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It is a well-known result that classical light (which I take here to mean mixtures of coherent states) cannot produce sub-Poissonian photon-counting statistics, with a single beam of coherent light corresponding to a Poissonian photon-counting statistics (as discussed for example here), and other kinds of non-quantum light corresponding to super-Poissonian statistics.

However, I have never seen this fact proven formally. Usually, texts show how some common kinds of classical light, such as thermal light, result in super-Poissonian statistics, and how quantum states can produce sub-Poissonian ones, but they do not tackle the general case.

More specifically, consider a state which is a mixture of coherent states. This corresponds to a photon counting probability $P(n)$ of the form $$P(n)=\sum_\lambda p_\lambda P_\lambda(n),$$ with $\sum_\lambda p_\lambda =1$, and $P_\lambda(n)$ being the Poisson distribution with expected value $\lambda$: $$P_\lambda(n)\equiv e^{-\lambda}\frac{\lambda^n}{n!}.$$ A super-Poissonian distribution is characterised by the property that the variance is greater than the expected value, that is, $\sigma^2\ge\mu$. More precisely, in the considered case this means $$\sum_n(n-\mu)^2P(n)\ge \mu,\quad \mu\equiv\sum_n nP(n).$$ Can this property be shown in full generality, without making reference to specific types of light?

glS
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2 Answers2

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Let us compute the first moments of $P(n)$: $$\mu\equiv\sum_n nP(n)=\sum_n n\sum_\lambda p_\lambda P_\lambda(n)=\sum_\lambda p_\lambda \lambda,$$ where I used the property of the Poisson distribution $\sum_n n P_\lambda(n)=\lambda$. Similarly, we have $$\sum_n n^2 P(n)=\sum_\lambda p_\lambda \lambda(\lambda+1),$$ where I used $\sum_n n^2 P_\lambda(n)=\lambda(\lambda+1)$.

The variance $\sigma^2$ of the distribution thus reads $$\sigma^2\equiv\sum_n (n-\mu)^2 P(n)=\sum_\lambda p_\lambda \lambda(\lambda+1)-\mu^2,$$ and finally the difference between variance and expected value, $\sigma^2-\mu$, is $$\sigma^2-\mu=\sum_\lambda p_\lambda\lambda(\lambda+1)-\mu(\mu+1).\tag1$$ Defining $f(\lambda)\equiv\lambda(\lambda+1)$, (1) can be written as $$\sigma^2-\mu=\sum_\lambda p_\lambda f(\lambda)-f\Big(\underbrace{\sum_\lambda p_\lambda \lambda}_{\mu}\Big).$$ The conclusion $\sigma^2-\mu\ge0$ now follows from $f$ being convex, together with Jensen's inequality.

This proves that an arbitrary mixture (convex combination) of Poissonians gives a super-Poissonian distribution satisfying $\sigma^2\ge\mu$.

glS
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For a single-mode thermal state, yes, we can, in general, show that the photon-counting statistics is super-Poisson using the method you proposed (as @glS has done it nicely). However, for a multi-mode thermal state, the photon-counting statistics is (surprisingly) Poisson, refer to https://physics.stackexchange.com/a/749705/396054 for more details.

Omid
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