# What is the Wigner function of a 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: $$$$\rho_\mathrm{th} = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^{n+1}}|n\rangle \langle n |$$$$ and use the Wigner Tranform: $$$$W_\mathrm{th}(x,p) = \int du \langle x- u/2 | \rho_\mathrm{th} | x + u/2 \rangle e^{\mathrm{i} p u/\hbar}$$$$

After inserting the density matrix into the Wigner transform I get: $$$$W_\mathrm{th}(x,p) = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^{n+1}} \int du \langle x- u/2 |n\rangle \langle n | x + u/2 \rangle e^{\mathrm{i} p u/\hbar} = \sum_n \frac{\bar{n}_\mathrm{th}^n}{(1+\bar{n}_\mathrm{th})^{n+1}} W_n(x,p),$$$$ where $$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. Commented Sep 14, 2019 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
Commented Sep 14, 2019 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.

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.

• This is such a very sweet solution. Commented Sep 15, 2019 at 2:52
• Thank you very much!
– Luke
Commented Sep 15, 2019 at 4:16
• in the last step of the last expression, could it be that the $1+n_{\text{th}}$ factor at the nominator is not supposed to be there? If I understand the notation and $z=2(x^2+p^2)$, the normalisation doesn't seem to match. Integrating in polar coordinates, we have $\int_0^\infty dr 2\pi r e^{-r^2/(1+2n)}= \pi(1+2n)$ (I'm defining $\zeta=x+ip$, so that $z=2|\zeta|^2$, and writing $r=|\zeta|$)
– glS
Commented Sep 28, 2020 at 20:01
• Sorry, I won't chase the normalizations... and I saw you altered the question's... Suffice it to say that the answer is the product of two Gaussians, in x and p respectively, so with exponent $-(x^2+p^2)/(1+2\bar n_{th})$. Commented Sep 28, 2020 at 20:27
• @CosmasZachos ah, yes, I see that's the reason. The probabilities for the thermal state were given incorrectly, so there is simply an additional $1+n_{\text{th}}$ factor in the denominator of the last equation.
– glS
Commented Sep 28, 2020 at 21:52

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 $$W_{\rho(\alpha,x)}(\nu) = \frac1\pi \operatorname{tr}[T(\nu,0) \rho(\alpha,x)] = \frac{2(1-x)}{1+x}\exp\left(-\frac{2(1-x)}{1+x}|\nu-\alpha|^2\right).$$