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\[ -\hbar \pfrac{}{\phi} Y(\theta, \phi) = \hbar
mY(\theta, \phi) \]
Find solutions
\[ \boxed{Y(\theta, \phi) = y(\theta) X(\phi)} \tag{3} \]
Plugging (3) into (2)
\[ -i\hbar \left( \pfrac{}{\phi} X(\phi) \right)
\cancel{y(\theta)} = \hbar m X(\phi)
\cancel{y(\theta)} \]
\[ X(\phi) = e^{im\phi)} \]
Given that wave function must be simple-valued in
$\RR^3$ $\implies X(\phi)$ must be invariant under
\[ \phi \to \phi + 2\pi \]
\[ \implies e^{i2m\pi} = 1 \implies m \in \ZZ
\tag{4} \]
Plug (4) into (1) and find
\[ \boxed{\frac{1}{\sin\theta} \pfrac{}{\theta}
\left( \sin \theta \pfrac{y(\theta)}{\theta}
\right) - \frac{m^2}{\sin^2\theta} y(\theta) =
-\frac{\lambda}{\hbar^2} y(\theta)} \tag{5} \]
This is the associated Legendre equation (IB
Methods) and it has solution
\[ y(\theta) = P_{l, m}(\cos\theta = (\sin
\theta)^{|m|} \dfrac[|m|]{}{(\cos\theta)}
P_l(\cos\theta) \]
(where $P_{l, m}$ is the associate Legendre
polynomial and $P_l$ is the ordinary Legendre
polynomial). Because $P_l(\cos\theta)$ is a
polynomial in $\cos\theta$ of degree $l$,
$\implies -l \le m \le l$ and (without proof) the
eigenvalues of $\hat{L}^2$ are
\[ \lambda = \hbar^2 l(l + 1) \]
($l = 0, 1, 2, \dots$) Put everything together:
\[ Y_{l, m}(\theta, \phi) = P_{l, m}(\cos\theta)
e^{im\phi} \]
$l = 0, 1, 2, \dots$, $-l \le m \le l$. Spherical
harmonics:
\[ \hat{L}^2 Y_{l, m}(\theta, \phi) = \hbar^2 l(l
= 1) Y_{l, m} (\theta \phi) \]
\[ \hat{L}_3 Y_{l, m}(\theta, \phi) = m\hbar Y_{l,
m}(\theta, \phi) \]
$l$, $m$ are quantum numbers that characterise:
\begin{itemize}
    \item $l \to$ total angular momentum
    \item $m \to$ azimuthal number, $z$-component
        of $L$.
\end{itemize}
In classical mechanics
\begin{center}
    \includegraphics[width=0.1\linewidth]
    {images/bfaf9ab26bf211ed.png}
\end{center}
\[ -|\bf{L}| \le L_z \le |\bf{L}| \leftrightarrow
-l \le m \le l \]
\[ Y_{0, 0}(\theta, \phi) = \frac{1}{\sqrt{4\pi}}
\qquad l = 0, m = 0 \]
\[ Y_{1, 0}(\theta, \phi) = \frac{3}{\sqrt{4\pi}}
\cos\theta \qquad l = 1, m = 0 \]
\[ Y_{1, \pm 1}(\theta, \phi) =
\frac{1}{\sqrt{4\pi}} \sin \theta e^{\pm i\phi}
\qquad l = 1, m = \pm 1 \]
All spherical harmonics are orthonormal (like all
eigenfunctions of Hermitian operators)
\[ (Y_{l, m}, Y_{l', m'}) =
\delta_{ll'}\delta_{mm'} \]
\[ \int_0^{2\pi} \dd \phi \int_{-1}^{1} \dd
\cos\theta Y_{lm}^* (\theta, \phi)
Y_{l'm'}(\theta, \phi) = \delta_{ll'} \delta_{mm'} \]

\subsection{The Hydrogen atom}

\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/e78d8a2e6bf411ed.png}
\end{center}
Model proton (nucleus) to be stationary at the
origin ($m_p \to \infty$, or equivalently $m_p \gg m_e$)
\[ F_{\text{coulomb}}(r) = -\frac{e^2}{4\pi
\eps_0} \frac{1}{r^2} = -\pfrac{U_{\text{coulomb}}}{r} \]
\[ U_{\text{coulomb}}(r) = -\frac{e^3}{4\pi\eps_0}
\frac{1}{r} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/fef9a7886bf411ed.png}
\end{center}
Bound states $E < 0$.
\[ -\frac{\hbar^2}{2m_e} \nabla^2 \chi(r, \theta
\phi) - \frac{e^2}{4\pi\eps_0} \frac{1}{r} \chi(r,
\theta, \phi) = E \chi(r, \theta, \phi) \tag{1} \]
Laplacian
\[ \nabla^2 = \frac{1}{r} \pfrac[2]{}{r} r +
\frac{1}{r^2 \sin^2\theta} \left( \sin\theta
\pfrac{}{\theta} \sin\theta \pfrac{}{\theta} +
\pfrac[2]{}{\phi} \right) \]
\[ \hat{L}^2 = \frac{\hbar^2}{\sin^2\theta} \left[
\sin\theta \pfrac{}{\theta} \sin \theta
\pfrac{}{\theta} + \pfrac[2]{}{\phi} \right] \]
\[ \implies -\hbar^2 \nabla^2 = -\frac{\hbar^2}{r}
\pfrac[2]{}{r} r + \frac{\hat{L}^2}{r^2} \tag{2} \]
Plug (2) into (1)
\[  -\frac{\hbar^2}{2m_e} \frac{1}{r} \left(
\pfrac[2]{}{r} r \chi(r, \theta, \phi) \right) +
\frac{\hat{L}^2}{2m_e r^2} \chi(r, \theta, \phi) -
\frac{e^2}{4\pi\eps_0 r} \chi(r, \theta, \phi) = E
\chi(r, \theta, \phi) \tag{3} \]
Because of eigenfunction of $\hat{H}$ are also
eigenfunction of $\hat{L}^2$ and $\hat{L}_3$
$\implies$ $\chi(r, \theta, \phi)$ must also be
eigenfunction of $\hat{L}^2$, $\hat{L}_3$.
\[ \implies \chi(r, \theta, \phi) = R(r) Y_{l, m}(\theta,
\phi) \]
\[ \implies \hat{L}^2 \chi = R(r) Lhar Y_{l, m}
(\theta, \phi) = \hbar^2 l(l + 1) R(r)Y_{l,
m}(\theta, \phi) \tag{4} \]
Plug (4) into (3)
\begin{align*}
    -\frac{\hbar^2}{2m_e} \left( \dfrac[2]{R(r)}{r}
    + \frac{2}{r} \dfrac{R(r)}{r} \right)
    \cancel{Y_{l, m}(\theta, \phi)}
    &+
    \frac{\hbar^2}{2m_e r^2} l(l + 1) R(r)
    \cancel{Y_{l, m}(\theta, \phi)} -
    \frac{e^2}{4\pi\eps_0}R(r) \cancel{Y_{l,
    m}(\theta, \phi)} \\
    &= ER(r) \cancel{Y_{l, m}(\theta,
    \phi)} \tag{5}
\end{align*}
We end up with a 1D equation for radial part
$R(r)$
\[ -\frac{\hbar^2}{2m} \left( \dfrac[2]{R}{r} +
\frac{2}{r} \dfrac{R}{r} \right) + \ub{\left(
-\frac{e^2}{4\pi\eps_0} \frac{1}{r} +
\frac{\hbar^2 l(l + 1)}{2m_e r^2}
\right)}_{V_{\text{eff}}(r)} R = ER
\tag{6} \]
($V_{\text{eff}}(r)$ is a bit like in classical
mechanics).

\subsubsection*{5.3.1 $l = 0$}

$V_{\text{eff}}(r) \to V_{\text{coulomb}}(r)$.
Rewrite (6) in terms of variables
\[ \nu^2 \equiv -\frac{2mE}{\hbar^2} > 0 \]
\[ \beta \equiv \frac{e^2m}{2\pi\eps_0\hbar^2} \]
In terms of $\nu^2$, $\beta$ (6) becomes
\[ \dfrac[2]{R}{r} + \frac{2}{r} \dfrac{R}{r} +
\left( \frac{\beta}{r} - \nu^2 \right) R = 0
\tag{7} \]
\begin{enumerate}[(i)]
    \item The asymptotic behaviour ($r \o \infty$)
        determined by
        \[ \dfrac[2]{R}{r} - \nu^2 R = 0 \]
        \[ R(r) \sim e^{\pm r\nu} \]
        as $r \to \infty$. Take $R(r) \sim
        e^{-r\nu}$ because of normalisability.
    \item At $r = 0$ eigenfunction has to be
        finite ($\sim A$).
\end{enumerate}
Exploiting (i) take
ansatz
\[ R(r) = f(r) e^{-\nu r} \tag{8} \]
Plug (8) into (7) and find
\[ f''(r) + \frac{2}{r} (1 - \nu r) f'(r)
+ \frac{1}{r} (\beta - 2\nu) f(r) = 0
\tag{9} \]
(9) is a homogeneous linear ODE with regular point
$r = 0$
\[ f(r) = r^c \sum_{n = 0}^\infty a_n r^n \]
\[ f'(r) = \sum_{n = 0}^\infty a_n(c + n) r^{c + n
- 1} \tag{10} \]
\[ f''(r) = \sum_{n = 0}^\infty a_n(c + n)(c + n -
1) r^{c + n - 2} \]
Plug (10) into (9):
\[ \sum_{n = 0}^\infty a_n(c + n)(c + n - 1) r^{c
+ n - 2} + \frac{2}{r} (1 - \nu r) a_n (c + n)
r^{c + n - 1} + (\beta - 2\nu) r^{c + n - 1}] = 0 \]
Constant power of $r$ has coefficient ($r^{c -
2}$)
\[ a_0 c(c - 1) + 2 a_0 c = 0 \]
\[ \implies a_0 c(c + 1) = 0 \]
$c = -1$ (then $X \sim \frac{A}{r}$) or $c = 0$
(then $X \sim A$). So $c = 0$ and the equation for
the other coefficients is
\[ \sum_{n = 1}^\infty a_n n(n + 1) a_{n - 1}
(\beta - 2\nu n)] r^{n - 2} = 0 \]
\[ \implies a_n = \frac{2\nu n - \beta}{n(n +
1)}a_{n - 1} \tag{11} \]
