As in, are there some parts of physics that aren’t as clear-cut as they usually are? If so, what are they?

  • FlowVoid@midwest.social
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    1 year ago

    There are even plenty of questions, like the delayed-choice quantum eraser, that have already been solved

    No, it has not been solved. At least not solved to the satisfaction of many physicists.

    In one respect, there is nothing to solve. Everyone agrees on what you would observe in this experiment. The observations agree with what quantum equations predict. So you could stop there, and there would be no problem.

    The problem arises when physicists want to assign meaning to quantum equations, to develop a human intuition. But so far every attempt to do so is flawed.

    For example, the quantum eraser experiment produces results that are counterintuitive to one interpretation of quantum mechanics. Sabine’s “solution” is to use a different interpretation instead. But her interpretation introduces so many counterintuitive results for other experiments that most physicists still prefer the interpretation that can’t explain the quantum eraser. Which is why they still think about it.

    In the end, choosing a particular interpretation amounts to choosing not if, but how QM will violate ordinary intuition. Sabine doesn’t actually solve this fundamental problem in her video. And since QM predictions are the same regardless of the interpretation, there is no correct choice.

    • TauZero@mander.xyz
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      1 year ago

      Have we watched the same Sabine video? Delayed choice quantum eraser has nothing to do with interpretations of quantum mechanics, at least in so far as every interpretation (Copenhagen, de Broglie-Bohm, Many-Worlds) predicts the same outcome, which is also the one observed. The “solution” to DCQEE is a matter of simple accounting. And every single popular science DCQEE video GETS IT WRONG. The omission is so reckless it borders on malicious IMO.

      For example, in that PBS video linked in this very thread, what does the host say at 7:07?

      If we only look at photons whose twins end up at detectors C or D, we do see an interference pattern. It looks like the simple act of scrambling the which-way information retroactively [makes the interference pattern appear].

      This is NOT WHAT THE PAPER SAYS OR SHOWS! On page 4 it is clear that figure R01 is the joint detection rate between screen and detector C-only! (Screen = D0, C = D1, D = D2, A = D3, B omitted). If you look at photons whose twins end up at detectors C inclusive-OR D, you DO NOT SEE AN INTERFERENCE PATTERN. (The paper doesn’t show that figure, you have to add figures R01 and R02 together yourself, and the peaks of one fill the troughs of the other because they are offset by phase π.) You get only 2 big peaks in total, just like in the standard which-way double slit experiment. The 2 peaks do NOT change retroactively no matter what choices you make! You NEED the information of whether detector C or D got activated to account which group (R01 or R02) to assign your detection event to! Only after you do the accounting can you see the two overlapping interference patterns within the data you already have and which itself does not change. If you consumed your twin photon at detector A or B to record which-way information, you cannot do the accounting! You only get one peak or the other (figure R03).

      It’s a very tiny difference between lexical “OR” and inclusive “OR”, but in this case it makes ALL the difference. For years I was mystified by the DCQEE and how it exposes the ability of retrocausality, and turns out every single video simply lied to me.

      • FlowVoid@midwest.social
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        1 year ago

        Right, but in order to get the observed effect at D1 or D2 there must be interaction/interference between a wave from mirror A and a wave from mirror B (because otherwise why would D1 and D2 behave differently from D3 and D4?).

        And that’s a problem for some interpretations of QM. Because when one of the entangled photons strikes the screen, its waveform is considered to have “collapsed”. Which means the waveform of the other entangled photon, still in flight, must also instantly “collapse”. Which means the photon still in flight can be reflected from mirror A or mirror B, but not both. Which means no interaction is possible at D1 or D2.

        • TauZero@mander.xyz
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          1 year ago

          It’s not a problem for Copenhagen if that’s the interpretation you are referring to. Yes, the first photon “collapses” when in strikes the screen, but it still went through both slits. Even in Copenhagen both slit paths are taken at once, the photon doesn’t collapse when it goes through the slit, it collapses later. When the first photon hits the screen and collapses, that doesn’t mean its twin photon collapses too. Where would it even collapse to, one path or the other? Why? The first photon didn’t take only one path! The twin photon is still in flight and still in superposition, taking both paths, and reflecting off both mirrors.

          • FlowVoid@midwest.social
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            1 year ago

            When the first photon hits the screen and collapses, that doesn’t mean its twin photon collapses too.

            Yes, it does. By definition, entangled particles are described by a single wave function. If the wave function collapses, it has to collapse for both of them.

            So for example, an entangled pair of electrons can have a superposition of up and down spin before either one is measured. But if you detect the spin of one electron as up, then you immediately know that the spin of the second electron must be down. And if the second electron must be down then it is no longer in superposition, i.e. its wave function has also collapsed.

            • TauZero@mander.xyz
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              1 year ago

              Ok, I thought about it some more, and I want to make a correction to my description! The twin photon does collapse, but it doesn’t collapse to a single point or a single path. It collapses to a different superposition, a subset of its original wavefunction.

              I understand it is an option even under Copenhagen. So in your two-electron example, where your have 1/sqrt(2)(|z+z-> + |z-z+>), when you measure your first electron, what if instead of measuring it along z+ you measure it along z+45°? It collapses into up or down along that axis (let’s say up), and the entangled second electron collapses too, but it doesn’t collapse into z-135°! The second electron collapses into a superposition of (I think) 1/2 |z+> + sqrt(3)/2 |z-> . I.e. when you measure the second electron along z+, you get 25% up and 75% down. The second electron is correlated to the first, but it is no longer the exact opposite to the first, because the axis you measured the first at was not z+ but inclined to it. There is exists no axis that you could measure the second electron at and get 100% up because it is not a pure state, it is still in superposition.

              So back to the quantum eraser experiment, when the first photon hits the screen D0 and collapses, say at position 1.5, the twin photon collapses to a sub-superposition of its state, something like 80% path A and 20% path B. It still takes both paths, but in such a manner that if you choose to measure which-path information at detector D3 it will be more strongly correlated with path A, and if you choose to measure the self-interference signal from the mirror at D1 or D2, it will still self-interfere and will be more strongly correlated with detector D1. What do you think?

              • FlowVoid@midwest.social
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                1 year ago

                In the electron example, if the two electrons are entangled then the wave functions must be the shared. The new superposition for the second electron would therefore be shared with the first electron. So if you measured the second electron along z+ and got up, then if you measured the first electron again, this time along z+, it would give down.

                Likewise if the twin photon is still in superposition, then the first photon is also in superposition. Which is hard to accept in the Copenhagen interpretation, given that the first photon has been absorbed. If absorption doesn’t completely collapse a wave function, then what does?

                • TauZero@mander.xyz
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                  1 year ago

                  So if you measured the second electron along z+ and got up, then if you measured the first electron again, this time along z+, it would give down.

                  Right! So what happens when you have two z+z- entangled electrons, and you measure one along z+45° and then the other along z+0°? What would happen if you measure the second electron along z+45° as well?

                  • FlowVoid@midwest.social
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                    1 year ago

                    Entangled electrons are entangled in all directions. If you measure one along any direction, you can completely predict the measurement of its pair in the same direction.

                    In other words, measuring one along X and its pair at Y is equivalent to measuring one along X and then measuring the same one again at Y (accounting for the sign shift in the pair, of course).