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?

  • $\begingroup$ Hint: use the binomial theorem. $\endgroup$ – knzhou Apr 19 '18 at 7:52
  • $\begingroup$ The final summation should be from $k$ to $\infty$. $\endgroup$ – Sunyam Apr 19 '18 at 8:17
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    $\begingroup$ If you want to ask a new question, please ask it as a new question rather than editing an existing one beyond all recognition. The aim of answering questions on this site is not only to help the person asking the question but also to help others who might have the same question. This question now has answers which bare no relation to what you are asking, making things harder for other people searching for answers. Please reverse your edit and ask your new question separately. $\endgroup$ – By Symmetry Apr 20 '18 at 17:24
  • $\begingroup$ I am sorry for this mistake. I will recollect the question and post it again. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Emilio Pisanty Apr 19 '18 at 8:51
  • $\begingroup$ Admittedly neither have I. I thought that by raising the subscript to a superscript, it could not be confused with the falling factorial. $\endgroup$ – Qmechanic Apr 19 '18 at 9:09
  • $\begingroup$ Fair enough - only now it's just confused with an ordinary power ;-). It's clearly labelled so it's not a problem. $\endgroup$ – Emilio Pisanty Apr 19 '18 at 9:10
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    $\begingroup$ Yeah, no notation is perfect :) $\endgroup$ – 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 $$\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}x_{}^{n-k}=(\frac{d}{dx})_{}^{k}\frac{1}{1-x}$$.

  • $\begingroup$ 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. $\endgroup$ – W. Voltera Apr 19 '18 at 16:55
  • $\begingroup$ 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 $n\geq k=l$). $\endgroup$ – Sunyam Apr 19 '18 at 19:16
  • $\begingroup$ I am still not getting the final answer. Have you reached to that $k! \langle n \rangle^k$? $\endgroup$ – W. Voltera Apr 20 '18 at 3:31
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    $\begingroup$ Yes. Use $\sum_{n=k}^{\infty}\frac{\langle n \rangle_{}^{n}}{(1 + \langle n \rangle)_{}^{n+1}}\frac{n!}{(n-k)!}=\frac{\langle n \rangle_{}^{k}}{(1 + \langle n \rangle)_{}^{k+1}}\sum_{n=k}^{\infty}(\frac{\langle n \rangle_{}^{}}{1 + \langle n \rangle_{}^{}})_{}^{n-k}\frac{n!}{(n-k)!}=\frac{\langle n \rangle_{}^{k}}{(1 + \langle n \rangle)_{}^{k+1}}(\frac{d}{dx})_{}^{k}(\frac{1}{1-x})|_{x=\frac{\langle n \rangle_{}^{}}{1 + \langle n \rangle_{}^{}}}=k!\langle n \rangle_{}^{k}$ $\endgroup$ – Sunyam Apr 20 '18 at 7:51

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