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Does an uncertainty in momentum of $\Delta p$ mean that the actual momentum is in the range $\langle p \rangle - \Delta p\space <p< \langle p\rangle+\Delta p\space$ ? Or does it mean that the actual momentum is in the range $p_{measured} - \Delta p< p< p_{measured}+\Delta p\space$ ?

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It means that if you measure the momentum of many copies of the same system you will get a distribution with the mean as $\langle p\rangle$ and the second moment as $\langle p^2\rangle$. Then the uncertainty $\Delta p $ is just the standard deviation of that distribution, i.e. $\Delta p = \sigma_p = \sqrt{\langle p^2\rangle - \langle p\rangle^2}$.

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    $\begingroup$ Just a nitpick: the mean square momentum is the second moment, not the first (the mean is the first moment). Also \langle \rangle brackets ($\langle \rangle$) look much better than angle bracksts (<>). $\endgroup$ – jacob1729 Oct 2 '19 at 19:29
  • $\begingroup$ But this means that possible momentums can be greater than $\langle p\rangle + \Delta p$ and $p_{measured} + \Delta p$ , right? $\endgroup$ – Brain Stroke Patient Oct 2 '19 at 19:31
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    $\begingroup$ Heads up, you can use \rangle for $\rangle$ and \langle for $\langle$ ;) $\endgroup$ – user2723984 Oct 2 '19 at 19:44
  • $\begingroup$ @Achilles'Advisor Yes, it's possible. $\endgroup$ – Mo Farzaneh Oct 2 '19 at 20:57
  • $\begingroup$ But, $p_{measured}$ is just a single measurement and is not necessarily equal to the mean of its distribution $\langle p\rangle$. $\endgroup$ – Mo Farzaneh Oct 2 '19 at 21:03

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