3
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ What is $d^2\lambda$ and what do the limits mean? Is it an integral over the entire complex plane? $\endgroup$ – Conifold Dec 1 '15 at 0:01
  • 1
    $\begingroup$ Products of polynomials and Gaussians can be integrated using integration by parts. See e.g. mathworld.wolfram.com/GaussianIntegral.html, starting at eqn (9). $\endgroup$ – Norbert Schuch Dec 1 '15 at 23:01
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.