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## Homework Statement

Hello, I need to show that the radial part of the hydrogen wave function has the form

[tex] \rho^{l+1} e^{-\rho} L_{n-l-1}^{2l+1} (2\rho) [/tex]

More specifically, I'm having trouble showing the [tex] L_{n-l-1}^{2l+1} (2\rho) [/tex] part because what I get is [tex] L_{n+1}^{2l+1} (2\rho) [/tex]. The derivation is relatively easy and it makes me furious that I'm unable to find the mistake.

## Homework Equations

[tex] L_{\alpha} (x) = e^x \frac{d^{\alpha}}{dx^{\alpha}} (e^{-x}x^\alpha) [/tex]

[tex] L_{\alpha}^\beta (x) = \frac{d^{\beta}}{dx^{\beta}} L_{\alpha} (x) = \frac{d^{\beta}}{dx^{\beta}} (e^x \frac{d^{\alpha}}{dx^{\alpha}} (e^{-x}x^\alpha)) [/tex]

## The Attempt at a Solution

This equation should be correct:

[tex] \rho \frac{d^2}{d \rho^2} f(\rho) + 2(l+1-\rho) \frac{d}{d \rho} f(\rho) + (2n-2(l+1)) f(\rho) = 0 [/tex]

Next I introduce the following parameters

[tex] \alpha = n+1 [/tex]

[tex] \beta = 2l+1 [/tex]

Then the last equation becomes

[tex] (2\rho) \frac{d^2}{d (2\rho)^2} f(2\rho) + (\beta+1-(2\rho)) \frac{d}{d (2\rho)} f(2\rho) + (\alpha-\beta) f(2\rho) = 0 [/tex]

First I deal with the case [tex] \beta=0 [/tex]

[tex] (2\rho) \frac{d^2}{d (2\rho)^2} f(2\rho) + (1-(2\rho)) \frac{d}{d (2\rho)} f(2\rho) + \alpha f(2\rho) = 0 [/tex]

I can show that the solution to this is [tex] f(2\rho) = L_\alpha (2\rho) [/tex]

Next I take the [tex] \beta [/tex]'th derivative of the last equation and get

[tex] (2\rho) \frac{d^\beta}{d (2\rho)^\beta} (\frac{d^2}{d (2\rho)^2} L_\alpha (2\rho)) + (\beta + 1-(2\rho)) \frac{d^\beta}{d (2\rho)^\beta} \frac{d}{d (2\rho)} L_\alpha (2\rho) + (\alpha - \beta) \frac{d^\beta}{d (2\rho)^\beta} L_\alpha (2\rho) = 0 [/tex]

Now by writing

[tex] f(2\rho) \equiv \frac{d^\beta}{d (2\rho)^\beta} L_\alpha (2\rho)) \equiv L_\alpha^\beta (2\rho) [/tex]

I arrive at

[tex] (2\rho) \frac{d^2}{d (2\rho)^2} f(2\rho) + (\beta+1-(2\rho)) \frac{d}{d (2\rho)} f(2\rho) + (\alpha-\beta) f(2\rho) = 0 [/tex]

which clearly has the solution [tex] L_\alpha^\beta (2\rho) = L_{n+1}^{2l+1} (2\rho) [/tex]

If you need me to show more thoroughly some of the steps then I can do this. I just didn't want to bother with too many details.

Of course, there's always the chance that I've made a typo somewhere, because the derivations of special functions are quite cumbersome. So far, I've been unable to find anything.

If anyone could take 2 minutes to analyze this I'd really appreciate it.