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

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?

• Just because two operators don't commute, doesn't mean you can't measure their expectation value. Feb 18, 2022 at 16:47
• That is correct, I should remove that part. Feb 18, 2022 at 16:49
• Presumably you want the expectation value of the electric field operator to equal the electric field you would measure classically.
– d_b
Feb 18, 2022 at 16:59