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.
Answer
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.
No comments:
Post a Comment