# Why does classical light always result in super-Poissonian statistics?

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?

• > "which I take here to mean mixtures of coherent states" This is quite important clarification of the notion of "classical light". I think in purely classical EM theory, for individual classical electromagnetic waves produced by molecules one can get sub-Poissonian statistics, since reflection and transmission of a polarized wave on a beam splitter is anticorrelated. – Ján Lalinský Jun 4 '19 at 17:50
• @JánLalinský I'm not sure I understand what you are saying. It is kind of the point here that coherent states cannot produce sub-Poissonian statistics. If the light produced by the molecule is not a coherent state (I don't know if that can happen or not) then it's fine that it can produce sub-Poissonian statistics. Do you have any reference do better understand your point? – glS Jun 4 '19 at 17:59
• My point is that you only get impossibility of sub-Poissonian statistics for 'classical light' if by 'classical light' you mean the thing defined by the concept of coherent states. In pure classical theory, classical light is something different and it is far from clear that one gets impossibility of sub-Poissonian statistics. But I think you understand this distinction, just wanted to remark that it is indeed important. – Ján Lalinský Jun 4 '19 at 18:02

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 $$\sigma^2\ge\mu$$.