vanesch said:
I will now address your criticisms verbally, but have a look at my small calculation in the other post, please ; it is the essence of what I had in mind.
I guess I'm more comfortable discussing it than dissecting the math.
vanesch said:
This shift comes about, if you trace it back, to a difference in optical pathlengths in the polarizing beamsplitter, and this shift is utterly important.
Schneibster said:
I see no reason to conclude that the optical path lengths are different between left and right (or, if you prefer, A and B); while the path lengths for the idler and signal photons might be (and in the realization are, by 8ns) different, the path lengths for left and right for either the idler or signal photons need not be so.
vanesch said:
I probably expressed myself badly: I was pointing to a half-wavelength pathlength difference due to the beamsplitter (sign difference between transmitted and reflected beam).
I guess I'm not quite clear on what component you were talking about with the "polarizing beamsplitter." I thought you meant the combination of the β-BBO crystal and the Glan-Thompson prism. I don't see any other component that could be described as a "polarizing beam splitter" in the optical path between the slits and the D0 signal photon detection area, though, so I'm still not clear on exactly what "half-wavelength pathlength difference" you are referring to.
I can see a half-wavelength pathlength difference between either signal path and its corresponding idler path; that is introduced by the β-BBO crystal, and accounts for the phase orthogonality between the signal photons and the idler photons (and thus the ability to sort them in a Glan-Thompson prism). But I cannot see any way that phase difference could make a difference in the positions of the interference patterns; because I also cannot see any difference in the idler photon paths between left and right (or, if you prefer, A and B).
I guess I want you to explain in precise terms what exact component you believe is introducing this difference, and how the introduced phase difference accounts for the "phase shift" between the two interference patterns.
vanesch said:
If the system is 100% efficient,
Schneibster said:
It is not and cannot ever be. D0 is driven back and forth across the field of the interference with a stepper motor. Think about it.
vanesch said:
That's an experimental detail: think of a position-sensitive detector which can measure the impact position of the photon at D0.
What I meant was: D1, D2, D3 (and D4) cover all cases, the second photon has to end up in one of those 4 detectors.
But because not every photon that is measured at D0 can be correlated with one measured at one of D1-D4, there is no reason to believe that there would be any visible interference anyway.
I guess I'm subconsciously making reference to a different state of the experiment (and I guess it's fair for me to do so, because it was explicitly mentioned in my OP): what if the BSA and BSB (which send idlers either to D1/D2 or D3/D4, randomly) were replaced with mirrors that always sent the idlers to the eraser D1/D2? Even in this case, the device cannot be anything like efficient, simply because we cannot detect corresponding signal photons for every idler; and conversely, even if we could, by using a CCD sensor with a maximum efficiency at 500nm of about 96%, unless D1-D4 have 100% quantum efficiency, we cannot guarantee to detect every corresponding idler photon for every signal photon that is detected. I'd estimate about 90% maximum efficiency, if we used state-of-the-art CCDs everywhere.
Your conclusion was, "It is only when the coincidence information is used to do subsample selection that you see subsamples with or without fringes." I have to agree with that; but I'm not sure why you felt you needed to precede it with, "If the system is 100% efficient, then every event is classified in one of the 3 subsamples, and so at D0, without coincidence, you don't have any interference pattern." I agree with it without agreeing with (what appeared to be) the premise; however, I also don't think it really
was a proper premise, and I still don't see why it is important that the interference is a subset. And last but not least, after the experimental modification I suggested with mirrors, there are only two possible alternatives, not three: no idler photons go to D3 or D4, so there are no non-interfering signal photons.
vanesch said:
The entire experiment is in fact based upon what I'd call a misunderstanding of quantum mechanics. It is assumed from the start that 'atom A OR atom B' emits a photon. But that's denying superposition !
Schneibster said:
In the non-interference case, that is exactly true. How is it a misunderstanding of quantum mechanics to state that if you can recover the which-path information, you see no interference?
vanesch said:
That's a correct statement. The less "correct" statement is that you have "information which you erase"... You only measure correlations, which themselves are the result of (superpositions of) product states.
OK, then explain the lack of visible interference in the D0 signal detection area when the mirrors are turned to send all the idler photons to D1 and D2. You can't see interference even then- without the correlation. But under what circumstances do you not see interference of two waves?
Only when you have which-path information. So I have to dispute your claim that there is no "information which you erase." It appears to me that in this experiment, we have disrupted the photon paths through obtaining which-path information and destroyed the interference- and that we get the interference back when we destroy the which-path information. But in order to protect causality, we cannot see that interference without looking at the correlation between the idler photon detection events and the signal photon detection events. Thus, we preserve both of:
1. Quantum mechanical waves always interfere, unless there is which-path information.
2. Obtaining which-path information always destroys quantum mechanical wave interference.
Both must be true; if there were no interference recoverable, then the first would be rendered untrue, and it would call everything all the way back to Young's original dual-slit experiment into question. If there were any way to see interference when which-path information had been obtained, then causality would be macroscopically violated.
vanesch said:
It is what I call "overselling". The only true information that is available is the correlations at the end of the experiment. I don't know what it means to have potentially available information which is then erased.
Then you haven't thought about the experiment when all the idler photons are sent to the eraser. Think about it again that way. Then you'll begin to see the very narrow tightrope that nature has to walk for this experiment to work exactly the way it does; causality must not be violated, but the wave nature must be visible whenever it should be seen.
vanesch said:
The incoming pump photon is put in an entangled state:
half goes to A, half goes to B.
Schneibster said:
Now, this is a misunderstanding of quantum mechanics. First, a single photon cannot be entangled with itself. Second, this is an all-or-nothing interaction here, and there are only three alternatives: the whole photon goes to A and generates an entangled pair, the whole photon goes to B and generates an entangled pair, or the whole photon cannot be shown to have gone to either A or B, and generates an entangled pair that cannot be shown to have come from either A or B.
vanesch said:
But that's denying the essence of quantum theory !
First, what you specifically said is shown above. Second, I still don't see any entangled state. Nothing you said following this has convinced me that this has anything to do with entanglement. Entanglement and superposition are not the same; they are different. Had you said that the incoming pump photon is superposed into an eigenstate whose eigenvalues are it goes to A or it goes to B, with 50% probability for each alternative, then I might have agreed with you. But I would also have pointed out that just because it is in such an eigenstate does not mean that that eigenstate ever has to collapse into a state, and realize one of the eigenvalues; and in fact, if it does not so collapse, if neither eigenvalue is realized, then we see interference! And
that, my friend, is the
essence of quantum mechanics!
Feynman once said that you can see everything there is to know about quantum mechanics in the dual-slit experiment with single electrons. I tend to agree with him strongly. He of course didn't mean that all the math is there; what he meant is that all the most important concepts are there.
vanesch said:
The exciting photon (pump) was in a superposition of positions at A and B. This superposition is maintained, and the excitations at A and B are then in an identical superposition. The whole story of "which way" information, which denies the superposition of these states, is what gives rise to all these "erasure paradoxes". Have a look at my very simple calculation.
Your third case is nothing else but the interference terms due to this superposition which you deny.
You apparently did not understand the implications of what I said. I do not deny that such a superposition exists; I was referring to it. It is patently obvious that it does. What I deny is that a superposition represents an entangled state
in this case. The entanglement does not occur until after the initial superposition.
What I was trying to indicate was exactly that superposition; my first case is the first eigenvalue of the eigenstate, my second case is the second eigenvalue of the eigenstate, and my third case is the uncollapsed eigenstate, which may or may not (depending on whether you erase or read the which-path information) collapse into a state of one of the two eigenvalues. If it does collapse, then the corresponding signal photon will not be part of an interference pattern; in other words, its probability distribution will be a smooth function of the distance from the center of the D0 detection area. On the other hand, if the eigenstate is never collapsed into a state, then the corresponding signal photon will be a member of one of two classes of photons that form interference patterns in the D0 detection area.
Now, represent me the possibility that that eigenstate will not collapse into a state of one of the eigenvalues. That is what is missing from your mathematics above. And that's why I don't critique them directly. I don't think you're applying them right.
vanesch said:
You missed what I said: a single photon can of course be in a superposition of 2 position states, A and B. That doesn't mean that I think that it is sliced in 2 half photons of course. It is this position superposition which is then propagated and gives rise to all the results.
No, actually, I have quoted exactly what you said above: "The incoming pump photon is put in an entangled state: half goes to A, half goes to B." And you didn't say "superposition;" you said "entangled." The two are different. Sometimes simultaneous, yes; but different. The "half" part I'll leave to anyone who cares to read it.
Schneibster said:
And the remarkable thing about this experiment is not that we cannot see interference directly; it is that we can recover interference at all by any means whatsoever.
vanesch said:
That is only remarkable if you think that the pump photon cannot be in a superposed state, half at A and half at B.
And again we have an apparent misunderstanding of the meaning of "superposition." It is
not "half at A and half at B;" it is at both, and neither. It
has a 50% probability of having been at A, and it
has a 50% probability of having been at B. But there is a third alternative: it was at
both, or
neither, and we see interference. And you have not addressed this third alternative. When your eigenstate collapses due to a measurement, then there will be no interference, whichever way the collapse occurs, because the eivenvalue of the state, whichever it is, will preclude seeing interference.
If and only if that eigenstate never collapses into a state will we see interference.
vanesch said:
But if you do accept that, by simply applying quantum mechanics (and no projections) nicely all the way, a lot of fuzz falls away.
Personally, I think that seeing interference is not "fuzz;" but you are welcome to your opinion.
vanesch said:
No, the projection postulate must be applied when you calculate your final quantities, which are here correlations between D0 and D1 or D2 or D3...
Sometimes, you can get away with intermediate use of the projection postulate, but not in these entangled cases.
I don't think that you can prove that, and I assert that the projection postulate can be used
any time there is a measurement. And producing the idler photons is a measurement; it measures which-path information. That is the reason that the interference cannot be seen. As I have said repeatedly, what I find remarkable, and a beautiful illustration of quantum logic, is that there is any way at all to recover the interference.
-Da Schneib