% vim: tw=50 % 18/10/2022 12PM \subsubsection*{Properties of inner product} \begin{enumerate}[(i)] \item $\langle \psi, \phi \rangle = \langle \phi, \psi \rangle^*$ \item antilinear in first entry, linear in second entry. So $\forall a_1, a_2 \in \CC$, \[ \langle a_1 \psi_1 + a_2 \psi_2, \phi \rangle = a_1^* \langle \psi_1, \phi \rangle + a_2^* \langle \psi_2, \phi \rangle \] \[ \langle \psi_1, a_1 \phi_1 + a_2 \phi_2 = a_1 \langle \psi, \phi_1 \rangle + a_2 \langle \psi, \phi_2 \rangle \] \item inner product of $\psi \in \mathcal{H}$ with itself is non-negative \[ \langle \psi, \psi \rangle = \int_{\RR^3} |\psi(\bf{x}, t)|^2 \dd^3 x > 0 \] \end{enumerate} \begin{definition*} The norm of wave function $\psi$ is the real function \[ \|\psi\| \equiv \sqrt{\langle \psi, \psi \rangle} \] \end{definition*} \begin{flashcard}[normalised-wavefunction] \begin{definition*} Wavefunction $\psi$ is normalised if \cloze{$\|\psi\| = 1$.} \end{definition*} \end{flashcard} \begin{definition*} Two wave functions $\psi, \phi$ are orthogonal if \[ \langle \psi, \phi \rangle = 0 \] \end{definition*} \begin{definition*} A set of wavefunctions $\{\psi_n\}$ is orthonormal if \[ \langle \psi_m, \psi_n \rangle = \delta_{mn} \] \end{definition*} \begin{definition*} A set of wavefunctions $\{\psi_n\}$ is complete if for all $\phi \in \mathcal{H}$ can be written as a linear combination of them \[ \forall \phi \in \mathcal{H} \qquad \phi = \sum_{n = 0}^\infty c_n \psi_n \qquad c_n \in \CC, \psi_n \in \mathcal{H} \] \end{definition*} \begin{lemma} If $\{\psi_n\}$ form a complete orthonormal basis of $\mathcal{H}$ then $c_n = \langle \psi_n, \phi \rangle$. \end{lemma} \begin{proof} \begin{align*} \langle \psi_n, \phi \rangle &= \left\langle \psi_n, \sum_{m = 0}^\infty c_m \psi_m \right\rangle \\ &= \sum_{m = 0}^\infty c_m \langle \psi_n, \psi_m \rangle \\ &= \sum_{m = 0}^\infty c_m \delta_{mn} \\ &= c_n \qedhere \end{align*} \end{proof} \subsection{Time-dependent Schr\"odinger equation} Recap: first postulate of quantum mechanics is Born's rule \[ P(\bf{x}, t) = \rho(\bf{x}, t) \dd^3 \bf{x} = |\psi(\bf{x}, t)|^2 \dd \bf{x} \] The second postulate is time dependent Schr\"odinger equation (TDSE): \begin{flashcard}[TDSE] \prompt{Time dependent Schr\"odinger equation?} \[ \boxed{\cloze{i\hbar \pfrac{\psi}{t} (\bf{x}, t)} = \cloze{-\frac{\hbar^2}{2m} \nabla^2 \psi(\bf{x}, t) + U(\bf{x}) \psi(\bf{x}, t)}} \] \end{flashcard} where $U(\bf{x}) \in \RR$ (potential). \begin{itemize} \item First derivative in $t$: once $\psi(x, t_0)$ is known, we can find out $\psi(x, t)$ at all times. \item asymmetry between $t$ and $x$, so time dependent Schr\"odinger equation is a non-relativistic equation. \end{itemize} \subsubsection*{Heuristic interpretation} $e^-$ diffraction (interference) $\to$ $e^-$ behaves like waves \[ \psi(x, t) \propto \exp[i(\bf{k} \cdot \bf{x} - \omega t)] \] almost describes the dynamics of $e^-$. Take De-Broglie \[ kbg = \frac{\bf{p}}{\hbar} \qquad \omega = \frac{E}{\hbar} \] for free particle \[E = \frac{|\bf{p}|^2}{2m} \implies \omega = \frac{|\bf{p}|^2}{2m\hbar} = \frac{\hbar}{2m} |\bf{k}|^2 \] dispersion relation for a particle-wave \[ \omega \propto |\bf{k}|^2 \] while for light-waves \[ \omega \propto |\bf{k}| \] if $\exp[i(\bf{k} \cdot \bf{x} - \omega t)]$ is a solution of the equation for the wave of $e^-$ and if $\omega = \frac{\hbar}{2m} |\bf{k}|^2$ then \[ \exp[i(\bf{k} \cdot \bf{x}) - i \frac{|\bf{k}|^2}{2m} \hbar t] = \exp[i(kx - \frac{k^2}{2m} \hbar t)] \] by dimensional analysis. \subsubsection*{Properties} \begin{enumerate}[(i)] \item $\int_{\RR^3} |\psi(\bf{x}, t)|^2 \dd^3 x = \mathcal{N} < \infty$. \begin{proof} \begin{align*} \dfrac{\mathcal{N}}{t} &= \frac{}{t} \int_{\RR^3} |\psi(\bf{x}, t)|^2 \dd^3 x \\ &= \int_{\RR^3} \pfrac{}{t} |\psi(\bf{x}, t)|^2 \dd^3 x \end{align*} but \[ \pfrac{}{t} (\psi^*(\bf{x}, t) \psi(\bf{x}, t)) = \psi^* \pfrac{\psi}{t} + \pfrac{\psi^*}{t} \psi \] Now TDSE gives \[ \pfrac{\psi}{t} = \frac{i\hbar}{2m} \nabla^2 \psi - i \frac{U}{\hbar}\psi \] and TDSE$^*$ gives \[ \pfrac{\psi^*}{t} = -\frac{i\hbar}{2m} \nabla^2 \psi^* + i \frac{U}{\hbar} \psi^* \] \[ \implies \pfrac{}{t} (\psi^* \psi) = \nabla \cdot \left[ \frac{i\hbar}{2m} (\psi^* \nabla \psi - \psi \nabla \psi^* ) \right] \] \[ \implies \dfrac{\mathcal{N}}{t} = \int_{\RR^3} \nabla \cdot \left[ \frac{i\hbar}{2m} (\psi^* \nabla \psi - \psi \nabla \psi^* \right] = 0 \] because $\psi, \psi^*$ are such that $|\psi|, |\psi^*| \to 0$ as $|\bf{x}| \to \infty$. \end{proof} \item Normalisation of wavefunction constant in time $\implies$ probability is conserved \begin{flashcard}[qm-continuity-equation] \prompt{QM continuity equation} \[ \cloze{\pfrac{\rho(\bf{x}, t)}{t} + \nabla \cdot \bf{J}(\bf{x}, t)} = 0 \] \end{flashcard} \[ \bf{J}(\bf{x}, t) = -[\cdots] = -\frac{i\hbar}{2m} [\psi^*(\bf{x}, t) \nabla \psi(\bf{x}, t) - \psi(\bf{x}, t) \nabla \psi^*(\bf{x}, t)] \] \begin{hiddenflashcard}[probability-current] \prompt{Probability current} \[ \bf{J}(\bf{x}, t) = \cloze{-\frac{i\hbar}{2m} (\psi^* \nabla \psi - \psi \nabla \psi^*)} \] \end{hiddenflashcard} \begin{hiddenflashcard}[probability-density] \prompt{Probability density} \[ \rho(\bf{x}, t) = \cloze{\psi^*(\bf{x}, t) \psi(\bf{x}, t)} \] \end{hiddenflashcard} (the conserved probability current of quantum physics states). \end{enumerate} \subsection{Expectation values and operators} How to extract info from $\psi$? \begin{definition*} Observable = any property of the particle describe by $\psi$ that can be measured. \end{definition*} \noindent In Quantum mechanics $\to$ operator acting on $\psi$, measurement $\to$ expectation value of an operator. \subsubsection*{2.5.1 Heuristic interpretation} From probabilistic interpretation, if want to measure the position of particle: \begin{align*} \langle x \rangle = \int_{-\infty}^\infty x |\psi(x, t)|^2 \dd x &= \int_{-\infty}^\infty \psi^*(x, t) x \psi(x, t) \dd x \end{align*} \[ O_x \to \hat{x} \to x \]