Solution of Hydrogen Radial Equation *

The differential equation we wish to solve is.

$$\bgroup\color{black}\left({d^2\over dr^2}+{2\over r}{d\over d... ...\ell(\ell+1)\hbar^2\over 2\mu r^2}\right)R_{E\ell}(r)=0 \egroup$$

First we change to a dimensionless variable $\rho$,

$$\bgroup\color{black}\rho=\sqrt{-8\mu E\over\hbar^2}r ,\egroup$$

giving the differential equation

$$\bgroup\color{black}{d^2R\over d\rho^2}+{2\over\rho}{dR\over ... ...rho^2}R +\left({\lambda\over\rho}-{1\over 4}\right)R=0 ,\egroup$$

where the constant

$$\bgroup\color{black}\lambda={Ze^2\over\hbar}\sqrt{-\mu\over 2E}=Z\alpha\sqrt{-\mu c^2\over 2E} .\egroup$$

Next we look at the equation for large $r$.

$$\bgroup\color{black}{d^2R\over d\rho^2}-{1\over 4}R=0 \egroup$$

This can be solved by $R=e^{-\rho\over 2}$, so we explicitly include this.

$$\bgroup\color{black}R(\rho)=G(\rho)e^{-\rho\over 2} \egroup$$

We should also pick of the small $r$ behavior.

$$\bgroup\color{black}{d^2R\over d\rho^2}+{2\over\rho}{dR\over d\rho}-{\ell(\ell+1)\over\rho^2}R=0 \egroup$$

Assuming $R=\rho^s$, we get

$$\begin{eqnarray*} s(s-1){R\over\rho^2}+2s{R\over\rho^2}-\ell(\ell+1){R\over\rho^2}=0 .\\ s^2-s+2s=\ell(\ell+1) \\ s(s+1)=\ell(\ell+1) \\ \end{eqnarray*}$$

So either $s=\ell$ or $s=-\ell-1$. The second is not well normalizable. We write $G$ as a sum.

$$\bgroup\color{black}G(\rho)=\rho^\ell\sum\limits_{k=0}^\infty a_k\rho^k=\sum\limits_{k=0}^\infty a_k\rho^{k+\ell} \egroup$$

The differential equation for $G(\rho)$ is

$$\bgroup\color{black}{d^2G\over d\rho^2}-\left(1-{2\over\rho}\... ...1\over\rho} -{\ell(\ell+1)\over\rho^2}\right)G(\rho)=0 .\egroup$$

We plug the sum into the differential equation.

$$\begin{eqnarray*} \sum\limits_{k=0}^\infty a_k\left((k+\ell)(k+\ell-1)\rho^{k+\e... ...}^\infty a_k\left((k+\ell)-(\lambda-1)\right)\rho^{k+\ell-1} \\ \end{eqnarray*}$$

Now we shift the sum so that each term contains $\rho^{k+\ell-1}$.

$$\bgroup\color{black}\sum\limits_{k=-1}^\infty a_{k+1}\left((k... ...fty a_k\left((k+\ell)-(\lambda-1)\right)\rho^{k+\ell-1} \egroup$$

The coefficient of each power of $\rho$ must be zero, so we can derive the recursion relation for the constants $a_k$.

$$\begin{eqnarray*} {a_{k+1}\over a_k}&=&{k+\ell+1-\lambda\over (k+\ell+1)(k+\ell)... ...+\ell+1-\lambda\over (k+1)(k+2\ell+2)}\rightarrow {1\over k} \\ \end{eqnarray*}$$

This is then the power series for

$$\bgroup\color{black}G(\rho)\rightarrow \rho^\ell e^\rho \egroup$$

unless it somehow terminates. We can terminate the series if for some value of $k=n_r$,

$$\bgroup\color{black}\lambda=n_r+\ell+1\equiv n .\egroup$$

The number of nodes in $G$ will be $n_r$. We will call $n$ the principal quantum number, since the energy will depend only on $n$.

Plugging in for $\lambda$ we get the energy eigenvalues.

$$\begin{eqnarray*} Z\alpha\sqrt{-\mu c^2\over 2E}=n .\\ E=-{1\over 2n^2}Z^2\alpha^2\mu c^2 \\ \end{eqnarray*}$$

The solutions are

$$\bgroup\color{black}R_{n\ell}(\rho)=\rho^\ell\sum\limits_{k=0}^\infty a_k\rho^k e^{-\rho/2} .\egroup$$

The recursion relation is

$$\bgroup\color{black}a_{k+1}={k+\ell+1-n\over (k+1)(k+2\ell+2)}a_k .\egroup$$

We can rewrite $\rho$, substituting the energy eigenvalue.

$$\bgroup\color{black}\rho=\sqrt{-8\mu E\over\hbar^2}r=\sqrt{4\... ...ar^2n^2}r ={2\mu cZ\alpha\over\hbar n}r={2Z\over na_0}r \egroup$$