Can expectation value be imaginary?

Question

I was solving a problem and the result of the expectation value of an operator came out to be $-\frac{\hbar}{4}$ $i$. Is this result possible? It seems counter intuitive.

Answer 1

If $A$ is self-adjoint, you can define $f(A)$ as a complex-valued observable, where $f: \mathbb R \to \mathbb C$ is a measurable complex-valued function: $$f(A) := \int_{\sigma(A)} f(x) dP^{(A)}(x)\:,$$ $P^{(A)}$ being the spectral measure (projector-valued) of $A$. $N= f(A)$ is a closed normal operator and admits a spectral decomposition $P^{(N)}$ supported on the spectrum $\sigma(N)\subset \mathbb C$ of $N$ and constructed out of $P^{(A)}$ and $f$. It is possible to define the expectation value of $N$ referred to a pure state represented by a normalized vector $\psi$ belonging to the domain $D(N)$ of $N$ $$\langle N\rangle_\psi := \int_{\sigma(N)} z d\langle \psi| P^{(N)}(z) \psi \rangle = \langle \psi | N \psi \rangle$$ since $\sigma(N)$ includes complex numbers, in general, $\langle N\rangle_\psi$ is a complex number.

Summing up, complex expectation values exist as soon as you define complex-valued observables. There is no mathematical obstruction in doing it. Nevertheless you can always decompose a complex observable $N$ into a pair of standard real (self-adjont) observables (mutually compatible), $(N+N^*)/2$ and $-i(N-N^*)/2$ and use the standard theory (paying attention to some subtleties concerning domains). This way is easier and, I guess, it is the reason why complex observables do not appear very often in the literature.

Answer 2

If the operator is not self-adjoint then this is a possibility. If you search on phys.SE you will find questions about the expectation value of xp in the case of the QHO, and this turns out to be imaginary