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Let us assume that one quantum particle is moving in some space (phase space). I appoint two observers to observe that particle. Now at any instant of time I ask the $1^{st}$ observer to measure the x-position and the y-component of linear momentum ; and I ask the $2^{nd}$ observer to measure the particle's y-position and x-component of linear momentum.

As $[x,p_y] = 0$ and $[y,p_x] = 0$ so H.U.P will not come into effect between these two pairs. And after they make their measurements if I combine their data and I will get the precise x-position , x-momentum , y-position , y-momentum of that particle at that moment. Seems like H.U.P failed. But pretty surely this is not the scenario. What is the discrepancy in this thought situation/experiment ?

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    $\begingroup$ You can ignore the y-direction and just consider the x-direction. Then you find that the second observer will have an uncertainty due to the measurement of the first observer that satisfies HUP. $\endgroup$ Mar 17, 2023 at 8:04
  • $\begingroup$ Thank you for your answer. I got it now. $\endgroup$
    – M Sagnik .
    Mar 17, 2023 at 8:06

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It looks like a version of Einstein-Podolsky-Rosen paradox, although the latter is done with spin and one is more concerned with the speed of information propagation than the HUP.

What has to be kept in mind in the experiment described in the OP, that the initial wave function collapses after the first measurement (into a state with definite values of $x$ and $p_y$) and the second measurement is done on this collapsed wave function, not on the initial state. SO one cannot really combine the data.

One could, of course, talk about a simultaneous measurement - in this case we have to resort to a bit of relativistic thinking to resolve the paradox: in this case the observers are not at the same point, and we can still reduce the situation to two consecutive measurements by switching to a different reference frame, e.g., that of the particle itself, since the particle meets one observer earlier than the other.

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Your formulation "$=0$" doesn't hold, as it's an inequality "you can't know better than ...". You can see easily if you note down $|\vec r \cdot \vec p| \le \frac{h}{2 \pi}$, so the first observer already spoiled it.

The philosophical insight was: any measurement spoils the experiment. (Ask Schroedingers cat ;)

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  • $\begingroup$ Yeah, but if we take only the first observer then he/she is measuring the x position and y momentum. As they commute , they won't have to satisfy the H.U.P. So I think there is no "know better than" condition for the quantities measured by the first observer. But I think the "know better than" condition arises when both the observers try to measure their assigned quantities simultaneously. And the H.U.P comes into play. Though the first observer already spoiled the experiment by making the observation, so the second observer most probably won't get the right values. $\endgroup$
    – M Sagnik .
    Mar 17, 2023 at 17:18

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