I'm currently taking a quantum mechanics course. We have proven that hermitian operators always have real eigenvalues, that we can choose the eigenvectors to be orthonormal, and that finite dimensional hermitian operators are diagonalizable (i.e., admit a complete basis of eigenvectors). Can this last result be generalized to the infinite dimensional case? The standard proof seems to use induction on the dimension of the operator, so this proof certainly doesn't carry over.


Yes and no. No, because hermitian operators need not have eigenvalues at all! However, if you assume that your hermitian operator is compact, then the result is true.

The correct approach to generalising this fact to infinite dimensions is by using the spectral theorem, where you replace summation with respect to eigenvalues corresponding to orthogonal eigenspaces by integration with respect to the spectral measure.

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    $\begingroup$ It's worth noting that in the case of operators in infinite-dimensional Hilbert spaces, the spectrum doesn't purely consist of Eigenvalues, but there's the continuous spectrum where you don't have a corresponding eigenvector (which is one of the reasons why the spectral theorem doesn't work with sums). $\endgroup$
    – Roland
    Dec 3 '15 at 9:40

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