# If momentum eigenstates are not in $L^2$, how can they form a basis for it?

When your matrices are all finite, the eigenvectors of a self-adjoint matrix $\mathbf{A}$ form an orthogonal basis for some space. It's almost too trivial to mention that each basis vector is, indeed, inside of the space that it is part of a basis for.

The momentum operator $\mathbf{\hat p}$ is self-adjoint, and also has a spanning basis of eigenfunctions that span Hilbert space. However, none of the eigenfunctions of $\mathbf{\hat p}$ are square-integrable. What's going on?

## 1 Answer

The momentum operator $-i\hbar \nabla$ is indeed an unbounded self-adjoint operator on the hilbert space $L^2(\mathbb{R}^d)$, with domain $H^1(\mathbb{R}^d)$, but it has purely continuous spectrum.

Operators with purely continuous spectra do not possess any eigenvector belonging to the hilbert space, and thus of course they do not possess a basis of eigenvectors. This is one of the differences between finite and infinite dimensional Hilbert spaces.

In the infinite dimensional case, only operators that are either compact (like density matrices) or with compact "inverse" (resolvent, more precisely, like the harmonic oscillator) have a basis of eigenvectors.