I How indistinguishable are photons?

[Killtech]

've lost track of your various examples. But if you are talking about the AB + A'B' setup I commented on in #12: There is no difference in observable outcomes regardless of ordering of observations. That is true of all EPR entanglement setups, especially swapping setups.

No, i have reduced the structure i am querying about to the following two states in a similar 4 beam setup but a different source emitting a different initial state instead of two independent Bell pairs.

1. a 2-photon 4 beam entanglement with same polarizations:
   |ϕ⟩=|HA,HA′⟩+|HB,HB′⟩
   So this case is a mirror experimental setup with a mirror photon state - whichever path one photon chose (A or B), his mirror photon had to do the same but mirrored.
2. a 4-photon 4 beam Bell state
   |ψ⟩=|HA,HA′,VB,VB′⟩+|VA,VA′,HB,HB′⟩
   Here the distinction is that the HOM interferences will differ when also taking into account polarization depending on the loss of coherence.

But you'll need to read a little more about the following discussion to follow the context, specifically to understand my statement about observable outcomes - since you need to know what what observables are being observed. The issue is that within the formalism the observable in question is technically non-local with the way no-signaling theorems use this word (and therefore failing to commute with observables in either part of the space), albeit it remains is a simple measurement.

The question moved on how physical such state would be since just because the formalism does allow to write them down doesn't make them necessarily realistic.

 

[DrChinese]

1. a 4-photon 4 beam Bell state
|ψ⟩=|HA,HA′,VB,VB′⟩+|VA,VA′,HB,HB′⟩2. "...For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations."

1. A 4 photon Bell state (presumably photons A A' B B') does not produce polarization as you have tried to write. There are a lot more outcomes to consider, such as |VA,HA′,HB,VB′⟩. To be clear: if these 4 are entangled together, there is no requirement that any pair subset follow perfect anti-correlations. On the other hand, if any 2 follow perfect anti-correlations, they are prohibited from being entangled to other particles due to monogamy of entanglement.

2. And again, I'll ask you to retract your statement per my challenge in #30. Or produce an appropriate reference. If you don't, I will pass this to the moderators as being personal speculation that is not generally accepted.

 

[PeterDonis]

For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations.

@DrChinese has already raised a valid challenge for this claim of yours:

I'll ask you to retract your statement per my challenge in #30. Or produce an appropriate reference. If you don't, I will pass this to the moderators as being personal speculation that is not generally accepted.

The moderators have already seen it.

@Killtech, I am closing this thread. If you are unable to supply a reference to back up your claim quoted above, the thread will remain closed. If you do have a reference, please PM me a link and the moderators will review it.