What is the Wigner function of a thermal state?

Question

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: \begin{equation} \rho_\mathrm{th} = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^{n+1}}|n\rangle \langle n | \end{equation} and use the Wigner Tranform: \begin{equation} W_\mathrm{th}(x,p) = \int du \langle x- u/2 | \rho_\mathrm{th} | x + u/2 \rangle e^{\mathrm{i} p u/\hbar} \end{equation}

After inserting the density matrix into the Wigner transform I get: \begin{equation} W_\mathrm{th}(x,p) = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^{n+1}} \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+1}} W_n(x,p), \end{equation} where $W_n(x,p)$ is the Wigner functions of the n-Fock state given by:

\begin{equation} W_n(x,p) = \frac{2}{\hbar \pi}(-1)^n e^{-2 \frac{H}{\hbar \omega} } L_n(4 \frac{H}{\hbar \omega} ), \end{equation} 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!

Answer 1

I believe the correct Wigner function for the eigenstates is half yours, so take it to be \begin{equation} W_n(x,p) = \frac{(-1)^n}{\hbar \pi} e^{- z/2} L_n(z ), \end{equation} 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.

These are the basic maneuvers in phase-space quantization, Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014.

Answer 2

You can directly derive $P,W$ and $Q$ functions of general displaced thermal states as shown in this other answer of mine. In brief:

1. Define displaced thermal states as $$\rho(\alpha,x) \equiv D(\alpha)\rho(x) D(-\alpha),\\ \rho(x)\equiv (1-x)x^{a^\dagger a}\equiv(1-x)\sum_{n=0}^\infty x^n |n\rangle\!\langle n|,$$ with $x=e^{-\beta}$, $0 \le x < 1$, and $D(\alpha)\equiv e^{\alpha a^\dagger-\bar\alpha a}$. Just set $\alpha=0$ to recover regular thermal states.

2. Consider for $s\le1$ the operators: $$T(\nu,s) \equiv \int \frac{\mathrm d^2\beta}{\pi} e^{\alpha\bar\beta-\bar\alpha\beta} e^{\frac s2|\beta|^2} D(\beta) =\frac{2}{1-s} D(\nu)\left(\frac{s+1}{s-1}\right)^{a^\dagger a} D(-\nu),$$ and observe that the Wigner can be written as $W_\rho(\nu)=\frac1\pi \operatorname{tr}[T(\nu,0)\rho]$. Displaced thermal states give $$\operatorname{tr}[T(\nu,s) \rho(\alpha,x)] = C\exp\left(-C|\nu-\alpha|^2\right), \qquad C\equiv \frac{2(1-x)}{x(s+1)-(s-1)},$$ with the expression valid for $s+\frac{x+1}{x-1}\le0$ (becoming a Dirac delta in the $0^-$ limit).

3. Thus the Wigner function, corresponding to $s=0$, has the form $$W_{\rho(\alpha,x)}(\nu) = \frac1\pi \operatorname{tr}[T(\nu,0) \rho(\alpha,x)] = \frac{2(1-x)}{1+x}\exp\left(-\frac{2(1-x)}{1+x}|\nu-\alpha|^2\right).$$