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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:

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

Peter Mortensen
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michael b
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2 Answers2

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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.

aghostinthefigures
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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.

nanoman
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