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

Saurabh Shringarpure
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    Related: https://physics.stackexchange.com/q/308929/2451 – Qmechanic Jan 06 '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 06 '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 Shringarpure Jan 06 '19 at 14:31
  • @Qmechanic In the link you provided, it was something like $d^2\lambda$ represents a single complex integral but two real integrals. Am I right? Could you explain it further? – Saurabh Shringarpure Jan 06 '19 at 14:50
  • @CosmasZachos What do you mean by complex integration measure? – Saurabh Shringarpure Jan 06 '19 at 16:25
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    @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 07 '19 at 00:11
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    @Qmechanic's answer should be sufficient, if i understand your question – Steven Sagona Jan 07 '19 at 00:12

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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).

Emilio Pisanty
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ZeroTheHero
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  • How do you go from $d^2\lambda=d\lambda^*d\lambda=dxdp$? – Saurabh Shringarpure Jan 06 '19 at 14:37
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    @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 06 '19 at 14:56
  • Thanks. Where can I find the intermediate steps with the Jacobian transformation? – Saurabh Shringarpure Jan 06 '19 at 15:18
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    @SaurabhShringarpure any book on multivariable calculus will do but see also https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Example_2:_polar-Cartesian_transformation as an example. – ZeroTheHero Jan 06 '19 at 15:20
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    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 07 '19 at 16:36