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(Edited according to the discussion with @naturallyInconsistent. The edited part is highlighted in italic.)

We have an experimental bench and we assign a coordinate system $(x,y)$ to it. We shall call $x$ the longitudinal coordinate and $y$ the transverse coordinate. We arrange a laser source at $(-1,0)$, a spectrometer at $(0,0)$, a detector $A$ at $(1,1)$, and another detector $B$ at $(1,-1)$. All of them cannot move longitudinally but can freely move transversely.

First, suppose that the laser source is super strong. Then the two detectors will move away from each other. Each of them will acquire a transverse momentum with the same magnitude but in the opposite direction. The total transverse momentum is always conserved to be $0$.

Now, suppose that the laser source can only emit a single photon at a time. If we wait for a sufficiently long time to accumulate enough photon emissions and measurement events, we should see the same phenomenon as above. Then, what happened during a single measurement? Without loss of generality, let us consider the very first measurement and suppose that detector $A$ detected the photon. Then there are two possible scenarios:
(I) Detector $A$ acquires a tiny transverse momentum while detector $B$ stays at rest. This violates momentum conservation.
(II) The two detectors acquire opposite transverse momenta with the same tiny magnitudes. This violates the locality.

Question: Which scenario describes the actual experimental result after a single measurement? or any other scenario?

Copenhagen's collapse picture seems to support scenario (I), because right before the measurement, according to the evolution equation, the quantum state $\left|{\Psi}\right\rangle$ seems to be a superposition of $\left|{\text{detect, move}}\right\rangle_{A}\otimes\left|{\text{not-detect, rest}}\right\rangle_{B}$ and $\left|{\text{not-detect, rest}}\right\rangle_{A}\otimes\left|{\text{detect, move}}\right\rangle_{B}$. $\left|{\Psi}\right\rangle$ has a zero expectation value of transverse momentum and thus satisfies momentum conservation. The measurement then violates momentum conservation.

Leo
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  • No, the Earth, due to fixing the spectrometer and longitudinal motion of the detectors, will absorb and cancel out the offending motion. It is a known issue that measurements give the illusion of violating conservation, but that is likely an illusion only due to neglecting the interaction of the system with the measurement apparatus. – naturallyInconsistent May 18 '23 at 13:31
  • No, measurement does not violate conservation laws. Some of the measured energy, momentum and angular momentum simply comes from the measurement apparatus. – FlatterMann May 18 '23 at 13:32
  • I also believe that conservation laws cannot be violated. – Leo May 18 '23 at 13:36
  • @naturallyInconsistent This thought experiment concerns the transverse motion. We can think that the two detectors are placed on an extremely smooth slide rail along the transverse direction. – Leo May 18 '23 at 13:40
  • @Leo that is implicit in my answer; the Earth will take away the transverse momentum because the spectrometer bounced the photon in that direction, and that bouncing has to obey conservation – naturallyInconsistent May 18 '23 at 13:42
  • @naturallyInconsistent Do you mean that, if we also allow the transverse motion of the spectrometer, then the spectrometer will move oppositely to detector $A$? – Leo May 18 '23 at 13:45
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    Yes, oppositely to the detector that moves. – naturallyInconsistent May 18 '23 at 13:46
  • I thought quantum mechanics is already known to "probably" violate locality? E.g. entangled particles having opposite measurements despite being outside each other's light cones – RC_23 May 18 '23 at 13:51
  • @naturallyInconsistent I see. I guess this is the most consistent scenario. I slightly modified my question according to our discussion. Are you interested in writing an answer based on our discussions so far? – Leo May 18 '23 at 14:06
  • @RC_23 Quantum mechanics doesn't violate locality. Entangled states are non-separable, though. The difference is that separability is a statement about Hilbert space while locality is about physical space. – FlatterMann May 18 '23 at 14:10
  • @Quillo Thanks, it indeed answers my question title. But that question and answers thereof are a bit conceptual. I'd like to see concretely how the experiment apparatus absorbs the momentum. – Leo May 18 '23 at 14:10
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    Then the answer is that all valid interpretations of QM has to satisfy that only one of the detector will move transversely, and the spectrometer moves oppositely, so that momentum is conserved in every scenario. It will never be both detectors moving. This is a bit too small to write as an answer. – naturallyInconsistent May 18 '23 at 14:12
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    @Leo The same way as in classical mechanics: it recoils. If we (willfully) neglect the momentum change of a wall in a simple bounce scenario, then the system seems to violate momentum conservation as well. If we do it right, then it doesn't. The violation comes from a false mental model and the math that we base on it. – FlatterMann May 18 '23 at 14:13

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A photon along the axis is deflected to one or the other detector by the spectrometer. The spectrometer acquires a tiny momentum.

Approximately equal numbers of photons are deflected up and down, so the average momentum of the spectrometer is $0$.

mmesser314
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