I earlier posted I question Representing a Bell measurement on non adjacent qubits for which I got an excellent answer. Now I want to build upon that and do further analysis which is where I am stuck. So here it goes. Consider the original state as $${|\psi\rangle} = s {\Bigl(|0\rangle_1|1\rangle_2-|1\rangle_1|0\rangle_2\Bigr)}\otimes{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)},$$ which is being shared by 3 parties where Alice get particle $1$ and $6$, Bob gets $2$ and $3$, Charlies gets $4$ and $5$. Now the sharing protocol is as follows:

  1. First, Alice performs a local unitary operation randomly on one of her two qubits 1 and 6, say, on qubit $1$.

  2. Then she performs a Bell-state measurement on qubits $1$ and $6$ and announces publicly her measurement outcome. After this, Bob and Charlie perform Bell-state measurements on their qubits, respectively, and record the measurement outcomes.

  3. If Bob and Charlie collaborate, according to their Bell-state measurement outcomes and Alice’s public announcement of the Bell-state measurement on qubits $1$ and $6$, they can deduce the exact local unitary operation that Alice performed on qubit $1$.

  4. For example, if Bob’s and Charlie’s outcomes are respectively $|\psi^{-}\rangle_{23}$ and $|\phi^{+}\rangle_{45}$, since the state Alice prepared initially in qubits $3$ and $4$ is $|\psi^{-}\rangle_{34}$, then from Eq.7, they can know that qubits $2$ and $5$ have projected to $|\phi^{+}\rangle_{25}$ after Alice’s Bell-state measurement on qubits $1$ and $6$. Since both the initial states of the qubit pair $(1,2)$ and the initial states of the qubit pair $(5,6)$ are $|\psi^{-}\rangle$, respectively, and Bob and Charlie know Alice’s Bell-state measurement outcome on qubits $1$ and $6$, $|\phi^{+}\rangle_{16}$ and they have already deduced the state $|\phi^{+}\rangle_{25}$ of qubits $2$ and $5$, then from Eq. $8$ they can determine that the local unitary the operation performed by Alice is $u_4$, that is, the secret message is the two classical bits “11.” What I have understood is, let's say the original state is $|\Psi\rangle$

  5. $${|\psi\rangle} = s {\Bigl(|0\rangle_1|1\rangle_2-|1\rangle_1|0\rangle_2\Bigr)}\otimes{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)},$$

  6. Now alice performs say $X$ gate on his qubit $1$, so the new state is ${|\psi\rangle} = s {\Bigl(|1\rangle_1|1\rangle_2-|0\rangle_1|0\rangle_2\Bigr)}\otimes{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_5|1\rangle_6-|1\rangle_5|0\rangle_6\Bigr)},$$

  7. Alice measures say the state $ |\psi^+\rangle_{16}$ , and tells both Bob Charlie. Now after his measurement Bob and Charlie receive the reduced state $${|\psi\rangle} = s {}{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_2|0\rangle_5+|1\rangle_2|1\rangle_5\Bigr)},$$

  8. Now, here things becoming complicated for me while reading, what happens when Bob measures $|\Psi^-\rangle_{34}$ and Charlie measures $|\Phi^+\rangle_{25}$. I am not able to understand the language the point $4$.

$u_3(|\Psi^-\rangle_{ab}) \otimes |\Psi^-\rangle_{cd}$= $ |\Phi^+\rangle_{ab} \otimes |\Psi^-\rangle_{cd} = |\Phi^-\rangle_{ac}|\Psi^+\rangle_{bd} - |\Psi^+\rangle_{ac}|\Phi^-\rangle_{bd}- |\Psi^-\rangle_{ac}|\Phi^+\rangle_{bd} + |\Phi^+\rangle_{ac}|\Psi^-\rangle_{bd}$ where $u_3=|1\rangle\langle 0|+ |0\rangle\langle 1|$. I understand all the operators except this equation and the theoretical explaination. Can somebody explain it in a simple manner?


1 Answer 1


What you should take from your previous question is a basic protocol. If you measure the ends of two Bell pairs using a Bell measurement, and if you apply the appropriate connection, what you are left with is a Bell state between the two unmeasured qubits. I could draw this diagrammatically as: enter image description here So, this is a repeating trick: enter image description here So far, this appears to ignore the unitary that Alice did. However, note that there is a general identity $$ U\otimes U(|01\rangle-|10\rangle)=|01\rangle-|10\rangle, $$ which I can rewrite as $$ U\otimes 1(|01\rangle-|10\rangle)=(1\otimes U^\dagger)(|01\rangle-|10\rangle). $$ Hence, Alice applying $U$ to qubit 1 is equivalent to Bob applying $U^\dagger$ on qubit 2. Furthermore, we can arrange it that (apart from Bob's final measurement, which is not part of the sequence of my diagram) Bob never does anything to his qubit 2 - the corrections can be performed upon qubits 5 and 3. So, it wouldn't matter if the unitary were applied by Bob before or after the sequence of measurements. Thus, we know that Bob has the Bell pair with $U^\dagger$ applied on qubit 2. Thus, if $U$ is one of the Pauli operators, Bob has one of the four Bell states, and his measurement will determine which one.

If you're looking for a more concrete mathematical statement, I suggest starting by finding exactly the structure of my diagrams and seeing within that the mathematics that you've already understood.


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