Eigenvectors of Hermitian operators are orthogonal ↩

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created: 2022-01-05 15:05:03
modified: 2022-01-10 04:13:04

So they can be made orthonormal by setting their magnitudes to 1:

Proof

Consider these eigenvalue equations:

Then take the inner product of the second equation with :

Here we used the Hermitievity of . Then from the first and last part:

If then the only possibility is:

Meaning that they are orthogonal.

In the degenerate case where there are multiple eigenvectors corresponding to the same eigenvalue, we investigate the degenerate subspace of the given eigenvalue. The operator then will of course be a Hermitian operator on this subspace as well. Hermitian operators can be brought into a diagonal form by choosing the right basis, in which these vectors will be orthogonal.

QED