bruce2g said:
I just want to thank Vanesch and all of the others who patiently contributed posts to clear up this question I had. I guess the short answer is that if the entanglement of the 'B' photons can reveal which slit the 'A' photons go through, then there will be no interference. If the 'B' photons do not reveal the which-path information, then there will be interference.
I think that's a fair summary, yes.
A slightly longer answer would note that you really need to do coincidence counting to be sure that you're dealing with entangled pairs, and that the geometry of the 'B' detector employed for this purpose can apparently also have an influence.
It seems pretty simple when you look at it this way. Thanks
again!
Yes, exactly. And you could even add yet another caveat: EVEN if you're using entangled pairs, and EVEN if the entanglement doesn't permit you *in principle* to deduce which-slit information using photon B (so that we are in the case where A can interfere), there might always be another, prozaic reason why we don't observe interference. The only thing we can say is that interference is potentially possible (in other words, that the entanglement is not forbidding an interference ; but maybe something else is screwing it up).
But the essential statement is indeed this: if the entanglement allows you, in principle, to use photon B to find out through which slit photon A went, in a specific interference setup, then this same entanglement will make it impossible for photon A to give rise to interference in this setup.
But to the person only looking at beam A (which might even ignore that beam A is entangled with something else), the photons of beam A do not look somehow "peculiar" in that they "refuse to interfere". He just sees beam A as a stochastic mixture of "photons in a mode that only go through slit 1" and "photons in a mode that only go through slit 2", and as such, he's not suprised not to see, from this mixture, any interference pattern.
this mixture is dictated by the reduced density matrix (reduced to beam A only) of the entangled state. The reason for that is that, if the entanglement allows you to find out "which slit" A went through, it means that the mode of A entangled with this tagging state for saying "slit 1" of photon B is a mode that only goes through slit 1, and the mode of A entangled with the tagging state for saying "slit 2" of photon B, is a mode that only goes through slit 2. As such, beam A, as seen alone, is a mixture of photons that only go through slit 1, or only go through slit 2. As such, it is not surprising that there is no interference from the mixture.
