# Quantum entanglement between 3 person

(This was a homework question, the professor gave the answer but not the actual work to get there, as it's considered very easy and purely 'work', but I can't figure it out.)

Alice shares a $|\Phi^+\rangle$ with Bob and another one with Charlie. What happens if Alice does a bell measurement on her two qubits? More precisely, in which entangled states is the pair of qubits owned by Bob and Charlie, in regards to Alice's result?

So basically, after creating the two $|\Phi^+\rangle$ states, Alice does a CNOT on her qubits followed by a Hadamard on the first and then measurement on both.

I know Bob and Charlie now are entangled in a Bell states, and I assume they get a $|\Phi^+\rangle$ if Alice measured 0 and 0, and a $|\Psi^+\rangle$ if she measured 0 and 1, a $|\Psi^-\rangle$ for 1 and 0, and finally a $|\Phi^-\rangle$ for 1 and 1.

My problem is : if she measures 0 and 0, doesn't that collapse the system into a simple $|00\rangle$ for Bob and Charlie? The math of the Bell measurement on Alice's qubits confuses me. How can I do a CNOT on her two qubits since they are entangled with Bob and Charlie? Hopefully I'm clear enough, this is my first question and my quantum knowledge (and latex) is sadly quite limited.

Thank you for any hints! I definitely do not need the actual states they end up with, as I simply want to know how to do the work to get there :)

• Do you know about teleportation? Nov 10 '16 at 18:33

Why would Alice measuring 0 and 0 collapse the remaining state into $|00\rangle$? She didn't measure the individual qubits, she measured how they relate to each. Bob and Charlie's qubits weren't measured to be zero, they were measured to agree.

Your guess that each measurement result gives you a different entangled state is correct. See this example circuit:

This process actually has a name: "entanglement swapping". We start with a situation where Alice is entangled with two unrelated parties:

A--B

A--C


Then Alice does her parity measurement thing and tells the other two the corrective operations to apply and we've "swapped" to this situation:

A  B
|  |
A  C


(Okay, there's not really A-A entanglement recovered from the protocol in a useful sense, but it's trivial to produce local entanglement and makes things look nice and symmetric.)

Another way to think about what's happening is that Alice is teleporting to C the qubit entangled with B. Or vice versa.

The way to compute these results is... basically, just do the math. Use the tensor product to expand gates so they apply to the 4-qubit system. Hit the state with each column of gates. Do the measurements. Look at the posterior state for each measurement outcome. Check that the given corrective operations reunite all the posterior states into a single entangled state.

• Ok thank you, it gives be better understanding of the whole. However I find myself still struggling at the maths, more precisely, after the CNOT on Alice's two qubits I arrive at the following : 1/2 ($|0000\rangle$ + $|0011\rangle$ + $|1110\rangle$ + $|1101\rangle$ ) (with Alice's qubits being the first and third.) Assuming this is right, then my confusion would come from the partial measurement. If I'm wrong here, well, any pointers? Thanks again!
– gasg
Nov 11 '16 at 5:28
• @gasg That looks correct to me. Nov 11 '16 at 11:46