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I'm using this resource along with Griffith's Introduction to Quantum Mechanics to try and reproduce the Ehrenfest theorem.

From equation $(176)$ in the link above, we have:

$$\frac{d\langle p\rangle}{dt}=\int_{-\infty}^{\infty}\left[\frac{-\hbar^2}{2m}\frac{\partial}{\partial x} \left( \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial x} \right) +V(x)\frac{\partial|\Psi^2|}{\partial x} \right] dx$$

I am able to get to here without issues, but next we have to show that:

$$\int_{-\infty}^{\infty}\left[\frac{-\hbar^2}{2m}\frac{\partial}{\partial x} \left( \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial x} \right) \right] dx = 0$$

Which would only be true if:

$$\left. \frac{\partial \Psi}{\partial x} \right|^{x=\infty}_{x=-\infty} = \left. \frac{\partial \Psi^*}{\partial x} \right|^{x=\infty}_{x=-\infty} = 0$$

Is there a way to know this generally? It's obviously true in certain cases of the wave function (e.g. $\Psi(x)=\exp[-x^2]$). In general, I thought the only condition for normalization was that:

$$\left. \Psi \right|^{x=\infty}_{x=-\infty} = \left. \Psi^* \right|^{x=\infty}_{x=-\infty} = 0$$

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2 Answers 2

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Fun fact; it's not true in general! For example, this answer lists an example of a function that is totally square-integrable and therefore viable as a wave-function but whose derivatives do not have a well-defined limit at infinity.

The real reason you can get away with doing this approximation is that we assume implicitly in quantum mechanics, perhaps with not enough forcefulness, that wave-functions have "compact support", i.e., the functions and their derivatives are only nonzero on a closed, bounded subset of space.

Some toy examples of wave-functions eschew this requirement, such as the quantum free particle with exact momentum, but this is not a true wave-function as it is not square-integrable.

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    $\begingroup$ I think the example that you link to is a smooth function whose derivative doesn't vanish asymptotically. (I deleted an answer that I wrote that claimed smoothness was enough to guarantee the derivative went to zero; I don't know if this is why you added the caveat about smoothness or not, but I think I was wrong based on that example). It seems that the vanishing of the derivatives at infinity is an additional assumption that's needed. $\endgroup$
    – Andrew
    Mar 13, 2022 at 23:28
  • $\begingroup$ Thanks for the link. I had searched around but didn't find anything - but as per usual this exact question had been explored before on Phys SE. $\endgroup$
    – michael b
    Mar 14, 2022 at 3:38
  • $\begingroup$ I agree with @Andrew -- the example doesn't support this answer because the wave function there is differentiable to all orders. $\endgroup$
    – nanoman
    Mar 14, 2022 at 8:38
  • $\begingroup$ It's differentiable (so it's smooth) but the derivative doesn't vanish at infinity. I think the real answer is that we don't care about any of the physics happening at a significant distance from the measurement. Unless it's a lattice... But that is a bit beyond me at the moment. $\endgroup$
    – michael b
    Mar 14, 2022 at 12:47
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    $\begingroup$ A common technique in analysis is to prove a certain identity first for "nice" functions (like $C_c(\mathbb{R}^n)$, i.e. smooth compactly support functions), and then to extend the identity to a larger space by an approximation argument. Here it is unreasonable to expect that Ehrenfest's theorem holds for all $L^2(\mathbb{R}^n)$-functions because the momentum operator is not defined on all square-integrable functions. But the result should be extendable to the Sobolev space $H^1(\mathbb{R}^n)$ which is the domain that makes the momentum operator self-adjoint. $\endgroup$
    – Janik
    Mar 14, 2022 at 19:02
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Another question's answer notes that the expectation value of kinetic energy is proportional to $$\int_{-\infty}^\infty dx\, \psi^* \left(-\frac{\partial^2\psi}{\partial x^2}\right) = \int_{-\infty}^\infty dx\, \left|\frac{\partial\psi}{\partial x}\right|^2 - \left.\psi^* \frac{\partial\psi}{\partial x}\right|_{-\infty}^\infty.$$ In the Ehrenfest derivation, you have already been willing to set boundary terms that include a factor of $\psi$ or $\psi^*$ (without derivatives) to zero. So the above reduces to $\int_{-\infty}^\infty dx\, (\partial\psi^*/\partial x)(\partial\psi/\partial x)$. Now, on physical grounds, this expectation value of kinetic energy should be finite. Thus, the integrand approaches zero as $x \to \pm\infty$, which suffices to show the vanishing of the boundary term you asked about.

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  • $\begingroup$ If the integrand were non-zero at infinity this would mean the the expectation of kinetic energy would be non finite. Is that correct? This is actually a really nice way to insert that step in Ehrenfests theorem. It seems pretty general too. So I'm curious what physical systems would fit some of the counter examples shown above. Maybe they are just mathematical figments $\endgroup$
    – michael b
    Apr 1, 2022 at 4:11

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