% 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 \[ \begin{cases} \hat{L}^2 Y(\theta, \phi) = \lambda Y(\theta, \phi) & \qquad (1) \\ \hat{L}_3 Y(\theta, \phi) = \hbar m Y(\theta, \phi) & \qquad (2) \end{cases} \]