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I have a normalized energy eigenfunction for the ground state of Hydrogen which is $$ \Psi(r) = \frac{1}{\sqrt{\pi a_0^3}}\exp\left(-\frac{r}{a_o}\right), $$ where $a_o$ is the Bohr radius,

I have been told that the expectation value of the electrons momentum, $ \left\langle\hat{\vec{p}}\right\rangle =\vec{0}$. I want to know why and I think I'm missing something obvious because my current understanding is not very great.

I'm thinking it has something to do with the electron being in a bound state and the orbit being spherically symmetric but I still think there should be a positive expectation value for momentum.

Could somebody throw me a hint here? I tried to search for answers but couldn't find.

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  • $\begingroup$ Did you notice that the position expectation value is also zero? $\endgroup$ – dmckee Mar 14 '15 at 1:50
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This is actually not surprising. Even an electron orbit in the Bohr atom has a zero expectation value for the momentum if you average over whole orbits.

Think of what it would mean for there to be a finite expectation value for momentum: you'd see an electron that's escaping from the nucleus!

Perhaps what you want to calculate is the root mean square speed of the electron. In that case, you want to find the expectation value of $\vec{p}^2$, not $\vec{p}$.

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  • $\begingroup$ Ah, I see. So the expectation value of momentum has to be zero because if it was not there would be a net average velocity. $\endgroup$ – user01520 Mar 14 '15 at 1:26
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Imagine you have the momentum operator act at a point, it's a vector operator so the result points in a direction, which direction? If the total is zero an easy way to get zero is if there is another place that gives an equal and opposite vector.

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  • $\begingroup$ I get it. That was really simple actually, I don't know how I didn't think of it. $\endgroup$ – user01520 Mar 14 '15 at 1:28
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I still think there should be a positive expectation value for momentum.

Momentum is signed, i.e., can take positive and negative values.

Consider the case of a state which is an equal superposition of momentum eigenstates with equal and opposite eigenvalues, $+p$ and $-p$

Since the probabilities of measuring either the positive or negative momentum are equal, the expectation value of the momentum, the average of an arbitrarily large number of momentum measurements, must be zero.

This doesn't mean that the outcome of a particular measurement will yield zero. Only that the average of the outcome of arbitrarily many measurements will yield zero.

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Although the physical answer is definitely rotational symmetry, maybe you'll find interesting that a real valued wave-function will always give zero for the expected value of $\mathbf p$.

PROOF. For a real valued function $f$, the Fourier transform $\tilde f$ satisfies $$\tilde f (\mathbf p )^* = \tilde f (-\mathbf p ). $$ Hence $$\vert \tilde f (\mathbf p)|^2=\tilde f (-\mathbf p )\tilde f (\mathbf p )$$ is even and $$\langle \mathbf p \rangle = \intop \text d ^3\mathbf p\vert \tilde f (\mathbf p )\vert^2\mathbf p=0,$$QED.

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