In a hypothetical scenario, if I were to measure the quantum spin of an electron and it showed "up," and then I traveled back in time without changing the initial conditions, would measuring the spin again still be probabilistic, or would it necessarily show "up" again? If its up does that mean its deterministic in backward time? If probabilistic does it indicate anything?
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$\begingroup$ When you measure the state of the system, you are necessarily a acting on it. Hence, after the measurement, in future if you measure the state of the system, you will necessarily get "up" because of your interaction with the system. However, if you traveled back in time, you would eliminate the effect of the measurement on the system. Hence, you wouldn't necessarily get "up". $\endgroup$– Karen BaghdasaryanCommented Jul 2 at 11:48
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$\begingroup$ Will you please confirm if the following statement is false or true? So, for example, a particle might be in a state where it will be found at a point A with probability 1/3 and B it will be at a point 2/3. The particle is not really localized before the measurement, but after the measurement, if we set up a bunch of identical copies of the system side by side, each in particular state psi, and then measure the position 1/ 3 of the time it will be in A and 2/3 of the time it will be in B. I heard this from some youtube video. $\endgroup$– VishnuCommented Jul 2 at 12:02
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1$\begingroup$ Which youtube video? $\endgroup$– Karen BaghdasaryanCommented Jul 2 at 12:29
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$\begingroup$ The symmetry at heart of conanical commutation relation, 5:01 time stamp. Channel name: physics with elliot $\endgroup$– VishnuCommented Jul 2 at 12:35
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If you have two identically prepared systems, say two copies of the state $ \alpha \vert \uparrow \rangle + \beta \vert \downarrow\rangle$, with $\vert \alpha\vert^2+\vert\beta\vert^2=1$, you might get "up" in the first experiment, and get "down" in the second. The respective probabilities are $\vert\alpha\vert^2$ and $\vert\beta\vert^2$ and you might get different outcomes even if the initial conditions and all other factors are identical. Of course, I am assuming that $\alpha\ne 0$ and $\beta\ne 0$, else the outcome is either "down" or "up" with certainty.
It's not clear what you mean by "travel back in time" in this context, since we can't travel back in time. But it is at the time of measurement that "God rolls the dice" so presumably if your second experiment is a copy of the first experiment that has been "travelled back in time" (how would you distinguish this from just another copy?) and you just let the experiment run its normal course then it is entirely possible that the outcome might be different.
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$\begingroup$ In GR it is theoretically possible to travel back in time, e.g., in a time loop where the past repeats again and again. What do you think would be the outcome of a repeated qm measurement in such a situation? $\endgroup$ Commented Jul 2 at 12:02
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$\begingroup$ @freecharly It's not clear if the rules of non-relativistic QM as we know them apply to GR in the speculative context you mention, but if they do then it does not change the current consensus that the "choice" of orientation is done at the measurement, and this choice is probabilistic so the outcome need not be the same. $\endgroup$ Commented Jul 2 at 12:04
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$\begingroup$ But then the closed causal chain could change for each closed loop, which would probably contradict the assumption of a time loop. $\endgroup$ Commented Jul 2 at 12:19
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$\begingroup$ @freecharly AFAIK, the answer to your query is yes, it would change the outcome. What you are pointing to is that more work needs to be done to understand measurements on highly curved spaces. I mean: QFT in these cases is complicated enough there could be all kinds of loopholes and tricks that avoid apparent or real contradictions. $\endgroup$ Commented Jul 2 at 12:25
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1$\begingroup$ @james I don't know how you can entangle dimensions, especially time. Time in QM is not an operator but a parameter, whereas (at least) $\hat x$ is an operator. In other words, I don't know what $\vert x,t\rangle$ means in your context. $\endgroup$ Commented Jul 2 at 13:37