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For a coherent state, write $a|\alpha ⟩=\alpha |\alpha ⟩$ and $|\alpha ⟩={\sum }_n{c}_n|n⟩$ 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 $⟨(a^{†}{)}^k{a}^k⟩={\mu }^k$. This factorization is the defining virtue of coherent states for optical calculations.^[ref]

As an original comparison, choose $\mu =2$. Then $p_0=e^{-2}$, $p_1=p_2=2e^{-2}$, $\mathrm{Var}N=2$, and $⟨N(N-1)⟩=4$. A two-photon Fock state also has mean two but has $\mathrm{Var}N=0$ and pair moment two. ConsequentlyIn particular, $|1⟩$ cannot supply a pair. Nevertheless, photon statistics alone does not determine a quantum state: averaging $|\alpha ⟩⟨\alpha |$ over a uniform quantum phase removes its off-diagonal entries while retaining exactly the same $p_n$ and these equal-time moments.

Photon statistics is the probability distribution of the number operator in specified field modes, or the distribution of registered counts in a specified gate. Those distributions agree only for an appropriate ideal measurement. Mode selection, losses, and detector dead time matter. For a one-mode density matrix, $p_n=⟨n|\rho |n⟩$ andThese two sums already distinguish preparations with the same expected value.

Quantum optics studies light using quantum states of the electromagnetic field and their interaction with matter. Photon statistics asks how detections fluctuate; the quantum theory of optical coherence relates those detections to field correlations. A coherent state, thermal light, and a Fock state can have the same mean photon number yet different pair counts. This distinction makes photon counting useful for testing sources intended to emit one photon at a time.