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Why is momentum expectation zero for gaussian wavepacket by Ehrenfest theorem?

Physics Asked on December 13, 2020

If $$psi(x)=Aexp(-x^2/a^2)exp(ikx)$$
$langle{p}rangle=0$ since
$langle{x}rangle=0$, since the integrand is an odd function and Ehrenfest theorem states $frac{dlangle{x}rangle}{dt}=frac{langle{p}rangle}{m}$.

But explicit calculation of $langle{p}rangle= int^{infty}_{-infty}psi^*(x) hat{p} psi(x)dx$ and using $hat{p}=-ihbar frac{partial}{partial{x}}$ gives $hbar k$. I think Ehrenfest theorem is giving the wrong result because of the $ikx$ term,how to correctly use Ehrenfest theorem in this case?

One Answer

Ehrenfest theorem should still works here. Your assertion that "$langle prangle=0$ since $langle xrangle = 0$" is wrong, as the relation is between $langle p rangle$ and the time-derivative of $langle x rangle$. You need to introduce dynamics via a Hamlitonian, and then you will be able to take the time-derivative of the expectation value. Assuming a free Hamiltonian of $H=p^2/2m$ you will get that Ehrenfest theorem holds.

Answered by yyy on December 13, 2020

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