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\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*}
\[ O_x \to \hat{x} \to x \]
