# The hydrogen atom, Legendre and Laguerre equations

I was looking for the solution of the Schrödinger equation for the hydrogen atom and I found that the angular component of the solution is:

$$\frac{1}{sen\theta} \frac{\textrm d}{\textrm d \theta} \left [ sen\theta \cdot \frac{\textrm d}{\textrm d \theta}\Theta(\theta) \right ] + \left [ l \cdot (l+1) - \frac{m^{2}}{sen^{2}\theta} \right ] \Theta(\theta) = 0$$

I did some research and I saw in many places that this expression is an Ordinary Differential Equation of the Legendre type and therefore can be solved by power series.

The problem is I can't see the relationship between the equation above and the usual Legendre ODE:

$$(1-x^{2}) \frac{\mathrm d^{2}{y}}{\mathrm d{x}^{2}} - 2x \frac{\mathrm d{y}}{\mathrm d{x}} + l(l+1)y = 0 \longleftrightarrow \frac{\textrm d}{\textrm dx} \left [ (1-x^{2}) \frac{\textrm dy}{\textrm dx} \right ] +l \cdot (l+1) y=0$$

I can't see how the $\Theta(\theta)$ equation can be transformed into something like this.

The same thing happened with the radial solution of the Schrödinger equation, which is:

$$\frac{1}{r^{2}} \frac{\textrm d}{\textrm d r} \left [ r^{2} \frac{\textrm d}{\textrm d r} R(r) \right ] + \left [ \frac{2 \mu}{\hbar^{2}} \left ( \frac{e^{2}}{4\pi \epsilon_{0}} \frac{1}{r} + E\right ) - \frac{l \cdot (l+1)}{r^{2}} \right ] R(r)= 0$$

This should be an ODE of the Laguerre type, but again I can't see how the equation above relates to the usual Laguerre ODE:

$$x\frac{\mathrm d^{2}{y}}{\mathrm d{x}^{2}} + (1-x) \frac{\mathrm d{y}}{\mathrm d{x}} yx + ny = 0$$

Can someone convince me that the $\Theta(\theta)$ and the $R(r)$ really are Legendre and Laguerre equations?

I am using this article and this presentation but I don't know if they're gonna be of help since they are in portuguese.

• Just make the substitution of $\cos(\theta)=x$. Then, the $m$ part corresponds to the "associated" Legendre equation. – Ruslan Jun 25 '17 at 17:14