 Coherent-state electric-field variance

For one optical mode, $E=E_0(a+a^{†})$ and $[a,a^{†}]=1$. A coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$, so normal ordering gives $⟨E^2⟩=E_0^2[(\alpha +{\alpha }^{\ast }{)}^2+1]$. 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.