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\begin{proposition*}
    If $f(r) = \sum_{n = 0}^\infty a_n r^n$ is
    infinite then $R(r)$ is not normalisable.
\end{proposition*}

\begin{proof}
    Asymptotic behaviour of $f(r)$ determined by
    \[ \frac{a_n}{a_{n - 1}} \stackrel{n \to
    \infty}{\longrightarrow} \frac{2\nu}{n} \]
    This is the same asymptotic behaviour as
    \[ g(r) = e^{2\nu r} = \sum_{n = 0}^\infty
    \frac{(2\nu)^n}{n!} r^n \]
    $b_n = \frac{(2\nu)^n}{n!}$, then
    \[ \frac{b_n}{b_{n - 1}} \stackrel{n \to
    \infty}{\longrightarrow} \frac{2\nu}{n} \]
    Asymptotically $f(r) \sim e^{2\nu r}$, $R(r) =
    f(r) e^{-\nu r} \sin e^\nu r$.
\end{proof}

\myskip
$\implies$ the series must terminate. $\exists N >
0$ such that
\[ a_N = 0 \quad \text{with} \quad a_{N - } \neq 0 \]
\[ \implies 2\nu N - \beta = 0\implies \nu =
\frac{\beta}{2N} \]
Substituting $\nu, \beta$,
\[ \boxed{E_N = -\frac{e^4 m_e}{32 \pi^2 \eps_0^2
\hbar^2} \frac{1}{N^2}} \]
with $N = 1, 2, 3, \dots$ same as Bohr's energy
spectrum. Eigenfunction $R_N(r)$, substitute $2N
\nu = \beta$ in (11) and find
\[ \frac{a_n}{a_{n - 1}} = -2\nu \frac{N - n}{n(n
+ 1)} \tag{12} \]
Can use (12) to find coefficient of $R_N(r)$.
\begin{itemize}
    \item[$N = 1$], polynomial of degree 0, set
        $a_0 = 1$ then normalise
        \[ R_1(r) = A_1 e^{-\nu r} \]
    \item[$N = 2$], polynomial of degree 1, set
        $a_0 = 1$,
        \[ \frac{a_1}{a_0} \stackrel{(12)}{=}
        -2\nu \frac{2 - 1}{2} \implies a_1 = -\nu
        a_0 = -\nu \]
        \[ R_2(r) = A_2(1 - \nu r) e^{-\nu r} \]
    \item[$N = 3$], polynomial of degree 2, $a_0 =
        1$, $a_1 = -2\nu$, $a_2 = \frac{2}{3}
        \nu^2$
        \[ R_3(r) = A_3(1 - 2\nu r + \frac{2}{3}
        \nu^2 r^2)e^{-\nu r} \]
\end{itemize}
In general
\[ R_N(r) = L_N(\nu r) e^{-\nu r} \]
where $L_n$ is the Laguerre polynomial of $O(N -
1)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9b4f5d3a6fe211ed.png}
\end{center}
$P(r) \propto r^2|R_N(r)|^2$. \\
Exercise: Compute $A_1$ and compare closest to
nucleus radius to Bohr radius
\[ \langle \hat{r} \rangle_{\chi_1 = R_1 Y_{00}} =
\frac{3}{2} a_0 \]
(Bohr radius is $\left. \dfrac{P(r)}{r} \right|_{r
= a_0} = 0$)

\subsubsection*{5.3.2 $l > 0$}

\[ \dfrac[2]{R}{r} + \frac{2}{r} \dfrac{R}{r} +
\left( \frac{\beta}{r} - 2\nu - \frac{l(l +
1)}{r^2} \right) R = 0 \tag{14} \]
Asymptotic behaviour:
\[ R(r) = f(r) e^{-\nu r} \tag{15} \]
\[ \implies \dfrac[2]{f}{r} + \frac{2}{r} (1 - \nu
r) \dfrac{f}{r} + \left( \frac{\beta}{r} - 2\nu -
\frac{l(l + 1)}{r^2} \right) f = 0 \tag{16} \]
Power series
\[ f(r) = r^\sigma \sum_{n = 0}^\infty a_n r^n
\tag{17} \]
Plug (17) into (16) and identify lowest power of
$r$ and set coefficient to zero
\[ a_0[\sigma(\sigma - 1) + 2\sigma - l(l + 1)]
r^{\sigma - 2} = 0 \]
\[ \implies \sigma(\sigma + 1) - l(l + 1) = 0 \]
So have $\sigma = -l - 1$ or $\sigma = l$. But if
$\sigma = -l - 1$ then $R(r) \sim \frac{1}{r^{l +
1}}$ as $r \to 0$, which is not integrable near $r
= 0$. But if $\sigma = l$, then $R(r) \sim 0$ as
$r \to 0$ which is fine. Now we know
\[ f(r) = r^l \sum_{n = 0}^\infty a_n r^n \tag{18} \]
Plug (18) into (16) and find
\[ \boxed{a_n = \frac{2\nu(n + l) - \beta}{n(n + 2l -1)}
a_{n - 1}} \tag{19} \]
As before easy to show that $R(r)$ would diverge
unless
\[ \exists n_{\text{max}} > 0 \quad \text{such
that} \quad a_{n_\text{max}} = 0,
a_{n_{\text{max}} - 1} \neq 0 \]
Plug $a_{n_{\text{max}}}$ in (19).
\[ 2\nu\ub{(n_{\text{max}} + l)}_{\equiv N} - \beta = 0 \]
\[ \implies 2\nu N - \beta = 0 \implies \nu =
\frac{\beta}{2N} \]
\begin{itemize}
    \item $E_N = -\frac{e^4 m_E}{32 \pi^2 \eps_0^2
        \hbar^2} \frac{1}{N^2}$, $N = 1, 2, \dots$
    \item Eigenvalues same but the degeneracy is
        larger $\forall N$, $N = n_{\text{max}} +
        l$. Can have $l = 0, 1, \dots, N - 1$.
        $-l \le m \le l$.
        \[ \ub{D(N)}_{\text{degeneracy}} = \sum_{l
        = 0}^{N - 1} \sum_{m = -l}^l 1 = \sum_{l =
        0}^{N - 1} (2l + 1) = N^2 \]
        energy level $N$ you have $N^2$ (linearly
        independent) states
        with same $E_N$.
    \item Eigenfunctions
        \[ \chi_{N, l, m}(r, \theta, \phi) = R_{N,
        l}(r) Y_{l, m}(\theta, \phi) = r^l g_{N,
        l} e^{-r/2N} Y_{l, m}(\theta, \phi) \]
        $g_{n, l}(r)$ polynomial of degree $(N - l
        - 1)$ defined by
        \[ g_{N, l}(r) = \sum_{n = 0}^{N - l - 1}
        a_k r^k \]
        with $a_k = \frac{2\nu}{k} \frac{k + l -
        N}{k + 2l + 1}$ (generalised Laguerre
        polynomials) quantum numbers $N = 0, 1, 2,
        \dots$ (principal quantum numbers), $l =
        0, \dots, N - 1$ (total angular momentum),
        $m = -l, \dots, l$ (azimuthal quantum
        number).
\end{itemize}
For $N = 4$ $l = 0$,
\[ R_{4, 0}(r) \propto (1 + c_{4, 0} r + d_{4, 0}
r^2 + e_{4, 0} r^2) e^{-r\beta/8} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c31342da6fe411ed.png}
\end{center}
$Y_{00}(\theta, \phi) = \frac{1}{\sqrt{4\pi}}$,
\begin{center}
    \includegraphics[width=0.4\linewidth]
    {images/d9c8b7586fe411ed.png}
\end{center}
For $N = 4$, $l = 1$,
\[ R_{4,1}(r) \propto r(c_{4, 1} + d_{4, 1} r +
e_{4, 1} r^2) e^{-r\beta/8} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/22febc246fe511ed.png}
\end{center}
$Y_{1,0}(\theta, \phi)$, $Y_{1,1}(\theta, \phi)$,
$Y_{1,-1}(\theta, \phi)$.
\begin{center}
    \includegraphics[width=0.4\linewidth]
    {images/3256935e6fe511ed.png}
\end{center}
For $N = 4$, $l = 2$,
\[ R_{4, 2}(r) \propto r^2(c_{4, 2} + d_{r, 2}r)
e^{-2\beta/8} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b4affa026fe511ed.png}
\end{center}
$Y_{2,0}(\theta, \phi)$, $Y_{2,\pm1}(\theta,
\phi)$, $Y_{2,\pm2}(\theta, \phi)$.
$N = 4$, $l = 3$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c11120826fe511ed.png}
\end{center}
\[ R_{4,3} = r^3(c_{4,3})e^{-r\beta/8} \]
$Y_{3,0}, Y_{3,\pm1}, Y_{3,\pm2},Y_{3\pm3}$.
\myskip
Bohr model:
\begin{itemize}
    \item $E_N$ was correct
    \item Bohr radius was sort of correct
    \item $L^2 = N^2\hbar^2$ wrong. Instead $L^2 =
        l(l + 1)\hbar^2$ with $l < N$.
    \item degeneracy wrong.
\end{itemize}

\subsection{Periodic table}

$z$, $e^-$,
\[ \chi(\bf{x}_1, \bf{x}_2, \dots, \bf{x}_z) =
\chi(\bf{x}_1) \cdots \chi(\bf{x}_z) \]
$E = \sum_{j = 1}^N E_j$. It's a poor
approximation.
