# Converting the complex Wigner function to its real form in terms of the quadrature operators

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 $$h=2$$

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

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