# Pure Point Spectrum implies Spanning Eigenfunctions

If $$H$$ is a self-adjoint operator on a Hilbert space $$\mathcal{H}$$, and the spectrum of $$H$$ is a pure point spectrum, i.e., the spectrum consists of discrete eigenvalues (perhaps with multiplicity $$>1$$), I wish to show that the eigenfunctions of $$H$$ span $$\mathcal{H}$$.

This seems like it should be simple, but I'm having a brain fart. I tried making some sort of degree argument, but this doesn't seem to work. Some of the books I've looked at even have this as the definition of an operator with pure point spectrum, so I must be missing something.

• I think this would more appropriately be asked on math.SE. – joshphysics Nov 19 '14 at 4:29

## 1 Answer

First of all "pure point spectrum", by definition just means that the eigenvectors form a Hilbert basis, so there is nothing to prove. Moreover, unfortunately, this does not mean that the spectrum includes only eigenvalues: There are, in principle, points of the continuous spectrum (the residual one is empty for normal operators as the case under consideration) which appear as limit points of the point spectrum. An example is the operator $$H^{-1}$$ where $$H$$ is the Hamiltonian of the harmonic oscillator. This operator admits $$0$$ as the unique point in the continuous spectrum.

The converse issue, whether or not, for a self-adjoint operator $$T$$ with spectrum made of eigenvalues only there is a Hilbert basis of eigenvectors, admits a positive answer.

The reason is the following. I assume that the Hilbert space $$\cal H$$ is separable. The self-adjoint operator $$T$$ admits a spectral measure $${\cal B}(\mathbb R)\ni E \mapsto P(E)$$, where $${\cal B}(\mathbb R)$$ are the Borel sets in $$\mathbb R$$. Consider the Borel set $$\mathbb R$$ itself. It can be disjointly decomposed as $$\mathbb R = A \cup_{n \in \mathbb N} \{\lambda_n\}$$ where the $$\lambda_n$$ are the distinct eigenvectors, forming a countable set at most since eigenvectors of different eigenvalues are orthogonal and $${\cal H}$$ is separable, thus allowing at most countable orthonormal sets. $$A := \mathbb R \setminus \cup_{n \in \mathbb N} \{\lambda_n\}$$.

Now we know that $$P(\{\lambda_n\}) \neq 0$$ is the orthogonal projector onto the eigenspace of $$\lambda_n$$, that $$P(A)=0$$ because $$A$$ has no intersection with the spectrum of $$T$$, and that the measure $$P$$ is $$\sigma$$-additive. Consequently $$P(\mathbb R) = P\left( A \cup_{n \in \mathbb N} \{\lambda_n\}\right) = \sum_{n \in \mathbb N} P(\{\lambda_n\})\:,$$ the sum is computed with respect to the strong operator topology. On the other hand, we know from the definition of spectral measure, that $$P(\mathbb R)=I$$. We end up with $$I = \sum_{n \in \mathbb N} P(\{\lambda_n\})$$ that is, for every $$\psi \in \cal H$$: $$\psi = \sum_{n \in \mathbb N} P(\{\lambda_n\})\psi$$ This is just one of the possible ways to say that a set of eigenvectors of $$T$$ forms a Hilbert basis of $$\cal H$$.