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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!

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  • $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Sep 14 '19 at 20:41
  • $\begingroup$ 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 :) $\endgroup$ – Luke Sep 14 '19 at 21:04
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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.

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    $\begingroup$ This is such a very sweet solution. $\endgroup$ – ZeroTheHero Sep 15 '19 at 2:52
  • $\begingroup$ Thank you very much! $\endgroup$ – Luke Sep 15 '19 at 4:16
  • $\begingroup$ 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 $r=|\zeta|$) $\endgroup$ – glS Sep 28 '20 at 20:01
  • $\begingroup$ 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 $-(x^2+p^2)/(1+2\bar n_{th})$. $\endgroup$ – Cosmas Zachos Sep 28 '20 at 20:27
  • $\begingroup$ @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. $\endgroup$ – glS Sep 28 '20 at 21:52

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