I noticed something that bugged me recently, the Wigner function which is defined for one mode in the complex plane as

$$W(\alpha)=\frac{1}{\pi^2}\int e^{\lambda\alpha^*-\lambda^*\alpha} \operatorname{Tr}\left[ \hat{\rho}e^{\lambda\hat{a}^\dagger} e^{-\lambda^* \hat{a}} \right] e^{-\frac{|\lambda|^2}{2}} \, d^2\lambda. $$

which for the vacuum state $\rho = |0 \rangle \langle 0|$ yields $W(\alpha)= \frac{2}{\pi} e^{-2|\alpha|^{2}}$ if we convert this complex Gaussian to real coordinates, we obtain a Gaussian whose covariance matrix diagonal with $1/4$ this is different than what I expected from which says the covariance matrix should be diagonal with entries $1/2$.

So I converted $W(\alpha)$ to its real form in the quadrature operators using the relations

$\hat{a} = \frac{\hat{x}+i\hat{p}}{\sqrt{2}}$ and $\hat{a}^{\dagger} = \frac{\hat{x}-i\hat{p}}{\sqrt{2}}$

and got for $\vec{\alpha} = [q,p]$,$\vec{\lambda}=[\xi,\eta]$, $\hat{r} = [\hat{x}, \hat{p}]$ and $\wedge$ denotes the symplectic inner product by the $\Omega = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$

$$W(q,p) = \frac{1}{\pi^{2}} \int \int d\eta d\xi e^{2i\alpha \wedge \lambda} tr(\hat{\rho}\exp(-\frac{-\sqrt{2}}{2}i(\lambda \wedge \hat{r})$$

If this is correct, is this off from the definition of the Wigner function using Plank's constant $h=1$ by some constant that accounts for the discrepancy with the vacuum state mentioned above?

Also, I actually can't seem to find a definition of Wigner function as the symplectic Fourier with $h =1$, I have this definition for $h=2$

$$W({\mathbf x}) = \int {d \xi ~d\eta\over (2\pi)^{2}} e^{-i(q\eta-p\xi)} \mathrm{Tr}(\hat \rho e^{i(\hat q \eta-\hat p \xi)} ). \tag{2} $$

  • 1
    $\begingroup$ Your definition (2) is correct and is not assuming fixing h. $\endgroup$ Mar 29, 2022 at 18:02
  • $\begingroup$ I see. So perhaps I can compare what I got with (2) $\endgroup$
    – user135520
    Mar 29, 2022 at 18:30
  • 1
    $\begingroup$ Recall the infinite dimensional Hilbert space, and trace, and odd normalization of ρ in (2)…. $\endgroup$ Mar 29, 2022 at 18:32
  • $\begingroup$ Oh to see why (2) doesn’t depend on $h$? Is that what you mean? $\endgroup$
    – user135520
    Mar 29, 2022 at 19:11
  • 1
    $\begingroup$ x,p,h, and W have dimensions you might keep track of, helpfully. $\endgroup$ Mar 29, 2022 at 19:20


Your Answer

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