I always heard the eigenfunctions of a self-adjoint operator form a complete basis. Where can I find a proof in infinite dimension space? Presumably readable for physicists.
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2$\begingroup$ This is a subtle question. Infinite-dimensional self-adjoint operators often have continuous spectrum so one has to generalize the notion of a basis - to a basis normalized to the Dirac delta-function - and they wouldn't be considered bases in most terminologies/axiomatic systems preferred by mathematicians. Physicists know how to use these things and all the rules, including the existence of the basis, follow from a careful generalization of the finite-dimensional case. $\endgroup$– Luboš MotlCommented Jun 22, 2013 at 17:10
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$\begingroup$ You may always regulate the system so that it becomes finite-dimensional, and then the proofs for the finite-dimensional case become usable - the essence of the proof, as accepted by physicists, is the same for the finite- and infinite-dimensional case. The limiting procedure of the regularization is able to produce new effects, but only when it comes to the number of eigenstates and eigenvalues which may get continuous (or mixed). $\endgroup$– Luboš MotlCommented Jun 22, 2013 at 17:12
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2 Answers
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The statement is simply false as it stands when adopting the standard Hilbert space formulation of QM. The true statement is that a self-adjoint operator with pure point spectrum admits a Hilbert basis made of eigenvectors. (It happens in particular, but not only, when either the operator is compact or its resolvent is.) The proof is not so simple and is a particular case of the general spectral decomposition theorem. A proof can be found in classical books on spectral theory like Rudin's functional analysis book, R&S or Prugovecki's book on mathematical foundation on QM. Even the original and always wonderful von Neumann's book contains a proof (I think the most readable for physicists since there all rigorous spectral theory was invented for the firs time).
The OP's statement admits another interpretation, in the sense of Gelfand's rigged Hilbert space theory. In that case the presence of a continuous part of the spectrum is admitted. That approach is much more intuitive than the pure Hilbert space one but, conversely, is extremely more technical mathematically speaking, also because it needs further topological hypotheses to be established than the standard ones of a self-adjoint operator in a Hilbert space. I do not know a book where a complete proof appears and I suspect that there is not (while I am sure that the statement is correct). The most complete proof I know can be grasped from several propositions in Gelfand Vilenkin's books on generalized functions theory. Perhaps the forth volume. However, some statements about some properties of some relevant measures necessary to achieve the final decomposition statement appear without proof therein.
answered Aug 24, 2014 at 9:58
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$\begingroup$ Yes, the theorem is in the fourth volume of Gelfand--Vilenkin as stated here: mathoverflow.net/a/126758. However, the proof there is not entirely satisfactory as stated there in the comments: mathoverflow.net/questions/125655/…. $\endgroup$ Commented Oct 24, 2023 at 6:13
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The operator has to be bounded and compact, or unbounded with compact resolvent for that to be true. You can consult the section on compact operators (chapter VI, Book 1) of Reed and Simon "Methods of modern mathematical physics".
