How is the complex integration done for the Wigner function in coherent state representation?

$$W(\alpha)=\frac{1}{\pi^2}\int e^{\lambda\alpha^*-\lambda^*\alpha} \operatorname{Tr}\left[ \hat{\rho}e^{\lambda\hat{a}^\dagger} e^{-\lambda^* \hat{a}} \right] e^{-\frac{|\lambda|^2}{2}} \, d^2\lambda.$$ (Ref: Eqn 3.136 on Page 67 in "Introductory quantum optics" by C. Gerry and P. Knight (2005))

My question is about $$d^2\lambda$$. Is it just $$d(\operatorname{Re}[\lambda]) \, d(\operatorname{Im}[\lambda])$$ or something else?

What are the techniques generally used to solve these integrals if they are something else?

• – Qmechanic Jan 6 '19 at 14:17
• Are you just asking about the Jacobian of the routine transformation of the complex integration measure, or something deeper about the evaluation of the trace? – Cosmas Zachos Jan 6 '19 at 14:18
• @CosmasZachos the question is about the complex double integration part only because the $d^2\lambda$ is not very insightful. – Saurabh Uday Shringarpure Jan 6 '19 at 14:31
• @SaurabhShringarpure Complex integration is a very involved topic, but it seems your question is just asking what this lambda^2 is defined as. Writing in that form is really just notational, and I think that you can simply think of it as d(Re[lambda] d(Im[lambda]). All of this stuff about jacobians is just details when you want actually perform the 2D integral (sometimes its easier to change the basis, so people are saying how to do that). – Steven Sagona Jan 7 '19 at 0:11
• @Qmechanic's answer should be sufficient, if i understand your question – Steven Sagona Jan 7 '19 at 0:12

Since $$\lambda=x+ip$$, $$d\lambda^* \,d\lambda$$ is basically $$dx \, dp$$ (as expected). Moreover, the factor $$e^{-\lambda \lambda^*/2}$$ is a Gaussian tail, which usually means integration by parts will come handy.
Unless we know more about $$\rho$$ there’s nothing else to say here. If $$\rho$$ is pure and a harmonic oscillator eigenstate $$\vert n\rangle$$, the integral is doable for any $$n$$ in terms of known special functions: you can find details in Schleich, Wolfgang P., Quantum optics in phase space (John Wiley & Sons, 2011).
• How do you go from $d^2\lambda=d\lambda^*d\lambda=dxdp$? – Saurabh Uday Shringarpure Jan 6 '19 at 14:37
• @SaurabhShringarpure $d^2\lambda$ is pretty much always defined as $d\lambda^* d\lambda$. Then its a Jabobian transformation to get to $dx dp$ given $\lambda=x+ip$. The details will include some factor of 2 or not depending on the precise relation between $\lambda$ and $x+ip$, v.g maybe $\lambda=x+ip$ or $\lambda=(x+ip)/\sqrt{2}$. – ZeroTheHero Jan 6 '19 at 14:56
• You can find this derivation in the book "Quantum optics" of Garrison and Chiao, even if they do the opposite way, from the x/p representation to the $\alpha$ plane. – steg Jan 7 '19 at 16:36