= Coherent-state photon statistics
{title2=$p_n=e^{-\mu}\mu^n/n!,\quad g^{(2)}(0)=1$}
For a <coherent state>, write $a|\alpha\rangle=\alpha|\alpha\rangle$ and $|\alpha\rangle=\sum_n c_n|n\rangle$ in <number states>. The <annihilation operator> gives $c_{n+1}\sqrt{n+1}=\alpha c_n$, so $c_n=c_0\alpha^n/\sqrt{n!}$. Normalization yields $|c_0|^2=e^{-|\alpha|^2}$. Thus <photon statistics> follows a <Poisson distribution> with $\mu=|\alpha|^2$, and the <normal ordering> of every moment gives $\langle(a^\dagger)^ka^k\rangle=\mu^k$. This factorization is the defining virtue of <coherent states> for optical calculations.<Coherent and Incoherent States of the Radiation Field by Roy Jay Glauber>[1963, §III, equations (3.1)–(3.7): coherent-state expansion; Universidade Estadual de Campinas mirror]{ref}
As an original comparison, choose $\mu=2$. Then $p_0=e^{-2}$, $p_1=p_2=2e^{-2}$, $\operatorname{Var}N=2$, and $\langle N(N-1)\rangle=4$. A two-<photon> <Fock state> also has mean two but has $\operatorname{Var}N=0$ and pair moment two. Consequently
$$
g^{(2)}_{\rm coherent}(0)=1,\qquad
g^{(2)}_{|n\rangle}(0)=1-\frac1n\quad(n\geq1).
$$
In particular, $|1\rangle$ cannot supply a pair. Nevertheless, <photon statistics> alone does not determine a <quantum state>: averaging $|\alpha\rangle\langle\alpha|$ over a uniform <quantum phase> removes its off-diagonal entries while retaining exactly the same $p_n$ and these equal-time moments.
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