= Coherent-state electric-field variance
{title2=$\operatorname{Var}E=E_0^2$}
For one optical mode, $E=E_0(a+a^\dagger)$ and $[a,a^\dagger]=1$. A <coherent state> obeys $a|\alpha\rangle=\alpha|\alpha\rangle$, so <normal ordering> gives $\langle E^2\rangle=E_0^2[(\alpha+\alpha^*)^2+1]$. Subtracting the squared mean leaves $E_0^2$, independent of the coherent displacement. The extra one is the <canonical commutation relation>, not noise from a random coherent amplitude. A rotated <optical quadrature> has the same coherent-state variance with the same amplitude normalization.
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