% vim: tw=50
% 17/11/2022 12PM

\newpage
\section{3D solutions of Schr\"odinger equation}

\subsection{TISE in 3D for spherically symmetric
potentials}

\begin{flashcard}[3D-TISE]
    \prompt{TISE in 3D?}
\[ \cloze{-\frac{\hbar^2}{2m} \nabla^2 \chi(\bf{x}) +
U(\bf{x}) \chi(\bf{x})} = \cloze{E \chi(\bf{x})} \]
\end{flashcard}
Laplacian operator $\nabla^2$
\begin{itemize}
    \item Cartesian coordinates $(x, y, z)$:
        \[ \nabla^2 \equiv \pfrac[2]{}{x} +
        \pfrac[2]{}{y} + \pfrac[2]{}{z} \]
    \item Spherical coordinates $(r, \theta,
        \phi)$
        \[ \nabla^2 = \frac{1}{r} \pfrac[2]{}{r} (R) +
        \frac{1}{r^2 \sin^2\theta} \left[ \sin\theta
        \pfrac{}{\theta} \left( \sin\theta
        \pfrac{}{\theta} \right) + \pfrac[2]{}{\phi} \right]  \]
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/2239675869f511ed.png}
        \end{center}
        \begin{align*}
            x
            &= r\cos\phi \sin \theta \\
            y
            &= r \sin \phi \sin \theta \\
            z
            &= r \cos\theta
        \end{align*}
        $0 \le r < \infty$, $0 \le \theta \le
        \pi$, $0 \le \phi \le 2\pi$. Reminder:
        \[ \int_{\RR^3} \dd V =
        \int_{-\infty}^\infty \dd x
        \int_{-\infty}^\infty \dd y
        \int_{-\infty}^\infty \dd z \]
        \[ \int_{\RR^3} \dd V = \int_0^{2\pi} \dd
        \phi \int_{-1}^1 \dd \ub{\cos
        \theta}_{\to \int_0^\pi \sin \theta \dd
        \theta}
        \int_0^\infty r^2 \dd r\]
\end{itemize}

\begin{definition*}
    Spherically symmetric potential
    \[ U(\bf{x}) = U(r, \theta, \phi) \equiv U(r) \]
    Clearly, even with a spherically symmetric
    potential $\phi(r, \theta, \phi)$.
\end{definition*}

\noindent
We start by focussing on a particular sub-class of
solutions of TISE, i.e. on Radial eigenfunctions
$\chi(r)$. If $\chi(r, \theta, \phi) = \chi(r)$
then
\[ \nabla^2 \chi(r) = \frac{1}{r} \pfrac[2]{}{r}
(r\chi(r)) \]
Plugging this into TISE in 3D:
\[ \boxed{-\frac{\hbar^2}{2m} \left(
\dfrac[2]{\chi}{r} + \frac{2}{r} \dfrac{\chi}{r}
\right) + U(r) \chi = E\chi} \tag{$*$} \]
Normalisation condition for $\chi \in
\mathcal{H}$:
\[ \int_{\RR^3}|\chi(r, \theta, \phi)|^2 \dd V <
\infty \]
\[ \implies \int_0^\infty |\chi(r)|^2 r^2 \dd r <
\infty \]
eigenfunctions $\chi(r)$ must go to 0 sufficiently
fast at $r \to \infty$ and behave well ($\sim
\frac{1}{r}$) (most singular behaviour) at $r \to
0$. \myskip
How to solve ($*$)? One way of doing it is to
define
\[ \sigma(r) \equiv r\chi(r) \]
\[ \implies -\frac{\hbar^2}{2m}
\dfrac[2]{\sigma(r)}{r} + U(r) \sigma(r) =
E\sigma(r) \tag{$**$} \]
This is like the 1D TISE defined only on $\RR^+$
and with usual normalisation condition on $\RR^2$:
\[ \int_0^\infty |\sigma(r)|^2 \dd r < \infty \]
We want $\sigma(r) = 0$ at $r = 0$, $\sigma'(r)$
finite at $r = 0$. \\
$\implies$ Solve ($**$) on $\RR$ and look for odd
solutions:
\[ \sigma(-r) = -\sigma(r) \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/ba35c9c469f611ed.png}
\end{center}

\subsubsection*{Example: Spherically symmetric
potential well}

\[ U(r) = \begin{cases}
    0 & r \le a \\
    U_0 & r > a
\end{cases} \qquad a \in \RR^+, U_0 \in \RR^+ \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/16d648b669f711ed.png}
\end{center}
TISE as ($**$) and solve it for $\sigma(r) =
r\chi(r)$ by analytically continuation on whole
$\RR$ and looking only for \emph{odd} solutions.
\[ -\frac{\hbar^2}{2m} \dfrac[2]{\sigma(r)}{r} +
U(r)\sigma(r) = E\sigma(r) \]
Look for odd parity bound states
\[ 0 \le E \le U_0 \]
\[ K = \sqrt{\frac{2mE}{\hbar^2}} \qquad \ol{k} =
\sqrt{\frac{2m(U_0 - E)}{\hbar^2}} \]
odd solutions:
\[ \sigma(r) = \begin{cases}
    A \sin(kr) & |r| \le a \\
    B e^{-\ol{k}r} & r > a \\
    -B e^{+\ol{k}r} & r < -a
\end{cases} \]
Boundary conditions for $\sigma(r)$:
\begin{itemize}
    \item continuity of $\sigma(r)$ at $r = a$
    \item continuity of $\sigma'(r)$ at $r = a$.
\end{itemize}
\[ \implies \begin{cases}
    A\sin ka = B e^{-\ol{k}a} \\
    kA \cos ka = -\ol{k} B e^{-\ol{k}a}
\end{cases} \]
\[ \implies -k\cot(ka) = \ol{k} \]
From definition:
\[ k^2 + \ol{k^2} = \frac{2mU_0}{\hbar^2} \]
Solve this graphically by defining
\[ \zeta = ka, \qquad \to \eta = -\zeta \cot\zeta \]
\[ \eta = \ol{k}a \qquad \to \eta^2 + \xi^2 =
r_0^2 \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/cfd3886e69f811ed.png}
\end{center}
If $r_0 < \frac{\pi}{2}$ ($\iff U_0 < \frac{\pi^2
\hbar^2}{3ma^2}$) then doesn't exist solution.
Two differences:
\begin{enumerate}[(1)]
    \item Below a given threshold for $U_0$ there
        does not exist bound state in 3D.
        (contrarily to 1D in which there exists
        even bound state)
    \item \[ \chi(r) = \begin{cases}
            A \frac{\sin(kr)}{r} & r < Q \\
            B \frac{e^{-\ol{k}r}}{r} & r \ge Q
    \end{cases} \]
\end{enumerate}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/13bd6f5469f911ed.png}
\end{center}

\subsection{Angular momentum in Quantum Mechanics}

Classical mechanics:
\[ \bf{L} = \bf{x} \times \bf{p} \]
When you have $U(r)$ then
\[ \dfrac{\bf{L}}{t} = \dot{\bf{x}} \times \bf{p}
+ \bf{x} \times \dot{\bf{p}} = 0 \]
In Dynamics and relativity the conservation of
angular momentum implies that 3D $\to$ 2D (once
take the plane $\bf{L} \cdot \bf{x} = 0$) $\to$ 1D
(solve Newton's second law on $\bf{e}_r$).

\begin{flashcard}[angular-momentum-operator]
\begin{definition*}
    Angular momentum operator
    \fcscrap{\[ \hat{\bf{L}} = \hat{\bf{x}} \times
    \hat{\bf{p}} \]}
    \[ \hat{\bf{L}} = \cloze{-i\hbar \bf{x} \times
    \nabla} \]
    \fcscrap{In 1D: $\hat{p} = -\hbar \pfrac{}{x}$ \\
    In 3D: $\hat{\bf{p}} = -\hbar\nabla$,
    $\hat{\bf{x}} = \bf{x}$.}
\end{definition*}
\end{flashcard}

\noindent
Write it in cartesian coordinates $(x_1, x_2,
x_3)$
\[ \hat{L}_i = -\hbar \eps_{ijk} x_j \pfrac{}{x_k}
\qquad \rightarrow (\eps_{ijk} \hat{x}_j \hat{p}_k) \]
$i = 1, 2, 3$.