= Solution
Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the <Fock states> gives
$$
1=|k|^2\sum_{n=0}^{\infty}\frac{|\alpha|^{2n}}{n!}=|k|^2e^{|\alpha|^2},\qquad\boxed{k=e^{-|\alpha|^2/2}.}
$$
Acting with the <annihilation operator> and shifting the summation index gives
$$
\hat a|\alpha\rangle=k\sum_{n\ge1}\frac{\alpha^n\sqrt n}{\sqrt{n!}}|n-1\rangle=\alpha k\sum_{m\ge0}\frac{\alpha^m}{\sqrt{m!}}|m\rangle=\boxed{\alpha|\alpha\rangle.}
$$
Thus a <coherent state> is an annihilation-operator eigenket. It is not also an eigenket of the <creation operator>: a common eigenket would make $[\hat a,\hat a^\dagger]$ annihilate it, contradicting $[\hat a,\hat a^\dagger]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relation
$$
\boxed{\langle\alpha|\hat a^\dagger=\alpha^*\langle\alpha|.}
$$
With $\hat n=\hat a^\dagger\hat a$, the <photon number> <expected value> is $\mu=|\alpha|^2$. Normal ordering $\hat n^2=(\hat a^\dagger)^2\hat a^2+\hat a^\dagger\hat a$ gives $\langle n^2\rangle=\mu^2+\mu$. Therefore
$$
\boxed{\langle n\rangle=\mu,\qquad\Delta n=\sqrt\mu=|\alpha|,\qquad\frac{\Delta n}{\langle n\rangle}=\frac1{\sqrt\mu}\quad(\mu>0).}
$$
At 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 distribution>
$$
\boxed{\Pr(n)=e^{-\mu}\frac{\mu^n}{n!},\qquad n=0,1,2,\ldots.}
$$
It depends on the single parameter $\mu=\langle n\rangle$, and has equal <expected value> and <variance>, consistent with the preceding moments.
For a <Fock state> $|n\rangle$, raising or lowering the number makes an orthogonal state, so the <optical quadrature> has $\langle\hat X\rangle=0$. Expanding its square gives
$$
\hat X^2=\frac12[\hat a^2+(\hat a^\dagger)^2+\hat a\hat a^\dagger+\hat a^\dagger\hat a],\qquad\boxed{\langle n|\hat X^2|n\rangle=n+\frac12.}
$$
The first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature <variance> $1/2$.
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