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I have a spatial wave function here $$\psi(x)=\frac{1}{\sqrt{a\sqrt{\pi}}}e^{-(x^2/2 a^2)+ikx} \quad .$$

I calculated its position expectation value, and it's zero, as expected since it's a stationary state as $|\psi|^2$ depends only on $x$.

Why is the Ehrenfest theorem not applicable here? That is, why $\langle p\rangle$ is non-zero if $\langle x\rangle$ is constant?

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    $\begingroup$ Regarding the question itself: Do you know what stationary state means (in mathematical terms)? $\endgroup$ Sep 10, 2023 at 17:23
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    $\begingroup$ You should edit the question. Write down why exactly you think what you think. $\endgroup$ Sep 10, 2023 at 17:35
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    $\begingroup$ Related. $\endgroup$ Sep 10, 2023 at 17:40
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    $\begingroup$ Note that the probability current can be non-zero even in a stationary state. See the section on plane waves in that article for your specific case. $\endgroup$
    – Andrew
    Sep 10, 2023 at 17:57
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    $\begingroup$ Hint: You have to first specify some time-dependence before speaking of stationary states. And second, you have to recall the underlying assumptions in Ehrenfest's theorem if you want to use it. In particular, there you assume that the wave function is a solution of the time-dependent Schrödinger equation for a Hamiltonian of a particular form. $\endgroup$ Sep 10, 2023 at 18:21

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Your wave function $$\psi(x)=\frac{1}{\sqrt{a\sqrt{\pi}}}e^{-(x^2/2 a^2)+ikx}$$ does not depend on time $t$ and as such is not a solution of the time-dependent Schrödinger equation $$i\hbar \frac{\partial\psi(x,t)}{\partial t}=H\psi(x,t).$$

But Schödinger's equation is a requirement of Ehrenfest's theorem. Therefore you cannot apply Ehrenfest's theorem with your time-independent wave function.

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  • $\begingroup$ Note that $H=P^2/2m+V(X)$ is required for the Ehrenfest theorem. Hence, this argument (or apparent contradiction) actually shows that $\psi(x)$ cannot be a stationary state of a Hamiltonian of this form, i.e. $\Psi(x,t)=e^{-iEt/\hbar}\psi(x)$ is not a solution of the TDSE for any $H$ as above, and hence $\psi(x)$ not an eigenstate of such a Hamiltonian. $\endgroup$ Sep 11, 2023 at 10:18
  • $\begingroup$ To add: At least if we ignore mathematical rigor for now (ignoring issues regarding domains, some conditions on the Hamiltonian, validity of the Ehrenfest theorem etc.). As I've linked under the question, the point is that (under some assumptions on the state), it can be shown that the expectation value of the momentum operator in any eigenstate of such a Hamiltonian vanishes. This thus means that $\psi$ as in the question is not such an eigenstate. $\endgroup$ Sep 11, 2023 at 11:50
  • $\begingroup$ Indeed, initially my assumption was that it's a solution to the time independent wave equation,thus it must satisfy the time independent wave equation,I tried to find the possible values of E and V for this presumed soln, I instead found out that (V-E)=constant times complex function of x(1)which can't be true since E has to be real for normalizable wave function and we assume V to be real in the ehrenfest derivation since it's conjugate remains V therefore (1) can't be hold true for the given spatial wave function for time independent wave equation $\endgroup$ Sep 11, 2023 at 12:50

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