What are the conditions that ensure that a linear operator on a separable Hilbert space has a discrete spectrum and its eigenvectors form a basis?

Question

I am particularly interesed in answers such as:

• symmetric, bounded, positive.

• self-adjoint and bounded

• essentialy self-adjoint and positive.

(Those were invented)

I'm interested in this question in the frame of Quantum mechanics.

Thanks a lot!

Answer 1

If the eigenvectors of $T$ form an orthonormal basis, the operator is necessarily normal. So, a necessary condition is that $T$ is densely defined and normal.

Since $Te_n-\lambda_ne_n$ for all eigenvectors $e_n$ and these form a basis, we have that $T=\sum_n\lambda _nP_n$, where the $P_n$ are projections. This doesn't characterize discrete spectrum, though, as the sequence $\{\lambda_n\}$ could have accumulation points.

For sufficient conditions, one could have

• $T$ is normal and finite-rank.

• $T$ is a linear combination of projections.

If we look for unbounded operators, the spectrum will have to be an unbounded sequence with no accumulation points. If $\lambda_0\in\mathbb C$ is not in $\sigma(T)$, then $\lambda_0 I+T$ is invertible, and the spectrum of $(\lambda_0 I+T)^{-1}$ consists of a sequence $\{\lambda_n\}$ with $\lambda_n\to0$. This gives us another sufficient condition:

• For any $\lambda_0\not\in \sigma(T)$ the operator $(\lambda_0 I+T)^{-1}$ is normal and compact.

The above condition is still not necessary: given a sequence of pairwise orthogonal infinite projections $\{P_n\}$, we can construct $$ T=\sum_n n\,P_n. $$

Answer 2

For Quantum, a common condition on the Hamiltonian $H$ would be selfadjoint with compact resolvent. The Hamiltonian is typically unbounded, but the resolvent $R(\lambda)=(H-\lambda I)^{-1}$ is bounded by definition, and defined for $\lambda\notin\mathbb{R}$. For $\mu,\lambda\notin\sigma(A)$, the resolvent equation holds: \begin{align} R(\lambda)&=R(\mu)+(\lambda-\mu)R(\lambda)R(\mu) \\ &=\{ I+(\lambda-\mu)R(\lambda)\}R(\mu) \end{align} Because of this, if the resolvent is compact for some $\mu$, then it is compact for all $\lambda$. A compact resolvent has eigenvalues that can cluster only at $0$, which translates to eigenvalues for $H$ that can cluster only at $\infty$.

The Hamiltonian for the reduced mass Hydrogen isotope does not have a compact resolvent because its eigenvalues cluster at $0$ from below, and above that there is continuous spectrum. The Harmonic oscillator Hamiltonian has compact resolvent.