# Determining the quantum secret

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?

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: So, this is a repeating trick: 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.