Problem involving entanglement swapping

Question

I'm trying to solve a problem. I have my own approach to it, as the one in the textbook seems overly complicated. I tried using my own approach, but I'm not sure if what I found was a coincidence or not.

Here's the problem (it's in French, I did my best to translate):

Alice and Bob are separated by a distance of $2L$. Suppose that Charlie is in the middle, i.e. he's at a distance $L$ from Alice and at a distance $L$ from Bob. Charlie produces two entangled pairs in the state $\left| B_{00} \right>_{12}$ and $\left| B_{00} \right>_{34}$. Therefore, there are 4 bits called $1,2,3,4$. Charlie keeps in his lab the particles $2$ and $3$ and sends particles $1$ and $4$ to Alice and Bob. Now, Charlie makes a measure in the Bell basis on particles $2$ and $3$ that he kept in his lab. Which are the possible results (with probabilities) of the measure in Charlie's Lab?

Here's my approach

Before the measure, we know that the initial state is : $$\left| \psi \right> = \left| B_{00} \right>_{12} \otimes \left| B_{00} \right>_{34}$$ $$=\frac{1}{2} \left( \left| 0 \right>_{1} \otimes \left| 00 \right>_{23} \otimes \left| 0 \right>_{4} + \left| 0 \right>_{1} \otimes \left| 01 \right>_{23} \otimes \left| 1 \right>_{4} + \left| 1 \right>_{1} \otimes \left| 10 \right>_{23} \otimes \left| 0 \right>_{4} + \left| 1 \right>_{1} \otimes \left| 11 \right>_{23} \otimes \left| 1 \right>_{4} \right)$$

Now, given that I know that $$\left| 00 \right>_{23} = \frac{1}{\sqrt{2}}(\left| B_{00} \right>_{23} + \left| B_{10} \right>_{23})$$, $$\left| 01 \right>_{23} = \frac{1}{\sqrt{2}}(\left| B_{01} \right>_{23} + \left| B_{11} \right>_{23})$$, and so on for the rest. I was able to rewrite the initial state as follows (although I'm not sure I can do this, given that I'm not sure I can do this):$$ \left| \psi \right> =\frac{1}{2\sqrt{2}} \left| B_{00} \right>_{23} \otimes (\left| 00 \right>_{14} + \left| 11 \right>_{14}) + \frac{1}{2\sqrt{2}} \left| B_{10} \right>_{23} \otimes (\left| 00 \right>_{14} - \left| 11 \right>_{14}) + \frac{1}{2\sqrt{2}} \left| B_{01} \right>_{23} \otimes (\left| 01 \right>_{14} + \left| 10 \right>_{14}) + \frac{1}{2\sqrt{2}} \left| B_{11} \right>_{23} \otimes (\left| 01 \right>_{14} - \left| 10 \right>_{14})$$ which reduces to: $$\left| \psi \right> =\frac{1}{2} \left| B_{00} \right>_{23} \otimes \left| B_{00} \right>_{14} + \frac{1}{2} \left| B_{01} \right>_{23} \otimes \left| B_{01} \right>_{14} + \frac{1}{2} \left| B_{10} \right>_{23} \otimes \left| B_{10} \right>_{14} + \frac{1}{2} \left| B_{11} \right>_{23} \otimes \left| B_{11} \right>_{14}$$

So, each state in each term of the sum is a possible state in Charlie's Lab after the measurement. In other words, measurements states are: $\left| B_{00} \right>_{23} \otimes \left| B_{00} \right>_{14}$ $\left| B_{01} \right>_{23} \otimes \left| B_{01} \right>_{14}$ $\left| B_{10} \right>_{23} \otimes \left| B_{10} \right>_{14}$ $\frac{1}{2} \left| B_{11} \right>_{23} \otimes \left| B_{11} \right>_{14}$

Therefore, the probability of measuring each state is $p=\frac{1}{4}$. I found this without using Born's rule. I used the fact that in another state, such as $\left| \phi \right>=\alpha\left| 0 \right> + \beta\left| 1 \right>$, the probability of measuring $\left| 0 \right>$ is $|\alpha|^2$(and the idea is the same for $\left| 1 \right>$). I used this fact, but I'm not sure why I would be allowed to do this. Now that I'm writing this question, I realize that perhaps, given that the states after the measure form a basis for our space, writing the initial state as a sum of basis vectors, I am allowed to do this. Would you say that I'm correct?

Answer 1

Judging from your computations, the Bell states are $$|B_{00}\rangle = |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ $$|B_{10}\rangle = |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$$ $$|B_{01}\rangle = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$$ $$|B_{11}\rangle = |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$$ I don't like changing order of qubits in tensor products, but it does affect the final result which I believe is correct: Charlie measurement outcomes are equally likely, and more: after the measurement Charlie knows the Bell state which is shared by Alice and Bob.

Answer 2

If you are interested only in the measurements of the system that comprises only 2nd and 3rd qubits, then you could compute the reduced density matrix $\rho_{23}$ on those qubits from the total density matrix $\rho_{1234}=|\psi\rangle \langle\psi|$, i.e. $$ \rho_{23} = \text{Tr}_{14}(\rho_{1234}) $$ Density matrix carry all information that is need to predict measurements outcomes.
Since in this case $$\rho_{23} = \frac{1}{4}I$$ the Bell measurement outcomes must be equiprobable.

To compute partial trace $\text{Tr}_{14}(\rho_{1234})$ note that the state $\rho_{1234}$ is disentangled relative to the decomposition $H_{12} \otimes H_{34}$, i.e. $\rho_{1234} = \rho_{12} \otimes \rho_{34}$. So it must be $$\text{Tr}_{14}(\rho_{1234}) = \text{Tr}_{14}(\rho_{12} \otimes \rho_{34}) = \text{Tr}_{1}(\rho_{12}) \otimes \text{Tr}_{4}(\rho_{34}) = \frac{1}{2}I\otimes\frac{1}{2}I = \frac{1}{4}I$$ Though, of course, partial trace could be computed directly.

As for your approach, it is correct, and also answers the question in what state qubits 1 and 4 will be after the Bell measurement of 2 and 3.