Normalization of radial Laguerre-Gauss

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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 normalization condition is established through the integral of the product of \Psi_m and \Psi_n over the range from 0 to infinity. A substitution of variables is suggested, changing from r to x using x = 2\pi r^2, which simplifies the integration process. The participant initially expresses confusion about the radial dependence but ultimately clarifies that the substitution resolves the issue without complications. The conversation concludes with the realization that the normalization can be achieved by correctly applying the variable change.
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Homework Statement



Normalization of radial Laguerre-Gauss

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

Homework Equations



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

The Attempt at a Solution



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

If I let x = 2\pi r^2, then I get dx = (4\pi r) dr. The radial dependence bothers me. I think there's a step I'm missing out.
 
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Just realized dx is not an issue. I don't need to substitute it by a dr, so there's no problem. All I need to do is replace 2\pi r^2 by x. I'm done.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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