 Solution

Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the Fock states givesActing with the annihilation operator and shifting the summation index givesThus a coherent state is an annihilation-operator eigenket. It is not also an eigenket of the creation operator: a common eigenket would make $[\hat{a},{\hat{a}}^{†}]$ annihilate it, contradicting $[\hat{a},{\hat{a}}^{†}]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relationWith $\hat{n}={\hat{a}}^{†}\hat{a}$, the photon number expected value is $\mu =|\alpha {|}^2$. Normal ordering ${\hat{n}}^2=({\hat{a}}^{†}{)}^2{\hat{a}}^2+{\hat{a}}^{†}\hat{a}$ gives $⟨n^2⟩={\mu }^2+\mu$. ThereforeAt large mean photon number the relative fluctuations vanish, giving a classical-field limit even though quantum fluctuations remain. For the vacuum $\mu =0$, the relative ratio itself is undefined.

The Born rule applied to the number-state coefficient gives the Poisson distributionIt depends on the single parameter $\mu =⟨n⟩$, and has equal expected value and variance, consistent with the preceding moments.

For a Fock state $|n⟩$, raising or lowering the number makes an orthogonal state, so the optical quadrature has $⟨\hat{X}⟩=0$. Expanding its square givesThe first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature variance $1/2$.