Past exam of the advanced quantum physics course of the University of Cambridge / 2012 / Paper 1 / 1 / c / Solution

The coherent state satisfies $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$ and $⟨a^{†}a⟩=|\alpha {|}^2$. The canonical commutation relation adds the vacuum term $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. Thereforeand the coherent-state electric-field variance isHere $E_0$ is taken positive as the field-amplitude scale. The coherent state displaces the mean field without changing this vacuum optical quadrature noise.