% vim: tw=50
% 22/11/2022 12PM

\subsubsection*{Recap of Quantum Mechanics in 3D
(Section 5)}

\begin{itemize}
    \item \[ -\frac{\hbar^2}{2m} \nabla^2
        \chi(\bf{x}) + U(\bf{x}) \chi(\bf{x}) = E
        \chi(\bf{x}) \qquad \bf{x} \in \RR^3 \]
        1D:
        \[ + \pfrac[2]{}{x} \]
        \[ \hat{p} = -i\hbar \pfrac{}{x} \]
        \[ \hat{p}^2 = -\hbar^2 \pfrac[2]{}{x} \]
        3D:
        \[ \nabla^2 = \pfrac[2]{}{x_1} +
        \pfrac[2]{}{x_2} + \pfrac[2]{}{x_3} \]
        \[ \hat{\bf{p}} = -i\hbar \nabla = \left(
        -i\hbar \pfrac{}{x_1} + -i\hbar
        \pfrac{}{x_2}, -\hbar \pfrac{}{x_3} \right) \]
        \[ |\hat{\bf{p}}|^2 = -\hbar^2 \nabla^2 \]
    \item Useful to write $\nabla^2$ in spherical
        coordinate $(r, \theta, \phi)$
        \[ \nabla^2 = \frac{1}{r} \pfrac[2]{}{r} r
        + \frac{1}{r^2 \sin^2\theta} \left[
        \sin\theta + \pfrac{}{\theta} \left(
        \sin\theta \pfrac{}{\theta} \right) +
        \pfrac[2]{}{\phi} \right] \]
    \item If $U(\bf{x}) = U(r)$ (spherically
        symmetric potential) we can find some
        special solutions of TISE $\chi(r)$
        (radial solutions).
    \item If take $(xhf) = U(r)$, $\chi(r, \theta,
        \phi_ = \chi(r)$
        \[ -\frac{\hbar^2}{2mr} \pfrac[2]{}{r} (r
        \chi(r) + U(r) \chi(r) = E \chi(r) \]
        if define $\sigma(r) = r \chi(r)$, TISE
        for $\chi(r)$ becomes
        \[ -\frac{\hbar^2}{2m}
        \dfrac[2]{\sigma(r)}{r} + U(r) \sigma(r) =
        E \sigma(r) \]
        in $\RR^+$, and with normalisation
        condition
        \[ \int_0^\infty |\sigma(r)|^2 \dd r <
        \infty \]
        because of normalisation conditions
        $\sigma(r) \to a$ as $r \to 0$. But we
        found $a = 0$. Why? If we allowed
        $\sigma(r) \approx a \neq 0$ as $r \to 0$
        (which means $\chi(r) \sim \frac{a}{r}$)
        then $\hat{H}$ would not be Hermitian.
        \begin{proof}
            For $\hat{H}$ to be Hermitian we need
            \[ (\phi, \hat{H} \chi) = (\hat{H}
            \phi, \chi) \qquad \forall \phi, \chi
            \in \mathcal{H} \]
            \begin{align*}
                (\phi, \hat{H} \chi)
                &=
                \int_0^\infty \dd r r^2 \phi(r)
                \hat{H} \chi(r) \\
                &= -\frac{\hbar^2}{2m}
                \int_0^\infty \dd r \phi
                \dfrac{}{r} \left( r^2
                \dfrac{\chi}{r} \right) \\
                &= -\frac{\hbar^2}{2m} \left[ r^2
                \phi \dfrac{\chi}{r} - r^2 \chi
                \dfrac{\phi}{r} \right]_0^\infty
                \ub{-
                \frac{\hbar^2}{2m} \dd r
                \dfrac{}{r} \left( r^2
                \dfrac{\phi}{r} \right)
                \chi}_{(\hat{H} \phi, \chi)}
            \end{align*}
            If $\phi(r) \sim B$ as $ \to 0$ with
            $B \neq 0$ then $\chi(r) \sim
            \frac{A}{r}$ as $r \to 0$ with $A \neq
            0$ then
            \[ r^2 \phi \dfrac{\chi}{r} - r^2 \chi
            \dfrac{\phi}{r} \not \to 0 \]
            as $r \to 0$.
        \end{proof}
        Due to Quantum Mechanics interpretation we
        classify $\chi(r) \sim \frac{A}{r}$ as
        unphysical, hence $\sigma(r) = 0$ at $r =
        0$.
\end{itemize}

\subsubsection*{Continuing from before the recap}

Properties:
\begin{itemize}
    \item $\hat{L}_i$ is Hermitian (Example sheet)
    \item $[\hat{L}_i, \hat{L}_j] \neq 0$ if $i
        \neq j$ (Example sheet). $\implies$
        different components
        of $\bf{L}$ cannot be determined
        simultaneously.
        \[ [\hat{L}_i, \hat{L}_j] = i\hbar
        \eps_{ijk} \hat{L}_k \]
        \begin{proof}
            \begin{align*}
                [\hat{L}_1, \hat{L}_2] \chi(x_1,
                x_2, x_3)
                &= -\hbar^2 \left[ \left( x_2
                \pfrac{}{x_3} - x_3 \pfrac{}{x_2}
                \right) \left( x_3 \pfrac{}{x_1}
                - x_1 \pfrac{}{x_3} \right) -
                \left( x_3 \pfrac{}{x_1} - x_1
                \pfrac{}{x_3} \right) \left( x_2
                \pfrac{}{x_3} - x_3 \pfrac{}{x_2}
                \right) \right] \chi(x_1, x_2,
                x_3) \\
                &= -\hbar^2 \left( x_2
                \pfrac{}{x_1} - x_1 \pfrac{}{x_2}
                \right) \chi(x_1, x_2, x_3) \\
                &= i\hbar \hat{L}_3 \chi(x_1, x_2,
                x_3) \qedhere
            \end{align*}
        \end{proof}
\end{itemize}

\begin{definition*}
    Total angular momentum operator $\hat{L}^2$
    \[ \hat{L}^2 = \hat{L}_1^2 + \hat{L}_2^2 +
    \hat{L}_3^2 \]
\end{definition*}

\noindent
Properties:
\begin{itemize}
    \item $[\hat{L}^2, \hat{L}_i] = 0$ (Example
        sheet)
    \item for $U(r)$ $[\hat{L}^2, \hat{H}] = 0$
        ($*$), $[\hat{L}_i, \hat{H}] = 0$.
        \begin{proof}
            \begin{itemize}
                \item \begin{align*}
                        [\hat{L}_i, \hat{x}_j]
                        &= [\eps_{imn} \hat{x}_m
                        \hat{p}_n, \hat{x}_j] \\
                        &= \eps_{imn} [\hat{x}_m
                        \hat{p}_n, \hat{x}_j] \\
                        &= \eps_{imn} (\hat{x}_m
                        [\hat{p}_n, \hat{x}_j] +
                        [\hat{x}_m, \hat{x}_j]
                        \hat{p}_n) \\
                        &= -i\hbar \eps_{imj}
                        \hat{x}_m \\
                        &= i\hbar \eps_{ijm}
                        \hat{x}_m
                \end{align*}
                \item \begin{align*}
                        [\hat{L}_i, \hat{x}_j^2]
                        &= [\hat{L}_i, \hat{x}_j]
                        + \hat{x}_j [\hat{L}_i,
                        \hat{x}_j] \\
                        &= i\hbar \eps_{ijm}
                        (\hat{x}_m \hat{x}_j +
                        \hat{x}_j \hat{x}_m) \\
                        &= 0
                \end{align*}
                \item $[\hat{L}_u, U(r)] = 0$ since
                    $r = \sqrt{\hat{x}_1^2 +
                    \hat{x}_2^2 + \hat{x}_3^2}$.
                \item $[\hat{L}_i, \hat{p}_j] =
                    i\hbar \eps_{ijm} \hat{p}_m$
                    (same proof as for $x_j$)
                \item $[\hat{L}_i, \hat{p}^2] = 0$
            \end{itemize}
            \[ \implies [\hat{L}_i, \hat{H}] = 0 \]
            and
            \[ [\hat{L}^2, \hat{H}] = 0 \]
            (trivially)
        \end{proof}
\end{itemize}
$\{\hat{H}, \hat{L}^2, \hat{L}_i\}$ set of
mutually commuting operators. Take $i =
3$. $\implies$
\begin{enumerate}[(1)]
    \item Can find joint eigenstates of these 3
        operators that form a basis of
        $\mathcal{H}$.
    \item eigenvalues of these 3 operators
        $|\bf{L}|$, $L_z$, $E$ can be
        simultaneously measured at an arbitrary
        precision.
    \item The set of operators is \emph{maximal}
        i.e. we cannot construct another
        independent operator (other than
        $\hat{I}$) that commutes with them.
\end{enumerate}
To find joint eigenfunctions of $\hat{L}^2$ and
$\hat{L}_3$ write $\hat{\bf{L}}$ in spherical
coordinates (appendix 7 of Maria Ubiali's notes)
\[ i\hbar \left( x_2 \pfrac{}{x_3} - x_3
\pfrac{}{x_2}, \dots, \dots \right) \]
\[ \pfrac{}{x_1} = \left( \pfrac{r}{x_1} \right)
\pfrac{}{r} + \left( \pfrac{\theta}{x_1} \right)
\pfrac{}{\theta} + \left( \pfrac{\phi}{x_1}
\right) \pfrac{}{\phi} \]
And put
\[ \hat{L}_3 = -i\hbar \pfrac{}{\phi} \]
\[ \hat{L}^2 = -\frac{\hbar^2}{\sin^2\theta}
\left[ \sin\theta \pfrac{}{\theta} \left(
\sin\theta \pfrac{}{\theta} \right) +
\pfrac[2]{}{\phi} \right] \]
Next time we will look for joint eigenfunction
\[ Y(\theta, \phi) \]
such that
\[ \begin{cases}
    \hat{L}^2 Y(\theta, \phi) = \lambda Y(\theta,
    \phi) & \qquad (1) \\
    \hat{L}_3 Y(\theta, \phi) = \hbar m Y(\theta,
    \phi) & \qquad (2)
\end{cases} \]
