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? 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!
 A: 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 
$$
T=\sum_n n\,P_n.
$$
A: 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.
