Can anyone give an non-trivial example of compact operators in quantum mechanics? Of course, any operator on a finite-dimensional Hilbert space is compact.

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    $\begingroup$ This seems like a list question; additionally, it is not clear what your notion of "non-trivial" is. E.g. projectors are clearly compact, as are all those finite-dimensional operators you already mention. Why do you seek such compact operators? $\endgroup$
    – ACuriousMind
    Jan 7, 2017 at 14:20
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    $\begingroup$ Projectors onto infinite dimensional subspaces are never compact! $\endgroup$ Jan 7, 2017 at 14:25
  • $\begingroup$ @ACuriousMind: This indeed seems like a list question. $\endgroup$
    – Qmechanic
    Jan 7, 2017 at 18:38

4 Answers 4


The point is that compact operators are first of all bounded and normal (self-adjoint in particular) bounded operators have bounded spectrum. In QM, the spectrum is the set of possible values of the observable represented by the operator (if self-adjoint). So the principal obstruction to find physically meaningful compact operators in QM is the fact that almost all important observables in QM may attain infinitely large values, just excluding observables related to the Lie algebra of compact groups (first of all $SU(2)$ and the spin observables) whose representations are always sums of finite dimensional irreducible representations in view of Peter-Weyl theorem.

The second obstruction is that the spectrum of these operators must be a point spectrum with $0$ at most as unique element of continuous spectrum part. Again, generators of the Lie algebra of compact Lie groups seem to be the only natural chance.

Looking for a compact Hamiltonian operator for instance, one should first of all looks for a physical system whose energies are bounded and discrete. It is not so easy for standard systems. A way out could be to separate the spectrum into two parts if this separation has some natural physical meaning, and to consider only the operator obtained by restricting the values to one part only. For instance a Hamiltonian which admits an infinite discrete set of negative eigenvalues $E_n$ accumulated by $E=0$ and bounded below and such that, for positive values the spectrum becomes continuous to accommodate scattering processes.

A candidate is the Hamiltonian of the hydrogen atom. Unfortunately the degeneracy of the negative levels of energy $E_n$ (due to the admitted increasing values of the eigenvalues of $L^2$ and $L_z$ for a fixed $n$) increases too rapidly to produce a compact operator when restricting the energies between the ground state $E_0$ and the threshold for scattering processes $E_\infty =0$. I believe that it should be however possible to construct such models where the restriction to the Hamiltonian to the discrete part of the spectrum is compact. In particular it should be possible dealing with one-dimensional systems.

Another strategy is to consider more complicated real functions of observables, which in a sense are observables as well. A typical example is the inverse $H^{-1}$ of the Hamiltonian operator $H$ of the harmonic oscillator. Here the spectrum is a pure point spectrum included in $(0, 1/E_0]$, the degeneracy of each eigenspace is absent (thus constant) and $H^{-1}$ can be approximated by a set of finite rank operators in the uniform norm just in view of the shape spectrum of the operator: $H^{-1}$ is compact.

Abandoning the request that the operator must represent an observable, there is an important case. Every density matrix (trace-class, unit-trace, positive operator) is always compact even if it is an incoherent superposition of an infinite class of pure states. This is just because trace-class operators are compact. The bound on the spectrum is given here by a pair of requirements: positivity (thus the spectrum is bounded below and the operator is self-adjoint) and finite trace (thus the spectrum is bounded above since the trace is the sum of eigenvalues).

The statistical operators describing quantum systems at thermal equilibrium confined in a box, e.g., $$\rho_\beta = \frac{e^{-\beta H}}{Z_\beta}$$ are in particular compact operators even if the Hamiltonian $H$ used to construct those statistical operators is not compact nor bounded above.

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    $\begingroup$ Congratulations on making what's possibly the mathsiest P.SE question to ever make the Hot Network Questions sidebar. $\endgroup$ Jan 7, 2017 at 21:01
  • $\begingroup$ @Emilio Pisanty Sorry, Emilio, I do not understand what you wrote: "Hot Network Question sidebar" ? $\endgroup$ Jan 7, 2017 at 21:12
  • $\begingroup$ That's that big list of questions on the right →, with posts from all SE sites (and also shown across all SE sites). $\endgroup$ Jan 7, 2017 at 21:16
  • $\begingroup$ I see, very funny! $\endgroup$ Jan 7, 2017 at 21:18
  • $\begingroup$ Thanks a lot! From your answer, I have the impression that compact operators are quite rare. This is annoying since in many books on Hilbert space, compact operators occupy quite a large portion. $\endgroup$
    – S. Kohn
    Jan 8, 2017 at 0:58

Compact operators often appear in integral equations and can be viewed as continuous generalizations of matrices, where the corresponding integral kernel must not be to singular and must decay fast enough at infinity.

An example is the Lippmann–Schwinger equation in quantum scattering theory, see https://en.wikipedia.org/wiki/Lippmann%E2%80%93Schwinger_equation .

  • $\begingroup$ Indeed I forgot to mention your case in my answer. The compact operator is assumed to be $(H-E)^{-1} \frac{V}{E-H_0 \pm i\epsilon}$ in the formula you quoted, where $H= H_0+V$, when dealing with "short range perturbations"... $\endgroup$ Jan 7, 2017 at 19:12

All the normal quantum states of a given W* algebra of quantum observables are represented as compact (actually trace class) operators on a given Hilbert space (where the algebra of observables is represented).

In other words all the usual states of quantum mechanics in the Schroedinger representation, i.e. density matrices (both pure and mixed) are compact operators.

The Hamiltonians describing confined particles are not compact, but have compact resolvent. However one might argue that the resolvent of an observable is not strictly a physical quantity.


For the Harmonic oscillator (or any system with a discrete spectrum), I am pretty sure the "resolvent" operator $$\hat R(E) = \frac{1}{E - \hat H}\;,$$ where $E$ is not one of the eigenvalues of $\hat H$, is compact. The reason is that the spectrum of this operator is bounded (take the two closest eigenvalues to $E$) and is therefore a compact set. The definition of a compact operator (that the sequence $\hat A f_n$ has a convergent subsequence) should hopefully follow.

I am willing to guess that for a continuous spectrum, the operator $$ \frac{1}{E-\hat H + i\epsilon}$$ (where $\epsilon$ needs to be taken to zero after the operation) is also compact, but I'm not sure how to prove it. I think it's common for Green's functions and integral kernel operators, etc. to be compact.

This is similar to jjcale's answer.


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