Average momentum of energy eigenstates is always zero?

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1. Jan 13, 2016

taishizhiqiu

Look at the following derivation:
$p=\frac{im}{\hbar}[H,r]$
if $H|\psi\rangle=E|\psi\rangle$, then
$\langle \psi|p|\psi \rangle = \frac{im}{\hbar}\langle \psi|Hr-rH|\psi \rangle = \frac{im}{\hbar}\langle \psi|r|\psi \rangle(E-E)=0$
What's wrong with my derivation or it is true that average momentum of energy eigenstates is always zero?

2. Jan 13, 2016

blue_leaf77

It's true.

3. Jan 13, 2016

king vitamin

It's certainly true for bound states. I can imagine situations where you can evade this - for example, a free particle on the infinite line (the energy eigenstates are not normalizable!) or on a ring (x is not a single-valued operator!).

4. Jan 13, 2016

hokhani

In general, eigenstates of Hamiltonian are not travelling Waves.

5. Jan 13, 2016

taishizhiqiu

what about box normalization of free electrons? That is, particles are confined in a region between $-L/2$ and $L/2$ and fulfill periodic boundary condition. This way, wave functions are normalizable.

6. Jan 13, 2016

blue_leaf77

Particles confined in a box are not free.

7. Jan 13, 2016

king vitamin

This is exactly what I meant when I said a free particle on a ring: the operator x is not single-valued. By the way, for the eigenfunctions of this problem, what is the variance of momentum? Notice what this means for the Heisenberg uncertainty principle (which you used in your original post).

8. Jan 13, 2016

Jilang

It is zero because there is equal probability of it travelling one way or the other. Try p^2, it won't be zero.

9. Jan 15, 2016

jfizzix

It is true; the average momentum of energy eigenstates is always zero.

True energy eigenstates are hard to find outside of bound states in most potentials.

Free particles are not found in energy eigenstates for the same reason that they are not found in position or momentum eigenstates. Such wavefunctions would be unphysical and non-normalizeable.

10. Jan 18, 2016

taishizhiqiu

Why x is not a single-valued operator will lead to breakdown of the claim in my original post? After all, we can define x to be between $-L/2$ and $L/2$.

11. Jan 18, 2016

Demystifier

Here you tacitly assumed that $\langle \psi|r|\psi \rangle$ is not infinite.There are energy eigenstates with non-zero momentum, but for such states your tacit assumption is not fulfilled. Indeed, for free Hamiltonian such a state is $e^{ipx/\hbar}$.