# Singular wave function

Given a wavefunction, $\psi(x)$, is it possible for $\psi$ to be singular at a point? Are there any rules against a wavefunctions having any singularities? For instance if

$$\psi(x) =\frac{\gamma(x)}{x},$$

where $\gamma$ is a continuous function of $x$. Is this a valid wave function?

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## 1 Answer

A wavefunction can be nowhere continuous. It is enough that it belongs to $L^2(\mathbb R)$, so, in general, no regularity conditions are imposed on values attained at every given point of $\mathbb R$. It is only required that $\int_{\mathbb R} |\psi(x)|^2 dx < +\infty$. (Regularity conditions can be imposed when the wavefunction is required to belong to the domain of some given operator representing an observable).

Your example however is not allowed as a good wavefunction if $\gamma(0) \neq 0$, since, in that case, $|\psi(x)|^2$ is not integrable (it can be if $\gamma(0)=0$ however there is no guarantee also in this case).

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What if the space was changed? Is $L^2(\mathbb{R}-\{0\})$ a valid space? – user119264 Apr 9 '14 at 9:00
@user119264: it's a valid space, but it won't help - the Lebesgue integral doesn't change if you just remove individual points (or any set of measure 0) – Christoph Apr 9 '14 at 9:20
As Christoph wrote, $L^2(\mathbb R - \{0\}) = L^2(\mathbb R)$ as Hilbert spaces: nothing changes. – Valter Moretti Apr 9 '14 at 11:04