# Eigenvalues, Hermitian operators and observables in quantum mechanics

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?

• With an infinite set of basis vectors, a gives vector is a linear combination of infinitely many basis vectors. I’m not sure whether that’s well-defined, see e.g. here en.wikipedia.org/wiki/Basis_(linear_algebra)#Definition Commented Oct 16, 2017 at 21:25

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.

• 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? ;) Commented Oct 16, 2017 at 22:31
• One might argue that "Hermitian" (what is actually meant here is probably "self-adjoint") already implies that we are talking about Hilbert spaces. Commented Oct 17, 2017 at 21:46