= Solution
{c}
For the normalized <coherent state>, $\langle a\rangle=\alpha$ and $\langle a^\dagger\rangle=\alpha^*$. The dimensionless <electric field> and <magnetic field> <optical quadratures> consequently have <expectation values>
$$
\boxed{\langle\mathcal E\rangle=\alpha+\alpha^*=2\operatorname{Re}\alpha,\qquad
\langle\mathcal B\rangle=-i(\alpha-\alpha^*)=2\operatorname{Im}\alpha.}
$$
The physical <electric field> and <magnetic field> strengths have the corresponding proportionality factors and mode functions. In free evolution replace $\alpha$ by $\alpha e^{-i\omega t}$ to recover the classical oscillation.
Expanding the squared operators cancels the $a^2$ and $(a^\dagger)^2$ terms:
$$
\mathcal E^2+\mathcal B^2=2(aa^\dagger+a^\dagger a)=4\hat N+2I.
$$
On the other hand, $(\alpha+\alpha^*)^2+[-i(\alpha-\alpha^*)]^2=4|\alpha|^2$. Hence
$$
\boxed{\langle\mathcal E^2+\mathcal B^2\rangle=4|\alpha|^2+2,\qquad
\langle\mathcal E\rangle^2+\langle\mathcal B\rangle^2=4|\alpha|^2.}
$$
The difference is the sum of the <variances>, not an extra mean field. Indeed $\operatorname{Var}(\mathcal E)=\operatorname{Var}(\mathcal B)=1$, independent of $\alpha$. The <commutator> $[\mathcal E,\mathcal B]=2iI$ requires $\sigma_{\mathcal E}\sigma_{\mathcal B}\geq1$, and the <coherent state> attains equality. The residual two units are the <vacuum fluctuations> of the two <optical quadratures>.
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