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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?

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  • $\begingroup$ You need a time-dependent wave function to apply the Ehrenfest theorem. $\endgroup$ – Vadim Sep 25 '20 at 8:25
  • $\begingroup$ Why that should be? Because in the derivation of Ehrenfest theorem we used Time independent Schroedinger equation only? $\endgroup$ – Manas Dogra Sep 25 '20 at 8:48
  • $\begingroup$ 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. $\endgroup$ – Vadim Sep 25 '20 at 9:02
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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.

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