# Delayed choice entanglement swapping: why are Alice & Bob's measurements useless without Victor's?

Here is the article by Ma et al.:http://arxiv.org/abs/1203.4834

I have read many explanations on this site and others that emphasize that Victor's data is needed to make Alice and Bob's usable... Folks frequently mention: "that Victor's data is needed to sort the data from Alice and Bob's trials into four subsets..."

I'm not a quantum physics native, so I do not understand why Victor's data is needed. Can somebody please explain in very simple terms:

1. why you cannot deduce correlation/entanglement from Alice and Bob's data alone?
2. whether it would make any difference to the above question if you can repeat the experiment (and collect many more data points from Alice and Bob) either through multiple parallel set ups of the same apparatus, or multiple measurements in a single apparatus, but before Victor does his thing?
• In a properly set up experiment Alice and Bob aren't seeing anything in their statistics that is due to entanglement, they would not even know if they are participating in an experiment with an entangled state or not. The correlation in a simple entanglement experiment is a pure two-point correlation of completely independent measurements. Multiple parallel measurements (i.e. ensemble averages) won't give you any more information than playing sixteen tables at the same time in a casino will let you win at roulette. Jun 19, 2015 at 5:48
• Anybody want to provide an actual answer? @CuriousOne: You didn't answer either question. You either haven't even skimmed that article, or are saying that I'm stupid (or maybe both). Jun 20, 2015 at 16:17
• Another way of asking my question: In this table, why is it that comparing Alice and Bob columns alone cannot help? What do we learn from the Victor column? nature.com/nphys/journal/v8/n6/fig_tab/nphys2294_F1.html Jun 20, 2015 at 16:21
• Cornelius, I was simply trying to make your life easier by giving you the skinny about the phenomenology of entanglement. I am sorry that you are taking this personally. Now that we all know that you are smarter than us we don't really need to help you anymore, do we? Have a good life, my friend and enjoy the wonders of quantum mysticism. Jun 20, 2015 at 18:02
• Anybody willing to help with this question? Anybody with an actual answer? Thanks in advance. Oct 8, 2015 at 15:22

Alice is the top wire, Bob is the bottom wire. They each create an EPR pair $|00\rangle + |11\rangle$ via H+CNOT operations and give them to Victor (the center two wires). Victor then does a Bell basis measurement (by performing operations that un-make an EPR pair then measuring each wire).
The really important thing to realize is that, after that Bell basis measurement, Alice and Bob could be in one of many possible entangled states. Their qubits might be entangled to agree along the X axis, or to disagree. They also might agree or disagree along the Z axis. Concretely speaking, they could be in the state $|00\rangle + |11\rangle$, or in the state $|01\rangle + |10\rangle$, or in the state $|00\rangle - |11\rangle$, or in the state $|01\rangle - |10\rangle$.
But then there's still Victor's measurement results. They reveal which entangled case Alice and Bob ended up in. Victor knows their state's X-parity and its Z-parity. All Victor has to do is tell Alice and Bob "Hey, the Z-parity came out 'disagree' so one of you flip your qubit 180 degrees around the X axis to fix that". Do the same for the X-parity, and the problem is solved: Alice and Bob will definitely end up in the state $|00\rangle + |11\rangle$ after the corrections. That's what the last two controlled operations are doing.