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\noindent
Expectation value of an observable is the mean
(average) of infinite series of measurements
performed on particles on the same state.
\begin{align*}
    \langle p \rangle
    &= m \dfrac{\langle x \rangle}{t} \\
    &= m \dfrac{}{t} \int_{-\infty}^\infty \psi^*
    x \psi \dd x \\
    &= m \int_{-\infty}^\infty x \pfrac{}{t}
    (\psi^* \psi) \\
    &= \frac{i\hbar m}{2m} \int_{-\infty}^\infty x
    \pfrac{}{x} \left( \psi^* \pfrac{\psi}{x} -
    \psi \pfrac{\psi^*}{x} \right) \dd x
    &&\text{(TDSE)} \\
    &= - \frac{i\hbar}{2} \int_{-\infty}^\infty
    \left( \psi^* \pfrac{\psi}{x} - \psi
    \pfrac{\psi^*}{x} \right) \dd x \\
    &= -i\hbar \int_{-\infty}^\infty \psi^*
    \pfrac{\psi}{x} \dd x \\
    &= \int_{-\infty}^\infty \psi^* \left( -i\hbar
    \pfrac{}{x} \right) \psi \dd x
\end{align*}
position $\to x$ \\
momentum $\to -i\hbar \pfrac{}{x}$

\begin{hiddenflashcard}[position-operator]
    Position operator? $\hat{x} = \cloze{x}$
\end{hiddenflashcard}

\begin{hiddenflashcard}[momentum-operator]
    Momentum operator? $\hat{p} = \cloze{-i\hbar
    \pfrac{}{x}}$
\end{hiddenflashcard}

\subsubsection*{2.5.2 Hermitian operators}

In $\CC^n$ linear map $\CC^n \to \CC^n$
\[ T : \ub{\bf{v}}_{\in \CC^n} \to \ub{\bf{w}}_{\in \CC^n}
\quad \bf{w} = T \bf{v} \]
In quantum mechanics linear maps $\mathcal{H} \to
\mathcal{H}$
\[ \hat{O} : \psi \to \tilde{\psi} \quad
\tilde{\psi} = (\hat{O} \psi) (x, t) \]

\begin{definition*}
    An operator $\hat{O}$ is any linear map
    $\mathcal{H} \to \mathcal{H}$ such that
    \[ \hat{O} (a_1 \psi_1 + a_2 \psi_2) = a_1
    \hat{O}(\psi_1) + a_2 \hat{O}(\psi_2) \]
    with $a_1, a_2 \in \CC$, $\psi_1, \psi_2 \in
    \mathcal{H}$. 
\end{definition*}

\subsubsection*{Examples}

\begin{itemize}
    \item finite differential operators
        \[ \sum_{m = 0}^N p_n(X) \pfrac[n]{}{x} \]
        with $p_n(x)$ a polynomial. In particular,
        $x$ and $-i \hbar \pfrac{}{x}$ are special
        cases.
    \item Translation
        \[ \hat{S}_a : \psi(x) \to \psi(x - a) \]
    \item Parity
        \[ \hat{P} : \psi(x) \to \psi(-x) \]
\end{itemize}

\begin{flashcard}[hermitian-conjugate]
\begin{definition*}
    The Hermitian conjugate $\hat{O}^\dag$ of an
    operator $\hat{O}$ is the operator such that
    \[ \cloze{\langle \hat{O}^\top \psi_1, \psi_2
    \rangle}
    = \cloze{\langle \psi_1, \hat{O} \psi_2 \rangle \quad
    \forall \psi_1, \psi_2 \in \mathcal{H}} \]
\end{definition*}
\end{flashcard}

\noindent
Verify (from the properties of the inner product)
that
\begin{itemize}
    \item $(a_1 \hat{A}_1 + a_2 \hat{A}_2)^\dag =
        a_1^* \hat{A}_1^\dag + a_2^*
        \hat{A}_2^\dag$ for any $a_1, a_2 \in \CC$
    \item $(\hat{A} \hat{B})^\dag = \hat{B}^\dag
        \hat{A}^\dag$.
\end{itemize}

\begin{definition*}
    An operator $\hat{O}$ is \emph{Hermitian} if
    \[ \hat{O} = \hat{O}^\dag \iff \langle \hat{O}
    \psi_1, \psi_2 \rangle = \langle \psi_1,
    \hat{O} \psi_2 \rangle \]
    All physics quantities in quantum mechanics
    are represented by Hermitian operators.
\end{definition*}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $\hat{x} : \psi(x, t) \to x\psi(x, t)$
        verify that $\hat{x}^\dag = \hat{x} \iff
            (\hat{x} \psi_1, \psi_2) = \psi_1
            \hat{x} \psi_2)$ for $\psi_1, \psi_2
            \in \mathcal{H}$
            \[ \langle x\psi_1, \psi_2 \rangle =
            \int_{-\infty}^\infty (x\psi_1)^*
            \psi_2 \dd x = \int_{-\infty}^\infty
            \psi_1^* x \psi_2 \dd x = \langle
            \psi_1, x \psi_2 \rangle \]
        \item $\hat{P} : \psi(x, t) \to
            -i\hbar \pfrac{\psi}{x}(x, t)$
            verify:
            \begin{align*}
                \langle \hat{P} \psi_1, \psi_2
                \rangle
                &= \int_{-\infty}^\infty \left(
                -i\hbar \pfrac{\psi_1}{x}
                \right)^* \psi_2 \dd x \\
                &= i\hbar [\psi_1^*
                \psi_2]_{-\infty}^\infty -i\hbar
                \int_{-\infty}^\infty \psi_1^*
                \pfrac{\psi_2}{x} \dd x \\
                &= \int_{-\infty}^\infty \psi_1^*
                \left( -i\hbar \pfrac{\psi_2}{x}
                \right) \dd x \\
                &= \langle \psi_1, \hat{P} \psi_2
                \rangle
            \end{align*}
        \item Kinetic energy
            \[ \hat{T} : \psi(x, t) \to
            \frac{\hat{P}^2}{2m} \psi(x, t) =
            -\frac{\hbar^2}{2m} \pfrac[2]{\psi}{x}
            \psi(x, t) \]
        \item potential energy
            \[ \hat{U} : \psi(x, t) \to U(\hat{X})
            \psi(x, t) = U(x) \psi(X, t) \]
        \item total energy
            \[ \hat{H} : \psi(x, t) \to (\hat{T} +
            \hat{U}) \psi(x, t) = \left(
            -\frac{\hbar^2}{2m} \pfrac[2]{}{x} +
            U(x) \right) \psi(x, t) \]
            Exercise: prove that $\hat{H}$ (the
            Hamiltonian operator) is Hermitian.
\end{enumerate}

\begin{theorem}
    The eigenvalue of Hermitian operators are
    real.
\end{theorem}

\begin{proof}
    Let $\hat{A}$ be a Hermitian operator with
    eigenvalue $a \in \CC$
    \[ \langle \psi, \hat{A} \psi \rangle =
    \langle \psi, a\psi \rangle = a\langle \psi,
    \psi \rangle = a \]
    But $\hat{A}$ Hermitian:
    \[ \langle \psi, \hat{A} \psi \rangle =
    \langle \hat{A} \psi, \psi \rangle = \langle a
    \psi, \psi \rangle = a^* \langle \psi, \psi
    \rangle = a^* \]
    $\implies a = a^*$.
\end{proof}

\begin{theorem}
    If $\hat{A}$ Hermitian operator, $\psi_1,
    \psi_2$ normalised eigenfunctions of $\hat{A}$
    with eigenvalues $a_1, a_2$ with $a_1 \neq
    a_2$ then $\psi_1$ and $\psi_2$ are
    orthogonal.
\end{theorem}

\begin{proof}
    We have
    \[ \hat{A} \psi_1 = a_1 \psi_1 \quad \hat{A}\
    \psi_2 = a_2 \psi_2 \qquad a_1, a_2 \in \RR \]
    Then
    \begin{align*}
        a_1 \langle \psi_1, \psi_2 \rangle
        &= a_1^* \langle \psi_1, \psi_2 \rangle \\
        &= \langle a_1, \psi_1, \psi_2 \rangle \\
        &= \langle \hat{A} \psi_1, \psi_2 \rangle
        \\
        &= \langle \psi_1, \hat{A} \psi_2 \rangle
        \\
        &= \langle \psi_1, \hat{A} \psi_2 \rangle
        \\
        &= \langle \psi_1, a \psi_2 \rangle \\
        &= a_2 \langle \psi_1, \psi_2 \rangle
    \end{align*}
    so $\langle \psi_1, \psi_2 \rangle = 0$ since
    $a_1 \neq a_2$.
\end{proof}

\begin{theorem}
    The discrete (or continuous) set of
    eigenfunctions of any Hermitian operator
    together form a complete orthonormal basis of
    $\mathcal{H}$.
    \[ \psi(x, t) = \sum_{i = 1}^N c_i \psi_i(x,
    t) \]
    $c_i \in \CC$, $\{\psi_i\}$ a set of
    eigenfunctions of $\hat{A} = \hat{A}^\dag$.
\end{theorem}