Average of momentum for stationary state

[CyberShot]

I know that the expectation value of momentum is always 0 for a stationary state.

But, is <p> always zero when the time-dependent parts of the wavefunction cancel out?

Is the following statement true?


<p> = 0 when the full wavefunction can be separated into a time component times the position component.





because the time parts of the conjugates will always cancel out.

 

[Bill_K]

No, because for example a plane wave ~ exp(i(kx-wt)) can be written as a product of a space component and a time component, and has nonzero momentum k.

 

[The_Duck]

Bill K's example is not a normalizable state. I think it is correct to say that any normalizable state for which you can write $$\psi(x, t) = X(x)T(t)$$ is a stationary state.

 

[Matterwave]

That can't be right, not all product states are eigenstates of the Hamiltonian...otherwise, why would we even bother to ever try to solve the TISE for X(x)? Am I missing something here...?

 

[The_Duck]

That can't be right, not all product states are eigenstates of the Hamiltonian...otherwise, why would we even bother to ever try to solve the TISE for X(x)? Am I missing something here...?

If $$\psi(x, t)$$ is a solution of the time-dependent Schrödinger equation and can be written as $$\psi(x, t) = X(x)T(t)$$ then X(x) is a solution of the time-independent Schrödinger equation, yes? That's what I was trying to say above.

 

[Matterwave]

Oh, because the way you phrased it, it seemed like you were suggesting the wave-function could be ANY normalizable product state.