The Radial Equation for $u(r)=rR(r)$ *

It is sometimes useful to use

$$\bgroup\color{black}u_{n\ell}(r)=rR_{n\ell}(r) \egroup$$

to solve a radial equation problem. We can rewrite the equation for $u$.

$$\begin{eqnarray*} \left({d^2\over dr^2}+{2\over r}{d\over dr}\right){u(r)\over r... ...left[E-V(r)-{\ell(\ell+1)\hbar^2\over 2\mu r^2}\right]u(r)=0 \\ \end{eqnarray*}$$

This now looks just like the one dimensional equation except the pseudo potential due to angular momentum has been added.

We do get the additional condition that

$$\bgroup\color{black}u(0)=0 \egroup$$

to keep $R$ normalizable.

For the case of a constant potential $V_0$, we define $k=\sqrt{2\mu (E-V_0)\over\hbar^2}$ and $\rho=kr$, and the radial equation becomes.

$$\begin{eqnarray*} {d^2u(r)\over dr^2}+{2\mu\over \hbar^2}\left[E-V_0-{\ell(\ell+... ...o)\over d\rho^2}-{\ell(\ell+1)\over \rho^2}u(\rho)+u(\rho)=0 \\ \end{eqnarray*}$$

For $\ell=0$, its easy to see that $\sin\rho$ and $\cos\rho$ are solutions. Dividing by $r$ to get $R(\rho)$, we see that these are $j_0$ and $n_0$. The solutions can be checked for other $\ell$, with some work.