[IX-103]

I'm not fond of superdeterminism since it's not that useful for making predictions. Any purely deterministic model has implications for free will, so that doesn't seem to be a legitimate criticism.

Actually I would like to know more about provable violations if Bell's theorem as I am somewhat attached to local determinism and haven't seen an experiment that I would consider convincing. I mean the theories behind the experiments are sound, but I'm not sure they're actually measuring what they think they are measuring due to limitations in the experiment setup -- in order to prove a violation of locality your system cannot be in a cyclostationary equilibrium.

In such an equilibrium the system state effectively becomes a standing wave so you risk measuring an effect that was actually a result of a previous cycle and mistakenly interpret it as being a result of the current cycle -- implying a violation in locality because the "cause" was outside of the light cone of the effect. Note that this is analogous to confusing the group and phase velocities of a radio wave (https://www.quora.com/What-is-the-difference-between-phase-v...).

[pdonis]

> In such an equilibrium the system state effectively becomes a standing wave so you risk measuring an effect that was actually a result of a previous cycle and mistakenly interpret it as being a result of the current cycle

I don't know where you're getting this from, but it doesn't describe quantum systems on which Bell inequality violations have been experimentally confirmed (such as photon pairs from parametric down conversion).

The only "loophole" that has not been completely closed at this point is that we don't have 100% efficient detectors, but we have detectors that are well over 90% efficient so the claim that somehow all the stuff that will "fix" the Bell inequality violations is hiding in the small percentage of photons not being detected isn't very compelling.

[IX-103]

In short the problem I'm addressing is that the interpretation of the experiment assumes that the system is memoryless, so that the only thing being measured is the interaction with the particles being measured.

In the experiments generating the photon pairs from parametric downconversion, for example, does the entire system start up, send 1 photon which gets split into the entangled photon pairs which then go to the detectors -- with no other photons generated?

If there is a warm-up period for the equipment or other photons are emitted or absorbed then there is the potential for memory effects that could interfere with the measurements.

For instance if we treat light as a wave then the cosine correlation with angle we see in the basic "two entangled photons with polarizing lenses experiment" is exactly what we would expect to see. The difficulty is simply resolving this with the particle nature of photons. If the experimental system has memory then it could easily have the phase of the effective wave or some other function of the history of photons encoded in the state of the system.

There are probably some ways to compensate for these memory effects and demonstrate their (non)existence, but I am not a physicist.

[pdonis]

> the problem I'm addressing is that the interpretation of the experiment assumes that the system is memoryless

That's easy to verify by testing the various components--parametric down conversion, prisms, beam splitters, etc.--and showing that if you shine repeated photons on them from the same source, prepared in the same state, they all come out in the same state, or more generally give the same results. All of the optical components involved in these experiments have been tested in this way: if they had failed such tests, they wouldn't be used in experiments because we wouldn't be able to be confident in their behavior.

> n the experiments generating the photon pairs from parametric downconversion, for example, does the entire system start up, send 1 photon which gets split into the entangled photon pairs which then go to the detectors -- with no other photons generated?

For current photon sources, it's impossible to control exactly when they emit a photon. The sources are so inefficient (in terms of converting input energy into photons that are useful for the experiment) that they end up emitting photons slowly enough that only one at a time is inside the apparatus. However, a typical experiment does not use just one photon. It has to take data from many photons because the results are statistical, so you need enough runs to do statistics.

> If the experimental system has memory then it could easily have the phase of the effective wave or some other function of the history of photons encoded in the state of the system.

We know how to design systems that do this: they're called "detectors" and "computers that store data". But such systems have to be carefully designed to do those jobs. Optical components like prisms and beam splitters are not designed to do that: they're designed to do exactly the opposite, to act the same way on every photon that comes into them in the same input state. As I noted above, those components have been extensively tested to make sure they do in fact do that; if they didn't, they wouldn't be used in experiments.

[IX-103]

>if you shine repeated photons on them from the same source, prepared in the same state, they all come out in the same state, or more generally give the same results.

Those kinds of measurements would violate the uncertainty principle. You can't know the complete state going in to the system or the complete state going out. You can run some tests and justify other assumptions based accepted theories. We generally have a good idea what happens when lots of photons pass through these components. We have some ideas of what happens to single photons (in a statistical sense), but the fundamental question we are investigating is whether there even is a local deterministic description of what happens to single photons passing through the component.

> The sources are so inefficient (in terms of converting input energy into photons that are useful for the experiment)... a typical experiment does not use just one photon. It has to take data from many photons

I was aware of that and it's part of my criticism. If the emitter were to only emit useable photons when it's "in the right state", what stops the "right state" for emitting photons to become correlated with the polarizers?

There are a bunch of "unusable" photons bouncing around interacting with everything and transporting global state. Then there are the "usable" photons that get reflected, absorbed and re-emitted by components of the test bed. Any time they interact with anything they modify the state of whatever they touch. What happens to the photons that reflect off of the polarizers and travel back into the emitter?

If a photon bounces off of a mirror it had to have 1. transfered momentum to whatever it hit, and 2. induced a sufficiently strong opposing electromagnetic field to cause the photon to be reflected or re-emitted. While these are tiny effects they are roughly the same order as the effects that caused the photon to be reflected in the first place, and they all require a change in state of the mirror so that momentum is conserved and Maxwell's laws are not violated (my guess would be that this could cause shifts in electron orbital or proton spin orientation, but that's a bit beyond me).

[pdonis]

> Those kinds of measurements would violate the uncertainty principle.

No, they don't. The uncertainty principle places limits on measurements of non-commuting observables on the same system. We are not talking about that here. See below.

> You can't know the complete state going in to the system

Sure you can: just prepare the system in a known state. For example, pass your photon through a vertically oriented polarizing filter: if it comes through, it must be vertically polarized, so you have complete knowledge of its polarization state. (You might have to try multiple photons to get one that passes through: that's why photon sources in these experiments are often inefficient.)

> or the complete state going out

Sure you can: you measure it. For example, you pass the vertically polarized photon that just came through your vertical polarization filter through a beam splitter, and you have detectors at each output of the beam splitter. Exactly one detector will fire for each photon.

> If the emitter were to only emit useable photons when it's "in the right state", what stops the "right state" for emitting photons to become correlated with the polarizers?

> There are a bunch of "unusable" photons bouncing around interacting with everything and transporting global state.

It looks like you don't have a good understanding of how the "emitter" works. What you are calling the "emitter" is really a filter, like the vertical polarizer described above: it throws away the photons coming from a source (like a laser) that don't meet a particular requirement (like vertical polarization). The thrown away photons are either absorbed (as in the case of the polarizer) or they just pass through the apparatus altogether and fly away (as in the case of parametric down conversion, for example: only a small percentage of the laser photons will be down converted, the rest just fly away and are gone).

In no case are the photons not used kept "bouncing around". They're gone. And the photons in the "right" state are just the ones that make it through the filter and are therefore in a known state when they come out, because that's how the filter works: the filter is uncorrelated with what's inside the experiment because, again, that's how the filter works (and it is tested to make sure it works that way).

> What happens to the photons that reflect off of the polarizers and travel back into the emitter?

There aren't any. See above.

> If a photon bounces off of a mirror it had to have 1. transfered momentum to whatever it hit, and 2. induced a sufficiently strong opposing electromagnetic field to cause the photon to be reflected or re-emitted.

1. Yes, but in these experiments the mirror is fixed to the Earth, so the momentum is transferred to the Earth, which means it's effectively gone. The entire Earth is not going to have a "memory" that can become correlated with the rest of the experiment.

2. No. You are thinking of it classically, but we are not talking about a classical process.