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The Ehrenfest theorem in quantum mechanics for a particle moving in one-dimension in an arbitrary nonuniform potential $V(x)$ is $$\frac{d}{dt}\langle p\rangle=-\left\langle\frac{\partial V(x)}{\partial x}\right\rangle,$$ and in three-dimension, it generalizes to $$\frac{d}{dt}\langle {\vec p}\rangle=-\left\langle\vec{\nabla} V({\vec r})\right\rangle.$$ This theorem holds whether or not the potential is even i.e. $V(-x)=V(x)$ or $V(-{\vec r})=V({}\vec r)$.

If the expectation values in the above equation are taken in an eigenstate of the Hamiltonian, $\langle p\rangle$ becomes completely time-independent which makes the left hand side zero. However, in general, the right hand side is nonzero. How is Ehrenfest theorem sustained here?

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    $\begingroup$ I think you need to rephrase the question a little bit or ask a new one. The answer to your question is simply that the Ehrenfest theorem holds (especially) for stationary states. From the comments on Zack's answer, I guess you want to know what the vanishing time derivative implies physically?! $\endgroup$ Commented Feb 15, 2021 at 11:30
  • $\begingroup$ @Jakob No. That's not what I am asking. I am simply asking if the LHS is zero for a stationary state, the RHS is nonzero, how is the theorem satisfied. See my comment at Zack. I want too see how will the theorem work out there. $\endgroup$ Commented Feb 15, 2021 at 12:26
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    $\begingroup$ As far as I can tell, you are concerned that $V'(x)$ does not seem like it should average out to zero for an arbitrary potential. You said in another comment that it works for the SHO because it is an even potential, but the same trick will not work in general -- of course, I completely agree. I suspect you are focusing too much on the potential and not enough on the stationary state. In a sense, you are hand-picking a probability distribution in such a way that the "net force" vanishes. $\endgroup$
    – Zack
    Commented Feb 15, 2021 at 15:52
  • $\begingroup$ @Zack Yes, I think I understand it now. Thanks $\endgroup$ Commented Feb 15, 2021 at 15:59

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Note that the RHS of your equation is indeed zero: In general, we evaluate the time derivative of the expectation value of an operator $A$ in the state $\Psi$ as $$ \frac{\mathrm{d}}{\mathrm{d}t} \langle A \rangle_{\Psi} = \frac{-i}{\hbar} \langle[A,H]\rangle_{\Psi} + \langle\partial_t A\rangle_{\Psi} \quad . $$

For a stationary state $\psi$, we have that $H\,\psi= E\,\psi$ and hence $\langle[A,H]\rangle_{\psi} = 0$. For $H = \frac{p^2}{2m} + V(x)$ it follows that $[p,H] = -i\hbar\, \partial_x V(x)$. Consequently, for a stationary state $\psi$ we find that $\langle \partial_x V(x) \rangle_{\psi} =0 $.

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    $\begingroup$ @mithusengupta123 The gradient of $1/r$ is a vector, not the scalar $-1/r^2$. If you want the $x$-component, then it's $-x/r^3$. $\endgroup$
    – J. Murray
    Commented Feb 15, 2021 at 12:42
  • $\begingroup$ @jacob I don't know why everybody is giving the proof of the Ehrenfest theorem. I am saying how it works out in each problem. You are saying that $<\partial_x V(x)>_\psi=0$ for a stationary state $\psi$. I can check this for the 1D harmonic oscillator potential explicitly. To be concrete, for $V(x)=x^2/2$, $\partial_x V(x)=x$ and since the eigenfunctions of the Hamiltonian are either even or odd, this turns out to be zero. But this works here because $V(x)$ is an even function and eigenfunctions in 1D are either even or odd. $\endgroup$ Commented Feb 15, 2021 at 13:04
  • $\begingroup$ This will not be obvious , if $V({\vec r})$ is not invariant under parity. Note that it is not obvious that the integral $\langle n\ell m|x/r^3|n\ell m\rangle=0$ (prior to the use of Ehrenfest theorem). Hope I am making sense now. $\endgroup$ Commented Feb 15, 2021 at 13:04
  • $\begingroup$ Okay, I can understand this. But then you have to calculate the integral and check the validity of the theorem! If you have problems with this, ask a new question. As I already stated in the comments on the OP: Your question is answered by the statement that the Ehrenfest Theorem holds for stationary states. As a side note, I might add that this is somehow part of the beauty of maths: If you have a general theorem which, under certain assumptions, guarantees you how to calculate things etc., then you do not have to check all special cases explicitly (given that the assumptions are fulfilled). $\endgroup$ Commented Feb 15, 2021 at 13:07
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    $\begingroup$ Thanks, now I understand it. My question was not worded correctly. :-) $\endgroup$ Commented Feb 15, 2021 at 13:12
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The answer is almost contained within your question itself: if the left-hand side is zero, then the right-hand side must be zero as well. One way you can understand why the right-hand side is zero from a stationary state would be to use a WKB-esque intuition: if we're in a steady state, then the probability density of particles is not building up or spreading out anywhere. That means, on average, the force pushing the particle in one direction or the other must on average be zero, or else the probability distribution would slosh around and change.

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  • $\begingroup$ Mathematically, for an arbitrary nonuniform potential I don't see the reason why the integral on the right hand side should vanish in a stationary state. $\endgroup$ Commented Feb 15, 2021 at 4:52
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    $\begingroup$ Well the mathematical reason is simply Ehrenfrest's theorem. If you want to see explicitly that $\langle V'(x) \rangle = 0$, you can always retrace the steps of the theorem. Of course, this is the most straightforward in the Heisenberg picture: $\dot{p} = -i [p,H] / \hbar = - V'(x)$, and taking expectation values with a steady state means $\langle V'(x) \rangle = 0$. But intuitively, this is exactly what you would expect to happen for any sort of potential: if it were not true, then there would (in a sense) be a net force on the probability distribution, and the state would not be stationary. $\endgroup$
    – Zack
    Commented Feb 15, 2021 at 5:50
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    $\begingroup$ @MarcoCiafa Yes, I checked it too. It worked, which is why removed my comment. Thanks anyway $\endgroup$ Commented Feb 15, 2021 at 15:42

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