The Wigner function can be computed as the Fourier transform of the Weyl-ordered characteristic function: $$ W(\alpha) = \frac{1}{\pi^2} \int e^{\lambda^* \alpha - \lambda \alpha^*} C_W(\lambda) d^2\lambda $$ as given by Eq. $(3.136)$ of "Introductory Quantum Optics" by C. Gerry and P. Knight, where the integral is over all space. When trying to compute $W$ for a given state I often encounter integrals of the form: $$ \int e^{a \lambda^* + b \lambda - c |\lambda|^2} d^2\lambda $$ for $a, b, c \in \mathbb{C}$ with $Re(c) > 0$. How does one compute such an integral?

Here is my attempt: Let $\lambda = x + ip$ such that the corresponding Jacobian determinant is $$ J = \begin{vmatrix} \dfrac{\partial \lambda^*}{\partial x} & \dfrac{\partial \lambda^*}{\partial p}\\[1em] \dfrac{\partial \lambda}{\partial x} & \dfrac{\partial \lambda}{\partial p} \end{vmatrix} = \begin{vmatrix} 1 & -i \\ 1 & i \end{vmatrix} = 2i $$ and hence \begin{align}\label{WFcnCIntIdent} \int e^{a \lambda^* + b \lambda - c |\lambda|^2} d^2\lambda &= |2 i| \int \int e^{a (x - ip) + b (x + ip)- c (x^2 + p^2)} dx dp \nonumber \\ &= 2 \int e^{-c x^2 + (a+b) x} dx \int e^{-c p^2 + (b-a) i p} dp \nonumber \\ &= 2 \sqrt{\frac{\pi}{c}} \exp \Bigg[ \frac{(a+b)^2}{4 c} \Bigg] \sqrt{\frac{\pi}{c}} \exp \Bigg[ \frac{-(b-a)^2}{4 c} \Bigg] \nonumber \\ &= \frac{2 \pi}{c} \exp \Bigg[ \frac{(a+b)^2}{4 c} - \frac{(b-a)^2}{4 c} \Bigg] \nonumber \\ &= \frac{2 \pi}{c} e^{ab/c} \end{align} as the integrals are Gaussian.

Is this the correct approach? When using the result I obtained my calculations are consistently off by factors of two so I think the correct result might be: $$ \int e^{a \lambda^* + b \lambda - c |\lambda|^2} d^2\lambda \overset{?}{=} \frac{\pi}{c} e^{ab/c}, $$ but I do not see where I went wrong. I might be misunderstanding the complex Jacobian or some other details. Any help would be appreciated, thanks!

  • $\begingroup$ Is the definition of integration measure $d^2 \lambda=d \lambda^*d\lambda$ you adopted alright? $\endgroup$ – Sunyam Jun 6 at 21:35
  • $\begingroup$ @Sunyam Looking back in the book I found that $d^2\lambda = d [Re(\lambda)] d [Im(\lambda)]$ so it seems that I was using the wrong integration measure which accounts for the factor of two discrepancy. Thanks! $\endgroup$ – Randolf Jun 6 at 21:53
  • $\begingroup$ Some conventions. $\endgroup$ – Cosmas Zachos Jun 7 at 1:00

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