# Justification of a common calculation

Hi,

isn't it a bit dangerous to claim that

$$\left[ x \cdot \left( \psi(x,t) \, \frac{\partial \psi^\ast (x,t)}{\partial x} + \psi^\ast(x,t) \, \frac{\partial \psi(x,t)}{\partial x} \right) \right]_{x=-\infty}^{x=\infty} = 0$$​

for example?

Expressions like this one are often found in popular quantum mechanics textbooks. How do you justify such expressions? I would prefer a mathematical- instead of a physical explanation...

With best regards

Very good question. I spent several days trying to figure it out when I first began learning QM (and haven't stopped since!). Here is my take on it.

It is implicitly assumed here that the wavefunction is square integrable meaning

$$\int_{-\infty}^{\infty} |\psi(x)|^2\,dx$$

exists and is finite. Books usually do not go further than this, but state conditions of "well behaviour" (but see * below). More rigorous books on QM will tell you that if the long-range behavior of the wavefunction is of the form $1/|x|^{\alpha}$ for $\alpha > 1$, then this condition is (usually) satisfied. [c.f. Schwabl, for instance.]

So in your expression, as $|x| \rightarrow \infty$, the terms involving the product of the wavefunction and its derivative will go to zero faster than $x$ tends to infinity, and hence 'overall' the expression will go to zero at $|x| = \infty$.

If you want to be more rigorous, you can define a subspace of the Hilbert space and regard only square integrable functions or functions satisfying some regularity conditions, as valid wavefunctions. That way, all such boundary terms automatically vanish.

* - Square integrability does not guarantee that the wavefunction goes to zero at infinity. For an explanation, see D.V. Widder, "Advanced Calculus" 2nd ed., Dover, New York, 1998, p. 325.

Thank you for your fast and detailed answer! That supports my assumption that several authors leave the reader deliberatly (or not) behind some serious issues :uhh: