I came across a problem in Griffiths where the derivative of the wave function (with respect to position in one dimension) evaluated at $\pm\infty$ is zero. Why is this? Is it true for any function that evaluates to zero at $\pm\infty$ or is there a special constraint on the wave function that I am forgetting?

  • $\begingroup$ $\uparrow$ Which problem? $\endgroup$
    – Qmechanic
    Commented Aug 16, 2015 at 20:56
  • $\begingroup$ The Problem 1.7 in Griffiths' second edition, p.18. The proof of momentum's compliance with Ehrenfest's theorem. $\endgroup$ Commented Jan 20, 2019 at 4:31

1 Answer 1


Not necessarily. Consider this function as an example:

$$\psi(x) = \frac{C\sin x^2}{\sqrt{x^2 + 1}}$$

This function is square-integrable and asymptotes to zero as $x\to\pm\infty$, but its derivative goes to $2\cos x^2$ in the same limit.

In quantum mechanics, we often assume that real systems are represented by wavefunctions which have no interesting features once you get far enough away from the origin. In practice, this means that the function and all its derivatives have "compact support:"

$$\lim_{x\to\pm\infty}\frac{\partial^n\psi}{\partial x^n} = 0\ \forall\ n\in\mathbb{Z}_{0,+}$$

This mathematical statement corresponds to the physical assumption that you can ignore anything happening far enough away from your experiment.

There are situations where it's useful to drop this assumption, though. For example, if you're analyzing a crystal lattice, it makes the calculations easier to assume the lattice extends infinitely far in all directions, and in that case you would use a wavefunction which is periodic all the way out to infinity. Of course, such wavefunctions usually have nonzero values in addition to nonzero derivatives at large $|x|$. I don't know of an example offhand which would use a wavefunction which asymptotes to zero but whose derivatives don't, though it wouldn't surprise me at all to learn of one.

  • $\begingroup$ Nice counterexample $\endgroup$ Commented Jun 17, 2012 at 5:40

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