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

    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).

                • TauZero@mander.xyz
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                  1 year ago

                  Hmm interesting. I may have been mistaken about the electrons only being entangled in a single direction. I thought that if you prepared a pair of electrons in state 1/sqrt(2) (|z+z-> + |z-z+>) and then measured it in y there would be no correlation, but based on: https://physics.stackexchange.com/questions/700218/intuition-for-results-of-a-measurement-of-entangled-spins
                  https://physics.stackexchange.com/questions/385477/what-is-the-quantum-state-of-spin-1-2-particle-in-y-direction
                  if I had done the 90° rotation properly, the math works out such that the electrons would still be entangled in the new y+ basis! There is no way to only entangle them in z alone - if they are entangled in z they are also entangled in x and y. My math skills were 20 years rusty, sorry!

                  I still think my original proposition, that in the DCQEE under Copenhagen, an observation that collapses one photon, collapses the other photon to a sub-superposition, can be salvaged. In the second stackexchange link we are reminded that for a single electron, the superposition state 1/sqrt(2) (|y+> - |y->) is the same as |z+> state! They describe the same wavefunction psi, expressed in different basis: (y+,y-) vs. (z+,z-). When we take a single electron in superposition 1/sqrt(2) (|z+> + |z->) and measure it in z, and it collapses to, say, z+, we know that it is a pure state in z basis, but expressed in y basis it is now a superposition of 1/sqrt(2) (|y+> - |y->)! Indeed if we measure it now in y, we will get 50% y+ and 50% y-.

                  So in DCQEE when you collapse the first photon into a single position on the screen, the twin photon does collapse, but its basis is not expressed in terms of single positions! It’s some weird agglomeration of them. If you were to take that “pure” state and express it in terms of position basis, you would get a superposition of, say, 80% path A and 20% path B.

                  • FlowVoid@midwest.social
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                    1 year ago

                    Well, if the second photon is in a new, weird superposition then the first photon must also be in the same new, weird superposition. Again, I don’t that’s compatible with Copenhagen given that the first photon no longer exists.

                    Note by the way that 50% y+ and 50% y- is how all photons start. So if that’s also the final state then there is no reason for it to prefer any detector over the others.