Normalization of radial Laguerre-Gauss

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SUMMARY

The discussion focuses on the normalization of the radial Laguerre-Gauss function, specifically the expression \(\Psi_n(r) = h_n L_n(2\pi r^2) e^{-\pi r^2}\). The integral for normalization is established as \(1 = \int_0^{\infty} \Psi_m^{\ast}(r) \Psi_n(r) dr\), leading to the substitution \(x = 2\pi r^2\) to simplify the integration process. The participant clarifies that the radial dependence does not complicate the substitution, confirming the normalization is achievable without further adjustments.

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Homework Statement



Normalization of radial Laguerre-Gauss

Normalize [tex]\Psi _n (r) = h_n L_n (2\pi r^2) e^{-\pi r^2}[/tex]

Homework Equations



[tex]\int _0 ^{\infty} e^{-x} \, x^k \, L_n ^{(k)} (x) \, L_m ^{(k)} (x) dx = \frac{(n+k)!}{n!} \delta _{m,n}[/tex]

The Attempt at a Solution



[tex]1 = \int _0 ^{\infty} \Psi _m ^{\ast} (r) \Psi _n (r) dr = \int _0 ^{\infty} h_m ^{\ast} L_m (2\pi r^2) e^{-\pi r^2} h_n L_n (2\pi r^2) e^{-\pi r^2} dr[/tex]

If I let [itex]x = 2\pi r^2[/itex], then I get [itex]dx = (4\pi r) dr[/itex]. The radial dependence bothers me. I think there's a step I'm missing out.
 
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Just realized [itex]dx[/itex] is not an issue. I don't need to substitute it by a [itex]dr[/itex], so there's no problem. All I need to do is replace [itex]2\pi r^2[/itex] by [itex]x[/itex]. I'm done.
 

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