# A summation in quantum optics paper

I am going through the paper "Moments of $P$ functions and nonclassical depths of quantum states", which contains the following passage:

### A. Thermal state

The density matrix of the thermal state can be written as $$\hat{\rho}_{\text{th}} = \sum_{n=0}^{\infty}\frac{\langle n\rangle^n}{\bigl(\langle n\rangle + 1\bigr)^{n + 1}}\lvert n\rangle\langle n\rvert.\tag{3.1}$$ So, we can obtain the moments as follows: \begin{align}\mu_{k,l} &= \operatorname{Tr}\bigl[(\hat{a}^\dagger)^l (\hat{a})^k \hat{\rho}_{\text{th}}\bigr] \\ &= \sum_{n=0}^{\infty}\frac{\langle n\rangle^n}{\bigl(\langle n\rangle + 1\bigr)^{n + 1}}\langle n\rvert(\hat{a}^\dagger)^l (\hat{a})^k\lvert n\rangle \\ &= \sum_{n=0}^{\infty}\frac{\langle n\rangle^n}{\bigl(\langle n\rangle + 1\bigr)^{n + 1}} \frac{n!}{(n - k)!} \delta_{k,l} \\ &= k!\langle n\rangle^k \delta_{k,l}.\tag{3.2} \end{align}

I want to understand the last step of Eq. 3.2. How is this summation carried out to reach the final answer?

• Hint: use the binomial theorem. – knzhou Apr 19 '18 at 7:52
• The final summation should be from $k$ to $\infty$. – Sunyam Apr 19 '18 at 8:17
• I am sorry for this mistake. I will recollect the question and post it again. – W. Voltera Apr 20 '18 at 17:44

Hint: Use the generalized binomial theorem $$\frac{1}{(1-x)^s}~=~\sum_{n=0}^{\infty} (s)^n\frac{x^n}{n!}, \qquad x,s\in\mathbb{C} , \qquad|x|<1 , \tag{1}$$ where $$(s)^n~:=~\frac{\Gamma(s+n)}{\Gamma(s)}~=~\frac{(n+s-1)!}{(s-1)!}\tag{2}$$ is the Pochhammer symbol/rising factorial.

• I'd never seen the notation $(s)^n$ for the Pochhammer symbol - normally I see it notated $(s)_n$. Is there some specific reason for the change? – Emilio Pisanty Apr 19 '18 at 8:51
• Admittedly neither have I. I thought that by raising the subscript to a superscript, it could not be confused with the falling factorial. – Qmechanic Apr 19 '18 at 9:09
• Fair enough - only now it's just confused with an ordinary power ;-). It's clearly labelled so it's not a problem. – Emilio Pisanty Apr 19 '18 at 9:10
• Yeah, no notation is perfect :) – Qmechanic Apr 19 '18 at 9:13

As noted in my comment summation ( over $n$) runs from $k$ to $\infty$ as you cannot destroy more that $n$ photons in a state $|n\rangle$.

Hint : Use $$\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}x_{}^{n-k}=(\frac{d}{dx})_{}^{k}\frac{1}{1-x}$$.

• Thanks @Sunyam. You mean the paper has a correction? But that is the way to define the trace. One needs to take the sum over all the states n. – W. Voltera Apr 19 '18 at 16:55
• Second step is perfectly alright, but third step can probably be a typo, sum should be from $k$ to $\infty$ (as $\langle n|(a_{}^{\dagger})_{}^{l}(a_{}^{})_{}^{k}|n\rangle \neq 0$ only for $n\geq k=l$). – Sunyam Apr 19 '18 at 19:16
• I am still not getting the final answer. Have you reached to that $k! \langle n \rangle^k$? – W. Voltera Apr 20 '18 at 3:31
• Yes. Use $\sum_{n=k}^{\infty}\frac{\langle n \rangle_{}^{n}}{(1 + \langle n \rangle)_{}^{n+1}}\frac{n!}{(n-k)!}=\frac{\langle n \rangle_{}^{k}}{(1 + \langle n \rangle)_{}^{k+1}}\sum_{n=k}^{\infty}(\frac{\langle n \rangle_{}^{}}{1 + \langle n \rangle_{}^{}})_{}^{n-k}\frac{n!}{(n-k)!}=\frac{\langle n \rangle_{}^{k}}{(1 + \langle n \rangle)_{}^{k+1}}(\frac{d}{dx})_{}^{k}(\frac{1}{1-x})|_{x=\frac{\langle n \rangle_{}^{}}{1 + \langle n \rangle_{}^{}}}=k!\langle n \rangle_{}^{k}$ – Sunyam Apr 20 '18 at 7:51