In quantum mechanics it's important that the wavefunction has to be finite everywhere, including infinity. It's a criteria for acceptable wavefunction. What exactly does this statement mean? And how it's associated with square integrability?

enter image description here

  • 1
    $\begingroup$ And the square integrability is related to the fact the probability of electron being somewhere is certainty. Integral of the $|\phi|^2$ over all the space is equal to 1. $\endgroup$
    – Poutnik
    Feb 2, 2021 at 12:22
  • $\begingroup$ The probability has to be 1 over all space i.e. $\int \psi^*\psi dv=1$ which means that the 'particle' has to exist somewhere, but if the wavefunction was infinite at any point this integral would not be possible and so the wavefunction would not make physical sense. $\endgroup$
    – porphyrin
    Feb 2, 2021 at 12:24
  • 5
    $\begingroup$ A function can indeed approach $\infty$ and still be square integrable, but the s orbital isn't an example of that! $\endgroup$ Feb 2, 2021 at 12:59

2 Answers 2


A broader point regarding functions that are finite:

A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.

This is a generalisation of the comments written above, not specifically quantum related, but a more general point on mathematical functions.


Your question is "Why does wave function must be finite"? Actually, that is not exactly true. Wave function can be infinite, as long as its follows this $$\int_{-\infty} ^\infty \psi^*(x)\psi(x) dx=1$$ What's the example? Dirac delta function is one of them.

For most cases, it is true that wave function of a bounded system must be finite for the simple reason that a wave function has to be normalizable ($\int_{-\infty} ^\infty \psi^*(x)\psi(x) dx=1$).

If it was infinite in some places (for example $\psi(x)=\frac{N}{x}$), we can try to integrate them over all space, and will get infinity $$\int_{-\infty} ^\infty \frac{N^2}{x^2} dx=\infty$$ And, that is not just make no sense, it is also not physical.

  • $\begingroup$ Aside the requirement for the normalization, is not a wavefunction having a local singularity potentially violating the Heisenberg uncertainty principle? $\endgroup$
    – Poutnik
    Dec 15, 2023 at 8:35
  • $\begingroup$ I am not sure, but that do make sense. But the Fourier transform of Dirac delta function is just complex exponential. So, basically, the "uncertainty" in the momentum wave function is infinite, making the $\Delta x \Delta p$ become an indeterminate form. $\endgroup$
    – Tensor
    Dec 15, 2023 at 8:40
  • $\begingroup$ And, eventually, are not there other mathematical requirements for wavefunction properties, like being differentiable in its domain? I guess that in context of being solution of differential equation, it looks like a mandatory requirement. Note that I am not skilled enough in math to decide myself. $\endgroup$
    – Poutnik
    Dec 15, 2023 at 10:09
  • 1
    $\begingroup$ (1) A distribution with $|\psi^2|=\delta(x)$ gives you an exact coordinate and a completely unknown momentum, so no violation of Heisenberg here. (2) A $\psi$ function with local singularity does not have to be like delta function. (3) There is no requirement for $\psi$ to be differentiable everywhere; indeed, $s$ orbital at $0$ is not. $\endgroup$ Dec 15, 2023 at 10:30
  • 1
    $\begingroup$ Indeed @Tensor . Commonly seen as a boundary condition when solving scattering problems. I didn't say the wavefunction had to disappear at very large r. I just said that it is a common boundary condition, being the appropriate one for when solving for bound states of the molecule. Other problems have different boundary conditions. $\endgroup$
    – Ian Bush
    Dec 16, 2023 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.