What does expectation value being equal to zero have to do with the condition of being a state of an operator

Question

In Gerry & Knight's Introduction to Quantum Optics book under the "Quantum fluctuations of a single-mode field" section it is claimed that the number state $|n>$ is not a state of well defined electric field. This claim makes sense to me since the number operator does not commute with the electric field operator
$$
\left[\hat{n}, \hat{E}_{x}\right]=\mathcal{E}_{0} \sin (k z)\left(\hat{a}^{\dagger}-\hat{a}\right) \neq 0
$$
as a result, these two operators are not simultaneously diagonalizable meaning there cannot share the same eigenvectors (hope this is indeed true). But then the book puts a reasoning which tells that number state is not a state of the well-defined electric field because the expectation value (or mean) is zero,
$$
\left\langle n\left|\hat{E}_{x}(z, t)\right| n\right\rangle=\mathcal{E}_{0} \sin (k z)\left[\langle n|\hat{a}| n\rangle+\left\langle n\left|\hat{a}^{\dagger}\right| n\right\rangle\right]=0
$$
I couldn't understand what does expectation value being equal to zero have to do with the condition of being a(n) (eigen?-) state of an operator?

Answer 1

You may be reading too much into this remark: the point is that, for any number state, the electric field that you measure will be zero. So then if you know you have a nonzero electric field, you shouldn't claim the field corresponds to a number state! This means that states with definite energy do not have definite electric fields, in contrast to the classical description in which you can specify the exact energy of an electric field.

The connection that you will see is that it is rather nice to have eigenstates of operators you're trying to measure, because then you're guaranteed to find a specific result for your measurement and so you can associate that measurement result with the value of the observable. So you will eventually look for eigenstates of the electric field operator (or really its "positive frequency" component) that you'll be able to say correspond to classical electric fields with some definite parameters.