I saw this integral in the book [Gerry C.C.,Knight P.L.] Introductory quantum optics: $$\frac{1}{\pi^2}\int_{-\infty}^{\infty}L_n(\lvert\lambda\rvert^2)e^{\lambda^*\alpha-\lambda\alpha^*-\frac{1}{2}\lvert\lambda\rvert^2}d^2\lambda$$ where $L_n(\lvert\lambda\rvert^2)$ is Laguerre polynomial and $\lambda$ and $\alpha$ complex

It is related to the Wigner distribution in Quantum mechanics. How can this be integrated?

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    $\begingroup$ What is $d^2\lambda$ and what do the limits mean? Is it an integral over the entire complex plane? $\endgroup$
    – Conifold
    Commented Dec 1, 2015 at 0:01
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    $\begingroup$ Products of polynomials and Gaussians can be integrated using integration by parts. See e.g. mathworld.wolfram.com/GaussianIntegral.html, starting at eqn (9). $\endgroup$ Commented Dec 1, 2015 at 23:01

1 Answer 1


To evaluate this integral, first separate it into its radial and angular dependences, by setting $\lambda = r e^{i\theta}$ and $\alpha=s e^{i\varphi}$, which produce \begin{align} I & := \frac{1}{\pi^2}\int_{-\infty}^{\infty}L_n(\lvert\lambda\rvert^2)e^{\lambda^*\alpha-\lambda\alpha^*-\frac{1}{2}\lvert\lambda\rvert^2}\mathrm d^2\lambda % \\& = % \frac{1}{\pi^2} \int_{0}^{\infty}\!\!\! \int_{0}^{2\pi} L_n(r^2) e^{rs e^{-i(\theta-\varphi)}-rs e^{i(\theta-\varphi)}-\frac{1}{2}r^2} r\,\mathrm d\theta\,\mathrm dr % \\& = % \frac{1}{\pi^2} \int_{0}^{\infty}\!\!\! \int_{0}^{2\pi} e^{-2i\,rs \sin(\theta-\varphi)} \mathrm d\theta \, L_n(r^2) e^{-\frac{1}{2}r^2} r\,\mathrm dr % \\& = % \frac{2}{\pi} \int_{0}^{\infty} J_0(2rs) L_n(r^2) e^{-\frac{1}{2}r^2} r\,\mathrm dr % \\& = % (-1)^n \frac{2}{\pi} e^{-2s^2} L_n\mathopen{}\left(4s^2\right)\mathclose{}, % \\& = % (-1)^n \frac{2}{\pi} e^{-2|\alpha|^2} L_n\mathopen{}\left(4|\alpha|^2\right)\mathclose{}, % \end{align} where the last step comes from standard integral tables, specifically integral (7.421.1) in Gradshteyn & Ryzhik, $$ \int_0^\infty x e^{-\frac12\alpha x^2} L_n(\tfrac12\beta x^2)J_0(xy) \mathrm dx = \frac{(\alpha-\beta)^n}{\alpha^{n+1}} e^{-\frac{1}{2\alpha}y^2} L_n\mathopen{}\left(\frac{\beta y^2}{2\alpha(\beta-\alpha)}\right)\mathclose{}, $$ which is referenced to Erdelyi's Tables of Integral Transforms vol II, p. 13 eq. (4), in case you don't have access to G&R.

Obviously, if you're going to depend on any of this, you should re-calculate everything to double-check I haven't messed up any details.


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