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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?

DanielC
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Non medical Roll-4 AYUSH Kumar
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    Regarding the question itself: Do you know what stationary state means (in mathematical terms)? – Tobias Fünke Sep 10 '23 at 17:23
  • Referring to Griffiths qm, stationary states are those in which the wave function can be written as product of two independent functions of time and space and the square of the wave function depends only on x and not on time – Non medical Roll-4 AYUSH Kumar Sep 10 '23 at 17:25
  • Using that logic, expectation value of position should be time independent, therefore it's mass times rate change should give momentum expectation value,which should be zero as is independent of t therefore it's d/dt should yield zero thus zero

    – Non medical Roll-4 AYUSH Kumar Sep 10 '23 at 17:30
  • Am I going wrong with the definition of

    =md/dt?

     – Non medical Roll-4 AYUSH Kumar Sep 10 '23 at 17:32
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    You should edit the question. Write down why exactly you think what you think. – Tobias Fünke Sep 10 '23 at 17:35
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    Related. – Tobias Fünke Sep 10 '23 at 17:40
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    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. – Andrew Sep 10 '23 at 17:57
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    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. – Tobias Fünke Sep 10 '23 at 18:21
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    Tobias is hinting at this and hoping you'll figure it out, but I'm just going to say it. Because I think this is just an issue misinterpreting the question. There's no indication, to me at least, that this is the full time-dependent state. If it was, it should have an $e^{-iEt}$. It doesn't say $\psi(x,t)=$, it says $\psi(x)=$, which indicates to me that this is the wave function at a particular time, not that it stays this way forever. There are two other hints - this wavefunction clearly has a momentum expectation value, so it should translate over time, and gaussian states expand over time. – AXensen Sep 11 '23 at 09:44
  • @AXensen yes, but this is not the full story. One could imagine that $\psi(x)$ is an eigenstate of a (time-independent) Hamiltonian (certainly it is an eigenstate of the operator $|\psi\rangle\langle \psi|$) and then consider its time evolution, which would result in a stationary state. And then you could have an apparent contradiction between the non-zero momentum expectation value and Ehrenfest's theorem. See also my comment under Thomas' answer. – Tobias Fünke Sep 11 '23 at 11:46

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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.

Thomas Fritsch
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  • 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. – Tobias Fünke Sep 11 '23 at 10:18
  • 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. – Tobias Fünke Sep 11 '23 at 11:50
  • 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 – Non medical Roll-4 AYUSH Kumar Sep 11 '23 at 12:50