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I was puzzled recently by what I've read in the internets about Heisenberg's uncertainty principle (one probably should never do this).

It claimed that in the usual relation $$ \overline{(x-\overline{x})^2} \cdot \overline{(p_x-\overline{p_x})^2} \ge \hbar/2 $$ uncertainties $\overline{(x-\overline{x})^2}$ and $\overline{(p_x-\overline{p_x})^2}$ correspond to the limits on precision of measuring instrument (e.g. we cannot build a device that would measure the momentum precisely).

But I always thought that the meaning of the above relation is as follows:

We have a large number $2N$ of "boxes" with identical systems, all in identical quantum state. We then measure coordinate of $N$ of them and momentum of the other half. We can make each of these measurements, both of coordinate and of momentum at the same time, as precise as possible, with instrumental error virtually equal to zero.

We then find that despite being in the same quantum state, results are not the same for different boxes. And after averaging over boxes $$ \overline{x} = \frac1N \sum_{box=1}^N x_{box} $$ $$ \overline{p} = \frac1N \sum_{box=N+1}^{2N} p_{box} $$ etc, we find the uncertainty relation. So each component in this sum is a precise value (in principle).

Is my understanding correct?

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    $\begingroup$ There is no limit on how precisely we can measure either position or momentum individually. The uncertainty principle simply says that there are no states for which the product of the variances of both measurements is smaller than some number which sets a physical scale. One can find plenty of states for which it is larger. For the single particle the best case scenario (for which the inequality turns into an equality) is that of gaussian (wave packet) states. I would warn you against a stochastic ensemble interpretation of this limit, as it leads down a semi-classical rabbit hole. $\endgroup$ – CuriousOne May 24 '15 at 16:44
  • $\begingroup$ The uncertainty principle isn't something that's coming out of the precision of our instruments. $\endgroup$ – Constantine Black May 25 '15 at 8:04
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    $\begingroup$ Related: physics.stackexchange.com/q/24068/2451 , physics.stackexchange.com/q/114133/2451 and links therein. $\endgroup$ – Qmechanic May 25 '15 at 9:36
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You understanding is correct. The Heisenberg Uncertainty Principle does not stem from the precision of our measuring apparatus; it is instead a fundamental property of the physical states which can be prepared. In essence it is the bandwidth theorem ($\Delta k\Delta x\geq 1/2$) translated into physical form by de Broglie's relation ($p=\hbar k$).

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