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Suppose we have a quantum state $\rho$ and let's denote the photon number operator $\hat{n}=\hat{a}^\dagger\hat{a}$ where $\hat{a}$ is the annihilation operator. Let mean photon number $\bar{n}=\operatorname{Tr}(\rho\hat{n})<\infty$ obviously be finite. Can $\rho$ have infinite photon number variance, which we define as follows:

$$\operatorname{Var}(\rho)=\operatorname{Tr}(\rho\hat{a}^\dagger\hat{a}^\dagger\hat{a}\hat{a})-\bar{n}^2$$

I ask this because it came up in a quantum information theory research problem that I am working on. Mathematically, it's easy to define $\rho$ with infinite photon number variance: let $\rho=|\psi\rangle\langle\psi|$ where $|\psi\rangle=\sum_{n=0}^\infty a_n|n\rangle$ such that $|a_n|^2=(n+1)^{-3}/\zeta(3)$ with $\zeta(3)=\sum_{k=1}^\infty1/k^3$ being the Reimann zeta function. That is, $|\psi\rangle$ is a superposition of number states which is prepared such that the photon number of the state is distributed according to the Zipf distribution.

However, I am wondering if such state (or one like it) can actually exist in nature. If yes, how would one construct it? If not, is there a concrete reason as to why? (ideally, I am looking for a mathematical proof, but would accept a generally accepted well-cited explanation.)

I am confused because it seems to me that the requirement if one applies a measurement $\sum_{m=k}^\infty|m\rangle\langle m|$ with a large $k$ to a coherent state there is a chance of it returning a positive answer, however small, as coherent states are not "peak photon number" limited (if they were, that chance would be zero starting at some fixed $k$). Here the difference is the heavy tail on the photon number.

This is my first question to this forum. I am not a physicist, I study information theory and have been looking at quantum information theory lately. I know a lot of math, but not necessarily how things work in the "universe we touch". I only have a rudimentary knowledge of quantum mechanics. Please forgive my ignorance and feel free to refer this question somewhere else if it's not suitable for this forum.

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