Franson interferometer: Entanglement?

[greypilgrim]

Hi.

As far as I understand the Franson interferometer, the photons are in an entangled state like
$$\left|\Psi\right\rangle=\frac{1}{\sqrt{2}}\left(\left|\text{short}\right\rangle\left|\text{short}\right\rangle+\left|\text{long}\right\rangle\left|\text{long}\right\rangle\right)$$
if the setup is symmetrical.

This kind of entanglement is new to me. For example entangled polarization or spins are created at the source, while above state is written in a basis that depends on the setup itself, which might be far away from the source. What happens if we remove the setup altogether after the photons have been emitted, but haven't entered the interferometer yet? Above expansion wouldn't make sense anymore. What is now entangled about the photons?

A second question: I read that there will be events where one photons takes the short and the other the long path. As this can clearly be detected by comparing the arrival times, it doesn't create interference, those measurements are excluded in postselection. Apparently this discards half of all events. But from the above state I don't see how this could even happen. Or does the source also emit different states that lead to this behaviour?

EDIT: Well now I'm completely confused. If we discard all events where the photons don't arrive at the same time, the ones remaining must obviously be short/short and long/long. What's quantum about this?

 

[jfizzix]

1.) If you remove the experiment, then the quantum state of the light will be different because different things would be happening to it. The entanglement that the experiment measures is still there (frequency-time entanglement), and the Franson interferometer is one way to measure a small part of that entanglement. It's much more difficult to measure the frequency and time correlations of the photon pairs directly to a high enough precision to prove that there's frequency-time entanglement, but the Franson interferometer allows us to witness at least some of the entanglement efficiently.

2.) The above state is not the full state of the photon pairs in the interferometer. By only looking at coincidence counts, you select only the subset of biphotons (photon pairs) that take either both the long paths, or both the short paths. The quantum state of the biphoton, relative to this post-selection, is the state given above.

 

[greypilgrim]

2.) The above state is not the full state of the photon pairs in the interferometer. By only looking at coincidence counts, you select only the subset of biphotons (photon pairs) that take either both the long paths, or both the short paths. The quantum state of the biphoton, relative to this post-selection, is the state given above.

But what role does the entanglement play in this case? Wouldn't a separable lead to the same statistics? Isn't it a triviality that two objects, leaving a source simultaneously at the same speed must have traveled the same distance when they're detected at the same time?

 

[jfizzix]

But what role does the entanglement play in this case? Wouldn't a separable lead to the same statistics? Isn't it a triviality that two objects, leaving a source simultaneously at the same speed must have traveled the same distance when they're detected at the same time?

Actually, no. If one were just producing pairs of photons at the same time that were otherwise uncorrelated, the resulting post-selected quantum state would be a statistical mixture of $|short\rangle|short\rangle$ and $|long\rangle|long\rangle$ and not a superposition of these two states. Because of this, you would not see an interference pattern in the coincidence counts as you change the length of the arms.