Show that R satisfies the radial Schrodinger equation

[warfreak131]

Homework Statement

Starting with $$\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$$ saubstitute into the Schrödinger equation and show (using the technique of separation of variables) that R satisfies:

$$(\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d}{dr})+\frac{C}{2mr^2}+V(r))R=ER(r)$$

Homework Equations

$$L^2 Y(\theta,\phi)=CY(\theta,\phi)$$
$$C=l(l+1)\hbar^2$$

The Attempt at a Solution

The way I wrote it above is exactly the way the teacher wrote it, and I'm assuming that the R on the left side of the equation is not a function of r. And that being the case, it passes through the derivatives on the left hand side and and are left with a function of r.

C is a constant, so that entire term is also just a function of r. And V(r) itself is a function of r, therefore I'm inclined to think that you can pull out some constant E and multiply it by some function R(r) to give it the required format.

 

[vela]

The R on the lefthand side is R(r).

The idea here is to plug the function $\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$ into the Schrödinger equation, expressing the Laplacian in spherical coordinates. After a little manipulation, you can separate the equation into two chunks, where one part depends only on r and the other depends only on θ and ϕ. The part that only depends on r is that equation you're trying to verify.

 

[warfreak131]

The R on the lefthand side is R(r).

The idea here is to plug the function $\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$ into the Schrödinger equation, expressing the Laplacian in spherical coordinates. After a little manipulation, you can separate the equation into two chunks, where one part depends only on r and the other depends only on θ and ϕ. The part that only depends on r is that equation you're trying to verify.

As the equation is right now, it has no theta or phi terms. Wouldnt I have to know the quantum numbers n, l, and m in order to introduce the other terms?

 

[vela]

You're supposed to start with the Schrödinger equation, which does have θ and ϕ terms.