# Uncertainty about the ensemble average?

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

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. $$\Delta p = \sigma_p = \sqrt{\langle p^2\rangle - \langle p\rangle^2}$$.
• 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 (<>). – jacob1729 Oct 2 '19 at 19:29
• But this means that possible momentums can be greater than $\langle p\rangle + \Delta p$ and $p_{measured} + \Delta p$ , right? – Brain Stroke Patient Oct 2 '19 at 19:31
• Heads up, you can use \rangle for $\rangle$ and \langle for $\langle$ ;) – user2723984 Oct 2 '19 at 19:44
• But, $p_{measured}$ is just a single measurement and is not necessarily equal to the mean of its distribution $\langle p\rangle$. – Mo Farzaneh Oct 2 '19 at 21:03