# Can an Hermitian unitary matrix in a Hilbert-rigged space be 3-dimensional?

While studying a couple of concepts, I've understood the following premises to be true:

1. Hermitian unitary matrices eigenvalues are unimodal (that's $$\pm1$$).

2. In physics, operators/observables are represented by Hermitian matrices, its eigen(functions/vectors) forming a complete orthonormal base.

3. In physics, each eigenvalue corresponds to a unique eigenfunction. If it happens to not be the case, it can be made so it is (according to Griffiths at least).

Therefore, it should follow that a Hermitian unitary matrix $$\hat{Q}$$ can only be 1 or 2-dimensional... right? If its eigenvalues can only be unimodal, and there can only be one of each, 2-dimensional self-adjoint unitary matrices are the only ones that can satisfy these constraints.

$$\hat{Q}(\varphi + \psi) = \pm \varphi + \mp \psi$$

for $$\varphi,\psi$$ orthonormal would be the only way in which $$\hat{Q}$$ could act.

Am I missing something? Or is this fact that makes spin operators so beautifully unique?

• premise 3 seems to me to be not-true, as it precludes all degeneracies. This would be a great problem not only for unitary Hermitian operators but also for operators like angular momentum $L^2$... Can you bring the relevant part in Griffiths?
– user275556
Commented Dec 7, 2021 at 11:34

You are misinterpreting point 3. Indeed, the idea that all eigenspaces are one-dimensional is contradicted by the existence of the identity operator (on any space), which is both Hermitian and unitary and whose $$+1$$ eigenspace is the entire Hilbert space on which it acts.