But in what substantial way do you think the results and what can be understood about QM are different?
Can’t they be equivalent but not identical?
Those are very good questions.
The delayed quantum erasure experiment and the delayed choice quantum eraser, both illustrate the same physical principles. The difference between the two experiments is how the ‘which path’ information is marked and that the pattern is observed depends on initially choosing one of a variety of different experimental set-ups in the first whereas the choice in the second is how the data is combined after the experiment has been performed.
This is one of the best summaries of quantum mechanics I have ever heard. And it’s really quite true. The world we perceive just does not match up with the quantum world. You have to accept that the math does; it expresses what normal human language and thought cannot.
I sort of agree up to a point,
It is very correct to think of the maths of being central importance. If you look at the postulates of QM they’re nearly all of them formulated mathematically. For example one postulate describes what happens when you combine two quantum mechanical systems together and references something called the “tensor product”. There’s no real interpretation for that except mathematical, though quantum entanglement comes directly and straightforwardly from this postulate.
Maths itself could be expressed purely in English without the need for symbols, but it would be extremely tedious and unwieldy and, beyond the very basics, understanding one mathematical definition requires you to understand several other more basic definitions.
However I would say the maths for basic quantum mechanics is not anywhere near as hard as many assume it is.
The way it is described, it sounds as if you can FORCE the result to be wrong.
They’re exactly equivalent. The detector on the p-path in the Walborn et al. version is equivalent to detectors 1 and 2 in the wiki version if the polarizer is placed in front of it, and thus, it doesn’t record which-path info, and equivalent to detectors 3 and 4 if there’s no polarizer and which-path info is present. The reason an interference pattern appears at the detector in the s-path is that the polarizer only permits about half of all photons to pass through, such that only half the photons arriving at the s-path detector are counted (because only they are in coincidence with counts in the p-path). If there were no coincidence counting, i.e. you were to just observer the photons incident on the detector in the s-path, you would never observe an interference pattern. You can see this by considering the ‘antifringe’ pattern that is generated when the polarizer is rotated by 90°: compared to the fringe pattern, it is slightly shifted, and effectively, it is the pattern generated when you reject the ‘other half’ of the photons. And adding fringe and antifringe patterns—counting all photons at s—gives again the interference-free distribution.
Oh, and thanks! (But really, I wouldn’t want anybody believing what I wrote just because of my perceived ‘authority’… It’s always better to ask if anything’s not clear.)
I’ve a question. In many of these experiments, theorists already know what to expect, and run the experiment in part just to demonstrate that reality really is as “spooky” as predicted.
Is that the case in the experiment of OP?
Yes. The behavior was exactly as predicted; anything else would have been surprising (the possibility of such a surprise being the reason one runs these experiments).
Or to quote Isaac Asimov yet again: The most exciting phrase to hear in science, the one that heralds new discoveries, is not “Eureka” but “That’s funny…”
In the OP’s link, there are two paragraphs that are at best misleading, that I think it’s important to understand:
This paragraph is true if the beam passing through both of the slits is x-polarized OR y-polarized OR left-circularly polarized OR right-circularly polarized. But in their setup, they have left-circularly polarized passing through one slit and right-circular through the other, destroying the interference pattern. They would have similarly not have had an interference pattern if they had x-polarized light passing through one slit and y-polarized light passing through the other.
The simple answer is that left-circular polarized photons won’t interfere with right-circular photons in that way. Nor will x-polarized photons interfere with y-polarized photons (x and y perpendicular).
Note that the authors Walborn et al in the paper Asympotically fat linked to understand this:
The polarization for the beam incident on the quarter wave plates could be at any angle, but’s helpful to consider four possibilities: x, y, x+y, and x-y (so every 45 degrees). Without the quarter wave plates, for any polarization, a fringe pattern would be observed, with the peaks and valleys in the same location. It doesn’t matter, then, if you know the polarization.
For x or y beams, after the quarter wave plate the two slits have orthogonal polarizations. For an x+y polarized beam, you get the fringing pattern, and for an x-y beam you get antifringing (see figures 4 and 5 in Walborn et all). The antifringing looks just like fringing, except that the locations of the peaks and valleys are swapped. Without the polarizer in beam p, you’d have all four cases equally likely, and since the fringing and anti-fringing have their peaks on the others valleys, you don’t see any fringing in their sum. When you put in the polarizer, you only have x+y (or x-y if you rotate it 90 degrees), so you do get the fringing.
You could also imagine measuring the polarization of beam p, and keeping track the s-beam pattern for all four cases separately. This ties in well with Half Man Half Wit’s conclusion above:
The “If only the counts at the detector D0 were analyzed” case corresponds to analyzing the pattern of the s-beam without including the polarization information. “brought into coincidence in order to observe the interference pattern” corresponds to looking at only one of the four polarization possibilities (or, in the actual experiment, adding the polarizer).
You have to be alive to be awarded the prize, so Schrödinger’s living cat is eligible, while the dead one is not.
I think you might be wrong about that. The point is that each photon interferes with ITSELF, not with other photons of different polarizations.
But even if we assume that were the case, I don’t think it would matter. I’m not sure I completely understand the optics but here are a couple of things maybe you would like to look at. The first is open access, the second you would probably have to get through a university library.
http://arxiv.org/pdf/0904.0204.pdf
This is correct. With the polarizer in place in the p beam to give x polarized (or y polarized) photons incident on the quarter wave plates, each photon is a superposition of left-circular through one slit and right circular through the other slit. Those two “components” of the photon are orthogonal to each other, so you don’t get the self-interference pattern.
x just means horizontally polarized and y just means vertically. Orthogonal just mean ‘at right angles’ and I don’t think that applies when comparing circular polarization to linear polarization - unless you can show me something that demonstrates otherwise.
Actually, what do you mean by “that”? One particular sentence? Every sentence in the entire post?
Orthogonal means much more than just “at right angles”. See especially Orthogonal Functions.
Even so, can you show me a cite for your position?
:dubious: Cite for what, exactly?
Cite that orthogonal means more than at right angles? Right there in my post that you quoted.
Cite that the wavefunctions going through each of the two slits with the quarter-wave plates in place are orthogonal? Right there in my post 52: