[Math] Do infinite dimensional Hermitian operators admit a complete basis of eigenvectors

eigenvalues-eigenvectorslinear algebraquantum mechanics

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.

Best Answer

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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