Average momentum of energy eigenstates is always zero?

[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?

 

[blue_leaf77]

it is true that average momentum of energy eigenstates is always zero?

It's true.

 

[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!).

 

[hokhani]

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?

In general, eigenstates of Hamiltonian are not traveling Waves.

 

[taishizhiqiu]

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!).

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.

 

[blue_leaf77]

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.

Particles confined in a box are not free.

 

[king vitamin]

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.

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).

 

[Jilang]

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

 

[jfizzix]

What's wrong with my derivation or it is true that average momentum of energy eigenstates is always zero?

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.

 

[taishizhiqiu]

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).

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 $L/2$.

 

[Demystifier]

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?

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 $e^{ipx/\hbar}$.