On the spectrum of a quantum mechanical system

Question

Can the spectrum of a quantum mechanical operator contain both real and complex numbers?

Answer 1

The spectrum of an operator can be complex but not the spectrum of an observable, pretty much by definition. Consider a two state system with the operator

$$A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$

This operator has eigenvalues $\pm i$, which obviously are complex. But it is not an observable, because an observable by definition has to be self-adjoint, and $A$ isn't; we demand that an observable be self-adjoint precisely because it guarantees that all eigenvalues will be real.

Edit: I just realized you might be asking whether a single operator can have both real and non-real eigenvalues. In that case,

$$B = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

has spectrum $\{1, i, -i\}$.

Answer 2

I would think the boson annihilation operator can be called a "quantum mechanical operator", and it has both real and complex eigenvalues.