Graduate Feynman solution for the radial wave function of the hydrogen atom

[emilionovati]

TL;DR

Problema in the interpretation of the Feynman solution of the Schrodinger equation for the hydrogen atom

Reading the classical Feynman lectures, I encounter the formula(19.53) that gives the radial component of the wave function:
$$
F_{n,l}(\rho)=\frac{e^{-\alpha\rho}}{\rho}\sum_{k=l+1}^n a_k \rho^k
$$
that, for $n=l+1$ becomes
$$
F_{n,l}=\frac{e^{-\rho/n}}{\rho}a_n\rho^n
$$
To find $a_n$ I use the recursive formula (19.50), but here I have problem. Using $k+1=n=l+1$ I find a division by zero.

$$
a_n =\frac{2\left( \frac{n-1}{n}-1 \right)}{(n-1)n-(n-1)n}
$$

so clearly I have a mistake. But where is it?

 

[anuttarasammyak]

I observe denominator of (19.50) is
n(n-1)-l(l-1)
for k=n-1. As $n \geq l+1$ it is not zero.

 

[emilionovati]

But, in the (19.50) the denominator is
$$
k(k+1)-l(l+1)
$$
that, for $k=n-1$ becomes
$$
(n-1)n-l(l+1)
$$
and, since $n=l+1$ we have
$$
(n-1)n-(n-1)n
$$

 

[anuttarasammyak]

Feynman says "This means that k must start at l+1 and end at n."
Your case, $k=n-1=(l+1)-1=l$, does not satisfy it.

 

[vanhees71]

The general formula is $n=n_r+l+1$, where the radial quantum number $n_r \in \mathbb{N}_0=\{0,1,2,\ldots\}$ and the orbital-angular-momentum quantum number $l \in \{0,1,2,\ldots \}$. The principle quantum number $n \in \mathbb{N}=\{1,2,\ldots \}$. Usually the hydrogen energy eigenstates (neglecting spin) are labeled by $(n,l,m)$ (where the magnetic quantum number $m \in \{-l,-l+2,\ldots,l-1,l\}$). The energy eigenvalues are $E_n=-\frac{m_e e^4}{2 (4 \pi \epsilon)^2 \hbar^2 n^2}=-1 \text{Ry}/n^2$ with $1 \text{Ry} \simeq 13.6 \; \text{eV}$; $n$ is thus called the principal quantum number (that the $E_n$ do not independently depend on $n_r$ and $l$ is due to the dynamical O(4) symmetry of the hydrogen bound states; $E_n$ is $n^2$ fold degenerate).

Since $n_r =n-l-1 \geq 0$ for a given $n$ the possible values for $l$ are $0,1,\ldots n-1$.