Consider a hermitian operator. So

a) in a space of infinite dimension its eigenvectors are a base.

b) in a finite-dimensional space the matrix that represents the hermitian operator is always diagonalizable.

c) 2 eigenvectors corresponding to different eigenvalues are collinear.

d) in a finite N-dimensional space there are N linearly dependent eigenvectors

I have to justify what is true and why it is true, and why the others are false. My teacher said that the true answer was b).

I learned in Quantum Mechanics that a Hermitian operator has always real eigenvalues. The operator is diagonalizable and the values of the diagonal are its eigenvalues.

An observable is a Hermitian operator whose eigenvectors constitute an orthonormal basis for the space E, even if it is of infinite dimension.

In my opinion both a) and b) are correct. I do not understand why a) is wrong.

What would be the answer if in the statement, instead of considering a Hermitian operator, we consider an observable?


1 Answer 1


First of all, you are correct that b) is correct, and c) is wrong.

d) is a bit weird - if $\vec v$ is an eigenvector, then so are $2 \vec v$, $3 \vec v$, ..., $N \vec v$ - those are $N$ linearly dependent eigenvectors. However, the intention of the questioner is obviously for you to realize that the "correct" statement would be "there are $N$ linearly independent eigenvectors".

The problem with a) is that operators can be a bit weird in infinite-dimensional spaces and sometimes it is not a good idea to think about them as matrices. The maybe easiest counterexample to a) is the position operator (on $L^2(\mathbb R)$) $$ (\hat x \psi)(x) = x \psi(x) . $$ $\hat x$ is Hermitian (with the right domain), but it does not have any eigenvectors: $x\psi_\lambda(x) = \lambda\psi_\lambda(x)$ implies that $\psi_\lambda(x) = 0$ whenever $x \neq \lambda$, so $\psi_\lambda(x) = 0$ almost everywhere.

In Physics lectures, you will work with "generalized eigenvalues" $\lambda \in \mathbb R$ that correspond to "generalized eigenfunctions" $\psi_\lambda(x) = \delta(x - \lambda)$, but note that $\psi_\lambda$ is not in the Hilbert space $L^2(\mathbb R)$. Mathematicians will tell you that the spectrum of $\hat x$ is not discrete, but continuous. In any case, the claim that "the eigenvectors are a basis of the Hilbert space" is wrong.

  • $\begingroup$ By a suitable extension of the "hermitian" notion to rigged Hilbert spaces, one can show that the spectral theorem of Gelfand-Kostyuchenko-Maurin provides a confirmation of: "Consider a hermitian operator. So // a) in a space of infinite dimension its eigenvectors are a base." P.S. Nowhere in the problem, nor in the text written by the user did she/he mention "Hilbert space", so why bring it up? ;) $\endgroup$
    – DanielC
    Oct 16, 2017 at 22:31
  • $\begingroup$ One might argue that "Hermitian" (what is actually meant here is probably "self-adjoint") already implies that we are talking about Hilbert spaces. $\endgroup$
    – Phoenix87
    Oct 17, 2017 at 21:46

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.