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

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

But explicit calculation of $$\langle{p}\rangle= \int^{\infty}_{-\infty}\psi^*(x) \hat{p} \psi(x)dx$$ and using $$\hat{p}=-i\hbar \frac{\partial}{\partial{x}}$$ 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?

• You need a time-dependent wave function to apply the Ehrenfest theorem. Sep 25 '20 at 8:25
• Why that should be? Because in the derivation of Ehrenfest theorem we used Time independent Schroedinger equation only? Sep 25 '20 at 8:48
• Yes. But more generally - you are looking for the time dependence of averages... while having neglected this very time dependence. It might be that you are also confused about Schrödinger vs. Heisenberg picture - either the wave function or the operators have to carry the time dependence. Sep 25 '20 at 9:02

Ehrenfest theorem should still works here. Your assertion that "$$\langle p\rangle=0$$ since $$\langle x\rangle = 0$$" 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.