= Solution
Put $b=\mathcal E e^{i(kr-\omega t)}$, allowing $\mathcal E$ to be complex. The Hermitian <electric field> is $\hat E=b\hat a+b^*\hat a^\dagger$. The <annihilation operator> and <creation operator> change <photon number>, so their diagonal matrix elements in a <Fock state> vanish. Since $\hat a\hat a^\dagger=\hat n+1$ and $\hat a^\dagger\hat a=\hat n$,
$$
\boxed{\langle\hat E\rangle_n=0,\qquad
\langle\hat E^2\rangle_n=(2n+1)|\mathcal E|^2=(\Delta E)^2_n.}
$$
The terms containing $\hat a^2$ or $\hat a^{\dagger2}$ connect orthogonal <number states> and contribute zero. Even as $n$ grows, there is no definite oscillating mean <electric field>; its quadrature fluctuations grow instead. A <Fock state> can represent fixed field <energy>, but not a classical sinusoidal wave of prescribed <wave phase>. This is distinct from representing a classical field with a randomly averaged phase.
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