these skills on board and use them to enhance the gender responsiveness of their Bioversity International projects. The video Revealing farmers' knowledge through research gives a window into the experiences of these Fellows as they took on the challenge. Two messages are evident in the film. First, participatory research can give a voice to those who are often not heard. Second, the Fellows themselves were surprised and transformed by facilitating this kind of co-learning process.
Researchers need to gain experience with participatory-research tools. Choosing the most appropriate tools is a challenge as there are many to choose between. A good researcher will choose a few tools designed to give a voice to those who don't know that they have knowledge to share or are not accustomed to being listened to.
Working with gender-differentiated groups brings out invisible, insider knowledge. Once men and women have worked in separate groups and they come together in plenary to share their results, it most often becomes evident that women and men have been looking at issues differently. This experience enriches and deepens knowledge and often results in concrete actions and changes.
Quantification of information takes creativity and the use of appropriate tools. When qualitative data, such as farmers' knowledge, is made visible using tools like scoring, it can be put into graphs. In addition to graphs there are a number of tested tools that can help represent qualitative information graphically.
Participatory research processes can increase self-esteem and build confidence when people (especially women) find out they know more than they thought they did, can do things they didn't think they could do and are being recognized for their knowledge. People who speak only indigenous languages, often women and marginalized groups, tend to be left out of development discussions. Research processes that give them space to express their knowledge can be empowering, as well as allowing scientists to learn from them.
Increased confidence and empowerment contribute to equality. When men and women from diverse social and ethnic groups are given access to knowledge about each others' skills, needs and experiences, it increases mutual respect and opens the door for innovation. They learn things about each other that they didn't know and talk about things they had never talked about. We found this with research in India bringing together diverse social and ethnic groups, and in Malaysia bringing together women and men.
Good gender-responsive participatory research requires a commitment to learning new skills and finding ways to listen to less articulate or more marginalized groups. It is a journey that changes us as professionals and makes our research more relevant to the future we are working towards.


Briggs & Forrester Living commence works on Kampus development.
Described as Manchester's own secret garden, the £250m mixed use development 'Kampus' is being developed by joint venture partners Capital & Centric and Henry Boot Developments to home a new vibrant cultural spot located in the heart of Manchester.
The next phase will include the refurbishment of two 3-storey Grade II listed buildings to create 58 one and two bedroom apartments.
Last year, Briggs & Forrester Living secured a contract on the Manchester development. Working in collaboration with contractors Mount Anvil, Briggs & Forrester Living will be responsible for the design and installation of services including thermal insulation and heating & cooling.
The former Manchester Metropolitan University site will be transformed into a distinctive new city centre neighbourhood offering apartments to rent, independent bars, restaurants and small- scale local retail units.
Work on the listed buildings is due this year and the Kampus project is due for completion in summer 2020.


The new products will officially be available on the MFT online store on October 4th at 10:00 pm EST (= Europe: October 5th at 4:00 am CET).
What a wonderful sentiment set! I love the font! The Moroccan Lattice cling stamp is perfect to create a sponged background. Of course… I chose rainbow colors! Ha ha! I first heat embossed the Moroccan Lattice Background with white embossing powder. I then sponged the panel with MFT dye inks: Grape Jelly, Blu Raspberry, Field Day, Sour Apple, Lemon Drop, Orange Fizz and Wild Cherry. I fixed the background on a top-folded card base made from Sweet Tooth cardstock. I die cut a black label from Blueprints 13 Die-namics and I heat embossed the sentiment with white embossing powder. The "Peace on Earth" sentiment is from the Hand-lettered Christmas set. To finish off my card, I fixed the sentiment to the card base with foam pads and I added three enamel dots.
Francine, This is so pretty! I love the white embossed stencil and the non-traditional Christmas colors! I need to step out and try these colors. They create such a WOW factor!
Beautiful use of the Moroccan Lattice that shows off the elegantly lettered sentiment!
The rainbow effect and colors are so beautiful with this sentiment. TFS!
Love your technique, I'm going to give it a try.
Beautiful rainbow background, Francine! I love how your sentiment pops against all the fun colors...perfect placement and gorgeous CAS design!
Francine, this is a very pretty card! I love the uniqueness of it versus the usual (but also pretty) or traditional Christmas cards. I love the affect the stencil made with the muted colors. The sentiment pops on the black cardstock. The enamel dots help to move the eye further down from the sentiment. Thanks for sharing!
Wow - you achieved your goal. A beautiful card.


This function links service accounts to subaccounts and creates a subaccount if one does not exist. To link multiple service accounts, the service accounts must share the same username and domain.
You can only link email, FTP, or Web Disk accounts.
The username for the service account.
The service account's associated domain.
Whether to link the email account to the subaccount.
0 — Do not merge.
Whether to link the FTP account to the subaccount.
Whether to link the Web Disk account to the subaccount.
An alternate email address for the subaccount's user.
The user's subaccount profile photo.
can_delete Boolean Whether the cPanel account user can delete the subaccount.
Whether the cPanel account user can change the subaccount's password.
The function returns a 0 value for subaccounts that inherit their password from the cPanel account.
can_set_quota Boolean Whether the cPanel account user can change the subaccount's disk usage quota.
0 — Did not dismiss.
The subaccount's username and domain name.
The system assigned unique value.
issues array Information about any issues or problems with the subaccount.
This array contains the type, area, service, message, used, and limit returns.
The affected section of cPanel & WHM.
A valid string. Your account is over quota.
The set quota Megabyte (MB) limit for the affected subaccount.
merge_candidates array An array of hashes that represents the service accounts that the system could merge for this subaccount. This array contains the alternate_email, avatar_url, can_delete, can_set_password, can_set_quota, dismissed, domain, full_username, guid, has_siblings, issues, merge_candidates, parent_type, phone_number, real_name, services, special, sub_account_exists, synced_password, type, and username returns.
hypothetical — A hypothetical subaccount that does not yet exist, but the user could create as part of a merge.
The name of the subaccount's user.
services hash This hash contains information about the subaccount's access to email, FTP, and Web Disk. This hash contains the email, ftp, and webdisk returns.
This hash contains the information that indicates the subaccount's email status.
The function returns this hash in the services hash.
This hash contains the enabled and quota returns.
Whether the subaccount can access Webmail, POP, and IMAP services.
The maximum amount of disk space allocated to subaccount's email account.
A positive integer that represents the maximum disk space that the subaccount may use, in Megabytes (MB).
In cPanel & WHM version 70 and later, this funciton will not return 0 as a value.
This hash contains the information that indicates the subaccount's FTP status.
This hash contains the enabled, quota, and homedir returns.
Whether the subaccount can access FTP.
The subaccount's FTP disk space quota.
A positive integer between 1 and 999999 that represents the maximum disk space that the subaccount may use, in Megabytes (MB).
The subaccount's FTP home directory.
This hash contains information that indicates the subaccount's Web Disk status.
This hash contains the enabled, homedir, perms, and private returns.
Whether the subaccount can access Web Disk.
The subaccount's Web Disk home directory.
Whether to grant write permissions to the subaccount.
sub_account_exists Boolean Whether a subaccount exists with the same username.
Whether the user has synchronized the passwords for each of the subaccount's service accounts.
hypothetical — A hypothetical subaccount that does not exist, but that the user could create as part of a merge.
cpanel — The cPanel account.
The username for the subaccount.
A username that meets the subaccount name limitations.
Characters — a-z, A-Z, 0-9, dot (.), hyphen (-), underscore (_).


text:
Q: link lists inside function in python I am trying to to connect the element of my lists in the function in way if I do any change on any element of
def itinerant(S1,S2,S3,S4,S5,S6,S7,S8):
    old_list=[S1,S2,S3,S4,S5,S6,S7,S8]
    outer=[S5,S6,S7,S8]
    outer_cube=[[S2,S3,S4],[S1,S3,S4],[S1,S2,S4],[S1,S2,S3]]

so for example if I make change on any of S3s on outer_cube in the 'outer_cube' list this change goes to the rest of S3s in this list and S3 'old_list' as well
my function returns
    new_list=[S1,S2,S3,S4,S5,S6,S7,S8]
with the updated elements.any suggestions would be appreciated.

A: You only have one instance of the s3 variable that is being passed in as a parameter. You assign it to a bunch of lists, but they all refer to the same s3. What is an s3? If it's a python object, you'll need to deep clone it.
import copy
news3= copy.deepcopy(s3) # deep (recursive) copy

If it's something like a dictionary, then you you can construct a copy like so dict(s3).
This means that in your code you'll need to make copies of all the the entities you want to manipulate in isolation.  The second line of your code becomes:
outer=[copy.deepcopy(S5),copy.deepcopy(S6),copy.deepcopy(S7),copy.deepcopy(S8)]

Same goes with your last line.

redpajama_set_name: RedPajamaStackExchange


SKY3DS can support the new Nintendo 3DS firm ware versions. You don't need very old firm ware to start using this new flash cart!
Easy to use – just drag and drop the your game backup ROMs to the MicroSD card using the Disktool software and enjoy it.
* Support all New3DS/3DSLL, 3DS/XL, 2DS consoles(US,EU,JP,KR,AU…).
• Support all New3DS/3DSLL, 3DS/XL, 2DS consoles(US,EU,JP) – don't need any emunand.
• Does NOT support multi region – it's region locking like a real game card, you can't play US or JP's games on your EU 3DS console.
* Thosis information e are the official specifications from the manufacturer. We can not guarantee if it is 100% working for ever on all consoles, make sure to download and install latest updates.


text:
Addicts Among Us Documentary
When: Thu., Oct. 3, 9 p.m. 2019
KEET-TV's new production, "Addicts Among Us," will dive into Humboldt County's addiction crisis by investigation how trauma experienced by the county's children has translated into oppressive rates of drug dependency. The hour-long film was produced and directed by local filmmaker Laura Montagna, whose previous work includes the annual production of TEDX-Humboldt. "Addicts Among Us" will premiere Thursday Oct. 3, at 9 pm on KEET-TV. The program will air again: ● 7 p.m. on Oct. 6; ● 2 p.m. on Oct.12; ● 8 p.m. on Oct. 22; ● 11 a.m. on Oct. 27; and ● 8 p.m. on Nov. 16.


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The first historically documented record of Katzelsdorf dates to the 1050. When Johann I Joseph Liechtenstein, prince of Mikulov and Valtice at that time, had purchased it, Katzelsdorf became part of the Valtice estate. During the adjustment of the borders between Czechoslovakia and Austria in 1925, the village lost its northern part, as well as so called "Little Katzelsdorf Chateau", nowadays non-existent salet, in the 1960s.
Neo-romanesque and neo-gothic St. Bartholomew's Church, built on the occasion of the 50thanniversary of the Johann II's rule by architect Karel Weinbrenner (1856–1929) between 1905–1908, stands as a dominant of the village. Built on a place of old and many times rebuilt gothic church, local people helped to build the new church by their hands and contribution of draft animals. Also, construction was financed by Johann II Liechtenstein himself.
There are one main and two adjacent altars inside the church. Above the main one, there are three stained-glass windows, depicting the patron saint St. Bartholomew in the centre, St. Florian on the left and St. Sebastian on the right. Eucharistic Heart of Jesus statue stands in the opposite of the pulpit, together with a memorial death plaque of World War I. Next to the plaque and statue, there are baptismal font, funeral altar on the left and St. Mary's altar on the right. Walls of the main nave are decorated with stone carvings of Stations of the Cross and stained-glass windows, which depict the Transfiguration of Jesus on the left wall and the Holy Family and Christ being baptized on the right one. Initiated by Magdalena Kersch, the sculpture of three archangels – Gabriel, Michael and Raphael – was built on the south side of the church as a reminiscence of World War I. Rieger company constructed the local organ in 1907. Christ the Redeemer statue rises high above the church portal.


text:
PM: New gov't committed to strengthening ties with China
Bernama | Published: 2019-05-26 10:36:00+08:00 | Modified: 2019-05-26 10:36:00+08:00
The new Malaysian government that came to power in May last year is committed to strengthening Malaysia-China relations with even closer and more meaningful co-operation, said Prime Minister Dr Mahathir Mohamad.
The trade and investment growth between the two countries in recent years had been meteoric and this trend would continue to generate more opportunities in the areas of tourism, science and technology as well as cultural exchanges, he said in a message in the souvenir book published by the Malaysia-China Friendship Association, better known as PPMC, to mark the 45th anniversary of diplomatic ties between the two countries.
Mahathir said if in the initial stages, it was more of government-to-government relations, the partnership had scaled numerous levels and chalked up milestones that today make China Malaysia's biggest trade partner.
"China's astonishing growth in the last few decades that catapulted it to become the world's second biggest economy, had contributed much to its extensive presence not only in Malaysia, but in other parts of the world.
"Our co-operation has grown at various levels and sectors. I believe apart from the roles played by government and political leaders, the various associations and non-governmental organisations from both nations, including PPMC, have contributed much to the strong and positive state of affairs enjoyed by both our countries," he said.
The prime minister said it was in June 1993 when he witnessed the signing of the memorandum of understanding between PPMC and its sister organisation, the China-Malaysia Friendship Association, in Beijing, during one of his many official visits to China.
"I wish to congratulate PPMC for being able to sustain the friendship through all these years and in fact, taking it to a higher level," he said.
Mahathir said despite changes of leadership in both countries, the partnership had remained constant through the years, thanks to their adherence to the principle of non-interference in each other's internal affairs and striving to be good neighbours.
"I have always believed in the principle of 'prosper thy neighbour' and as China prospered, it has in turn prospered Malaysia as well," he added.
On China's Belt and Road initiative, the prime minister said Malaysia had taken the position of supporting it for the benefits were mutually shared and distributed.
He added that for Malaysia, it was a continuation of the ancient trade that started way back during the Malacca empire and trading between both countries had never diminished for centuries.
PPMC was formed in 1992 with members comprising entrepreneurs, investors, academicians, artists, educationists, politicians and professionals, as well as government officials sharing mutual interest in wanting to see a strong and sustained friendship, understanding and partnership between Malaysia and China.
In April 2005, veteran diplomat and former Malaysian ambassador to China Abdul Majid Ahmad Khan was elected PPMC president and he still holds the office till today.
M'sia keen for more 'fair' deals with China - Saifuddin
US-China trade war will benefit Malaysia
Mahathir's China visit has been a diplomatic disaster
M'sia to enter 10 new halal markets via China's Obor initiative
malaysia-china friendship association


The Electronic Resources and Libraries Conference tracks are updated by our Program Planning Committee. The 2019 program will span these tracks and subtopics and related areas of interest to our community.
Managing electronic resources is a challenge, whether you're new to it or have been engaged in the endeavor for years. Which systems and tools can be used to manage electronic resources more effectively? How are we successfully handling new content formats? What can we achieve through more thoughtful licensing? How can standards and best practices assist our efforts?
How do we demonstrate value to our larger organizations? Are our collection analysis processes efficient and effective enough? How can collecting and using data be easier? How can we best assess our online collections?
Our organizations regularly evolve due to changes in leadership and strategic vision, budget constraints, user needs, and the simple fact that so many of our resources are now online. All of this impacts the way we manage e-resources. What type of leadership has helped create positive change? What are some examples of beneficial organizational shift and improved communication? Where are there opportunities for internal collaboration? Where do we still need to improve the way we work?
In the digital world, libraries don't stand alone. They work closely with consortia, vendors, other libraries, and intermediaries. How are these relationships working? Are we getting the most out of our partnerships with other organizations or groups? How can we improve our external relationships?
Libraries exist in large part to support our users. How can we better serve our user populations? What kinds of communications will help us reach them when so much is competing for their attention? What tools, activities, or methods can help libraries better track user needs? How can we demonstrate to users the value of libraries and create a better user experience? How do we integrate library content and services into online spaces users already inhabit?
How do we deal with new models of scholarship that are emerging? How do we accommodate new forms of content? What can we do to facilitate knowledge sharing and access? What role can the library play in the creation and distribution of the products of scholarship and creativity?
So much of what we do in libraries today is driven by technology, and many of the problems we face can be addressed by employing, customizing, or developing technologies. How are current technologies being used? What emerging technologies are on the horizon? How we can we employ them effectively to meet the information needs of the library, internally and externally?
Data science is the process of gaining insight from data to inform decision making. In libraries, there are a wide gamut of 'data savvy' roles that orbit within and around the world of data science. From the metadata librarians automating curation workflows to the collection managers gaining insights into their collections, to the subject librarians working closely with researchers, the roles in libraries exist more or less on a data-savvy spectrum to solve internal challenges and to improve services. How has your library made sense of complex datasets to make better decisions? How have data science skills within the library improved services with user communities? What unique partnerships formed around teaching or using data science on campus? What are the unique roles libraries can have in data science and management?
These sessions span all of the tracks, introducing an attendee to a foundational area of librarianship (such as licensing, ebook management, troubleshooting) or an emerging but important topic starting at an introductory level. These courses would not require background knowledge of the topic and would be appropriate for a practitioner new to eresources or as an introduction to a new topic.


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