@DrChinese, I don't even want to respond to most of what is in your posts #107, #108, and #109, because until you have read through all of the previous posts of mine that I referenced, I think discussion is premature. I have no problem with it taking some time for you to work through those posts, they took me a fair bit of time to write and writing them is of course going to be easier for me than working through them will be for you. Take all the time you need, I'll still be here.
That said, there are a few things that it might help to clarify now:
DrChinese said:
If there is decoherence upon measurement of Photon 1 (or at the BSM), which I have no particular objection to assuming for discussion purposes: then you are saying that the entangled state of Photon 2 is over.
Not in the MWI, no. In the MWI, Photon 2 is still entangled after the Photon 1 measurement: it's just that the entanglement is no longer with Photon 1, but with all of the degrees of freedom that got involved in the Photon 1 measurement (and the degrees of freedom in the environment that that entanglement spreads to). What
is true is that Photon 2 is no longer
maximally entangled with any one of those individual degrees of freedom. But the overall entanglement of Photon 1 is still there. Note that none of the states I wrote down in my previous posts are product states of Photon 2 with something else; all of them are entangled.
In a collapse interpretation, yes, measuring one of a pair of entangled particles ends the entanglement, because the collapse forces the state to be a product state. But there is no collapse in the MWI, and the math in my posts reflects that.
DrChinese said:
Under your line of thinking, order *should* matter.
This is simply wrong, and is one of the things I really wish you would postpone discussion of until you have worked your way through all of my previous posts.
DrChinese said:
Further, we know that the total entanglement cannot exceed a certain threshold due to Monogamy considerations, right?
I already addressed this in post #103.
DrChinese said:
in reality: the swap actually occurs at the BSM at the precise moment (or small time window) in which Photons 2 & 3 become indistinguishable, right? After all, that's an irreversible operation. And yet... what really happens is that they become entangled.
If we want to break down the "swap/no swap decision" process, which I didn't do in my earlier posts, here is what I gather from your earlier posts and the papers you referenced. I am treating the idealized version where if a swap is possible at all, it always happens, i.e., the sole relevant variable is the experimenter's decision.
(1) The experimenter makes a decision that determines whether or not a swap occurs. We model this in the math as there being some amplitude ##s## for a swap to occur, and a corresponding amplitude ##n## for no swap to occur, such that ##|s|^2 + |n|^2 = 1##. The operator that I called ##U_{S/N}## in my earlier posts can then be expressed as ##s U_S + n I##, where ##U_S## is the unitary swap operator and ##I## is the identity.
(2a) If the experimenter decides that a swap will occur, photons 2 & 3 arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and one photon is detected in each output arm of the BSM. This provides the "event ready" indication that a swap has taken place. The state after the swap is given by the unitary operator ##U_S## applied to the state before the swap.
(2b) If the experimenter decides that a swap will not occur, photons 2 & 3 do not arrive at the BSM within a short enough time window to be indistinguishable, they go through the BSM, and a detection occurs either in just one output arm of the BSM, or no detection occurs at all in either output arm of the BSM. For our purposes we combine all of those possibilities into the "no swap" result. The state is unchanged in this case because the operator involved is just the identity.
In the short time between photons 2 & 3 going through the BSM and the detections (if any) in the output arms of the BSM, yes, photons 2 & 3 will be entangled if there is a swap. Once the detections take place, that entanglement spreads to all the degrees of freedom involved in the detections, and their environment. If there is no swap, the previous entanglements of photons 2 & 3 get transferred to either the detector degrees of freedom (if the photons are detected) or directly to the environment (if they aren't detected and just decohere naturally because of their finite coherence time).
DrChinese said:
It does if we want the results to be irreversible. Without decoherence, we could imagine, for example, recombining photons, as in a Mach-Zehnder interferometer, and undoing the swap operation.
DrChinese said:
If deterministic, then it is fair game to assert that the ordering of events either local or nonlocal will change outcomes.
In general it might, but in the case under discussion, it doesn't. You can't just assert that the ordering
will change outcomes based on "determinism". You have to actually do the math and see. That's what I've done.
DrChinese said:
QM Collapse = MWI branching? What's the difference here?
MWI branching is unitary. Collapse is not.
For example, if I have two entangled photons in the singlet state, ##H_1 V_2 - V_1 H_2##, and I measure Photon 1, collapse says the state becomes either ##\bar{H}_1 V_2## or ##\bar{V}_1 H_2##; those are both non-unitary transformations from the original state. The MWI says the state becomes ##\bar{H}_1 V_2 - \bar{V}_1 H_2##, which is a unitary transformation from the original state (the bars just mean spreading entanglement among the degrees of freedom in the Photon 1 detector and its environment). Those are different states. They are indistinguishable experimentally because the bars, indicating decoherence, mean that you can't interfere the terms any longer, so there is no way, for example, to build a Mach-Zehnder interferometer that undoes the Photon 1 measurement and allows us to distinguish the MWI state from either of the two collapse states.