# Wigner function of thermal state

I am wondering how you would compute the Wigner Function of a Thermal State with average phonon number $$\bar{n}_{\mathrm{th}}$$. I know the result should be a Gaussian with variance in position $$\langle x^2\rangle = (2 \bar{n}_\mathrm{th}+1) x_\mathrm{zp}^2$$ and in momentum $$\langle p^2\rangle = (2 \bar{n}_\mathrm{th}+1) \hbar/(4x_\mathrm{zp}^2)$$.

But how do I show that?

I write the thermal density matrix in the Fock basis: $$$$\rho_\mathrm{th} = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^n}|n\rangle \langle n |$$$$ and use the Wigner Tranform: $$$$W_\mathrm{th}(x,p) = \int du \langle x- u/2 | \rho_\mathrm{th} | x + u/2 \rangle e^{\mathrm{i} p u/\hbar}$$$$

After inserting the density matrix into the Wigner transform I get: $$$$W_\mathrm{th}(x,p) = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^n} \int du \langle x- u/2 |n\rangle \langle n | x + u/2 \rangle e^{\mathrm{i} p u/\hbar} = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^n} W_n(x,p),$$$$ where $$W_n(x,p)$$ is the Wigner functions of the n-Fock state given by:

$$$$W_n(x,p) = \frac{2}{\hbar \pi}(-1)^n e^{-2 \frac{H}{\hbar \omega} } L_n(4 \frac{H}{\hbar \omega} ),$$$$ with $$H= \frac{1}{2} m \omega^2 x^2 + \frac{p^2}{2 m}$$, and $$L_n$$ the nth Laguerre poloynomial.

Everything correct until here?

Now I am too stupid to do the last sum. Was looking for identities and what not half of this day.

Any ideas? I would also appreciate a simpler solution. Thanks for any Help!

• You do know that the suitably scaled Laguerre polynomials, like their cousins Hermite, are a complete set, no? You are comfortable with their generating function? I believe you should write it in your question itself. – Cosmas Zachos Sep 14 at 20:41
• Yeah, unfortunately, their property of being a complete set doesn't help me, since I have these n dependent prefactors. I will write down what I did in more detail. Honestly, I hoped that there would be an easier solution :) – Luke Sep 14 at 21:04

I believe the correct Wigner function for the eigenstates is half yours, so take it to be $$$$W_n(x,p) = \frac{(-1)^n}{\hbar \pi} e^{- z/2} L_n(z ),$$$$ where $$z=4 H/\hbar\omega$$.
You know that, since the resolution of the identity must be $$\sum_n W_n= \frac{1}{2\pi \hbar}=1/h,$$ which, indeed, holds (trivially checkable) by dint of the standard generating function of the Laguerre polynomials, $$\sum_n t^n L_n(z)= \frac{e^{-tz/(1-t)}} {1-t} ~~.$$
Your sum then readily collapses to $$\sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^n} W_n(x,p)= \frac{e^{-z/2}}{\pi \hbar} \sum_n \left (\frac{- \bar{n}_\mathrm{th}}{1+ \bar{n}_\mathrm{th}} \right )^n L_n (z) = \frac{ (1+\bar{n}_\mathrm{th})}{\pi \hbar (1+2\bar{n}_\mathrm{th})} ~ e^{-z / 2 (1+2 \bar{n}_\mathrm{th}) } ,$$ a gaussian in x and p with the requisite widths.