When do the eigenkets of an operator form a(n) (orthogonal) basis for the Hilbert space? Is it always the case when the operator is Hermitian? Does the operator need to be Hermitian?

  • $\begingroup$ Not necessarily. Eigenvectors belonging to different eigenvalues are orthogonal, i.e., eigenvalues can be degenerate. But they can be make orthogonal. Hermitian implies the eigenvalues are real. What does "a(n) (orthogonal)" mean? $\endgroup$ – Cinaed Simson Apr 5 at 1:30
  • 1
    $\begingroup$ @CinaedSimson "a basis" or "an orthogonal basis" $\endgroup$ – PiKindOfGuy Apr 5 at 11:37
  • 1
    $\begingroup$ While this question might technically be more suited for Math Stack Exchange, I think a Physics-oriented answer is what the OP is looking for and it is not in dire need of being closed as off-topic. $\endgroup$ – Dvij Mankad Apr 8 at 0:44

Restricting to finite-dimensional Hilbert spaces, you have a complete basis of orthonormal eigenkets if and only if the operator is normal, meaning that it commutes with its adjoint, $[A, A^\dagger] = 0$.

In particular, Hermitian operators are normal since they are equal to their adjoints, anti-Hermitian operators are normal, and unitary operators are normal since $[A, A^\dagger] = [A, A^{-1}] = 0$. That covers most options you see in quantum mechanics classes.

In the infinite-dimensional case, there are many more subtleties, which I honestly don't have the slightest understanding of. However, I've never seen a realistic setup where the subtleties cause one to get the wrong physical answer, as long as you follow your nose.

  • 1
    $\begingroup$ What about an infinite-dimensional Hilbert space? Also, you gave a sufficient condition, but you didn't specify if it was necessary. $\endgroup$ – PiKindOfGuy Apr 4 at 23:39
  • $\begingroup$ Furthermore, you didn't address the orthogonality condition. $\endgroup$ – PiKindOfGuy Apr 4 at 23:40
  • $\begingroup$ @PiKindOfGuy I edited! $\endgroup$ – knzhou Apr 4 at 23:44

Does the operator need to be Hermitian?

No, the operator doesn't need to be Hermitian. Here is an example.

Is it always the case when the operator is Hermitian?

No, the operator also needs to be compact. In the finite case, however, you don't need this extra hypothesis.

When do the eigenkets of an operator form a basis for the Hilbert space?

When is this basis orthogonal?

In general, the Spectral Theorem (at least, the version useful to the study of Hilbert spaces in quantum mechanics) tells us that if an operator is compact and normal, its eigenkets will form an orthonormal basis for the Hilbert space (see Kowalski's Spectral Theory, p. 20). I don't know if one could find a different version of the Spectral Theorem, with weaker hypotheses, that doesn't give you orthonormality, but I can't see why that would be useful.

  • $\begingroup$ In mathematical terms, one would say that it is a sufficient condition for a linear operator on a Hilbert space to be normal and compact in order to have orthogonal eigenvectors spanning as a basis the Hilbert space, but it not a necessary one. For example, the 1D harmonic oscillator has an unbounded self-adjoint Hamiltonian with orthonormal eigenvectors spanning the Hilbert space, so it is not a compact one, albeit a normal one. $\endgroup$ – DanielC Apr 11 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.