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Are there some criteria to tell when the point spectrum of a linear operator is discrete? In general it is not the same (take the spectrum of the "annihilation" operator). More specifically, what are the conditions that should be satisfied by a symmetric (or even self-adjoint) operator to have "point spectrum" = "discrete spectrum"?

glS
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Aleksandr
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When dealing with differential operators, the most common condition is where the resolvent operator is compact. Resolvents of differential operators are often compact on bounded domains because they map uniformly bounded sets of continuous functions to sets of functions with uniformly bounded derivatives, which means the image is an equicontinuous set of functions. Some argument of that type is usually given to prove a resolvent is compact. Showing compactness is trickier on infinite domains, and the resolvent need not be compact, which is the case for the simplest differential operator, the momentum operator.

Not all operators with point spectrum have some compact resolvent, though. When considered on $L^{2}$, the Fourier transform has 4 eigenvalues and a complete orthonormal basis of eigenfunctions (the Hermite functions.) If the dimensions of any of the eigenspaces is infinite, then an operator cannot be compact, unless the eigenvalue is $0$. Or, if the eigenvalues cluster somewhere other than at $0$, the operator cannot be compact.

If the resolvent $R(\lambda) = (A-\lambda I)^{-1}$ of an operator $A : \mathcal{D}(A)\subset \mathcal{H}\rightarrow\mathcal{H}$ on a separable infinite-dimensional Hilbert space $\mathcal{H}$ is compact for some $\lambda$, then the eigenvalues cluster only at $0$, which forces the eigenvalues of $A$ to cluster only at $\infty$.

Disintegrating By Parts
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    An even easier example of an operator with discrete point spectrum with no compact resolvent is $I$. – Nate Eldredge Mar 16 '15 at 16:22
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    @NateEldredge : Even easier, and can't be simpler. :) – Disintegrating By Parts Mar 16 '15 at 16:35
  • Thanks! I kind of knew about the criterion based on compactness, but the answer was still very useful. But let me make sure that I got it right: There is no if and only if condition for the discrete spectrum and one has to analyze the different classes of operators? – Aleksandr Mar 16 '15 at 17:56
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    For Quantum non-Relativistic Hamiltonians, there are quite a few results, and some of them very simple because pure discrete just means that all states are bounds states. You can pretty much guess what potentials would give you only bound states. Continuous spectrum occurs when the energies are high enough to escape all bound states. General conditions leading to discrete spectrum only are not going to be easy once you move beyond the low-hanging fruit of compactness or move beyond looking at specific applications. I'm not aware of anything that matches what you want. – Disintegrating By Parts Mar 16 '15 at 18:38