0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Hint: use the binomial theorem. $\endgroup$
    – knzhou
    Commented Apr 19, 2018 at 7:52
  • $\begingroup$ The final summation should be from $k$ to $\infty$. $\endgroup$
    – Sunyam
    Commented Apr 19, 2018 at 8:17
  • 3
    $\begingroup$ 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. $\endgroup$ Commented Apr 20, 2018 at 17:24
  • $\begingroup$ I am sorry for this mistake. I will recollect the question and post it again. $\endgroup$
    – W. Voltera
    Commented Apr 20, 2018 at 17:44

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Commented Apr 19, 2018 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
    Commented Apr 19, 2018 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. $\endgroup$ Commented Apr 19, 2018 at 9:10
  • 3
    $\begingroup$ Yeah, no notation is perfect :) $\endgroup$
    – Qmechanic
    Commented Apr 19, 2018 at 9:13
0
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
    Commented Apr 19, 2018 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
    Commented Apr 19, 2018 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
    Commented Apr 20, 2018 at 3:31
  • 1
    $\begingroup$ 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
    Commented Apr 20, 2018 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.