% vim: tw=50
% 10/11/2022 12PM

\subsubsection*{Recap of chapter 2}

Hermitian operators $\leftrightarrow$ observables
\[ \hat{O}^+ = \hat{O} \iff (\hat{O} \psi, \phi) =
(\psi, \hat{O}, \phi) ~\forall \psi, \phi \in
\mathcal{H} \]
Have:
\begin{itemize}
    \item Real eigenvalues (Theorem 2.1)
    \item If $\hat{O} \psi_1 = a \psi_1$, $\hat{O}
        \psi_2 = b \psi_2$ with $a \neq b$ then
        $(\psi_1, \psi_2) = 0$ (Theorem 2.5)
    \item Eigenstates of Hermitian operator form a
        complete basis of $\mathcal{H}$. (Theorem
        2.6)
\end{itemize}
Quantum measurement:
\begin{itemize}
    \item Eigenvalues of $\hat{O}$ are possible
        outcomes of measurement of the observable $O$.
    \item If $\psi = \sum_i a_i \psi_i$, $\psi_i$
        eigenstates of $\hat{O}$ then $P(O =
        \lambda_i) = a_i^2 = |(\psi_i, \psi)|^2$
    \item Immediately after a measurement with
        outcome $\lambda_i$, the wave function
        becomes $\psi_i$.
\end{itemize}

\newpage
\section{Simultaneous measurements in Quantum
Mechanics}

\subsection{Commutators}

\begin{flashcard}[commutator-of-operators]
\begin{definition*}
    Commutator of two operators $\hat{A}, \hat{B}$
    is the operator
    \[ [\hat{A}, \hat{B}] = \cloze{\hat{A} \hat{B} -
    \hat{B} \hat{A}} \]
\end{definition*}
\end{flashcard}

\noindent
Properties:
\begin{itemize}
    \item $[\hat{A}, \hat{B}] = -[\hat{B},
        \hat{A}]$
    \item $[\hat{A}, \hat{A}] = 0$
    \item $[\hat{A}, \hat{B} \hat{C}] = [\hat{A},
        \hat{B}] \hat{C} + \hat{B} [\hat{A},
        \hat{C}]$
    \item $[\hat{A}, \hat{B}, \hat{C}] =
        \hat{A}[\hat{B}, \hat{C}] + [\hat{A},
        \hat{C}] \hat{B}$.
\end{itemize}
Exercise: Compute $[\hat{x}, \hat{p}]$ in 1
dimension. \\
Take $\psi \in \mathcal{H}$
\[ \hat{x} \hat{p} \psi = x \left( -i \hbar
\pfrac{}{x} \right) \psi(x) = -i\hbar x
\pfrac{\psi}{x}(x) \]
\[ \hat{p}\hat{x} \psi = -i\hbar \pfrac{}{x}(x
\psi(x)) = -i\hbar \psi(x) - i\hbar x
\pfrac{\psi}{x} \]
\[ \implies [\hat{x}, \hat{p}]\psi = i\hbar \psi
\implies [\hat{x}, \hat{p}] = i\hbar \hat{I} \]
Canonical commutator relation.

\begin{definition*}
    Two Hermitian operators $\hat{A}$ and
    $\hat{B}$ are \emph{simultaneously} diagonalisable in
    $\mathcal{H}$ is it exists a complete basis of
    joint eigenfunctions $\{\psi_i\}$ such that
    \[ \hat{A} \psi_i = a_i \psi_i \]
    \[ \hat{B} \psi_i = b_i \psi_i \]
    with $a_i, b_i \in \RR$.
\end{definition*}

\begin{theorem}
    Two Hermitian operators $\hat{A}$ and
    $\hat{B}$ are simultaneously diagonalisable
    \[ \iff [\hat{A}, \hat{B}] = 0 \]
\end{theorem}

\begin{proof}
    \begin{enumerate}
        \item[$\Rightarrow$] If $\hat{A}, \hat{B}$
            simultaneously diagonalisable then
            $\{\psi_i\}$ set of joint
            eigenfunctions that is a complete
            basis of $\mathcal{H}$.
            \[ \forall \psi_i \quad [\hat{A},
            \hat{B}]\psi_i = \hat{A}\hat{B} \psi_i
            -\hat{B} \hat{A} \psi_i = (a_i b_i -
            b_i a_i)\psi_i = 0 \]
            Take $\psi \in \mathcal{H}$.
            \[ [\hat{A}, \hat{B} = \sum_i c_i
            [\hat{A}, \hat{B}] \psi_i = 0 \]
            \[ \implies [\hat{A}, \hat{B}] = 0 \]
        \item[$\Leftarrow$] If $[\hat{A}, \hat{B}]
            = 0$ and $\psi_i$ eigenfunction of
            $\hat{A}$ with eigenvalues $a_i$.
            \[ 0 = [\hat{A}, \hat{B}] \psi_i =
            \hat{A} \hat{B} \psi_i - \hat{B}
            \hat{A} \psi_i = \hat{A} \hat{B}
            \psi_i - a_i \hat{B} \psi_i \]
            so
            \[ \hat{A}(\hat{B} \psi_i) = a_i
            (\hat{B} \psi_i) \]
            $\hat{B}$ maps the eigenspace $E_i$ of
            $\hat{A}$ with eigenvalue $a_i$ into
            itself so $\hat{B} \mid_{E_i}$ is an
            Hermitian operator of $E_i$. Since
            this holds for all eigenspace $E_i$ of
            $\hat{A}$, we can find a complete
            basis of simultaneous eigenfunctions
            of $\hat{A}$ and $\hat{B}$.
    \end{enumerate}
\end{proof}

\subsection{Heisenberg's Uncertainty Principle}

\begin{flashcard}[uncertainty-in-measurement-of-observable]
\begin{definition*}
    The uncertainty in a measurement of an
    observable $A$ on a state $\psi$ is defined as
    \[ \Delta_\psi A = \cloze{\sqrt{(\Delta_\psi
    A)^2}} \]
    where
    \begin{align*}
        \cloze{(\Delta_\psi A)^2}
        \fcscrap{&= \langle (\hat{A} - \langle \hat{A}
        \rangle_\psi \hat{I})^2 \rangle_\psi \\}
        &= \cloze{\langle \hat{A}^2 \rangle_\psi -
        (\langle \hat{A} \rangle_\psi)^2}
    \end{align*}
\end{definition*}
\end{flashcard}

\noindent
The two definitions are equivalent:
\begin{align*}
    \langle (\hat{A} - \langle \hat{A}
    \rangle_\psi \hat{I})^2 \rangle_\psi
    &= \int_{\RR^3} \psi^* (\hat{A} - \langle
    \hat{A} \rangle_\psi \hat{I})^2 \psi \dd^3 x
    \\
    &= \int_{\RR^3} \psi^* \hat{A}^2 \psi \dd^3 x
    + (\langle \hat{A} \rangle_\psi)^2
    \int_{\RR^3} \psi^* \psi \dd^3 x - 2 \langle
    \hat{A} \rangle_\psi \int_{\RR^3} \psi^*
    \hat{A} \psi \dd^3 x \\
    &= \langle \hat{A}^2 \rangle_\psi + (\langle
    \hat{A} \rangle_\psi)^2 - 2(\langle \hat{A}
    \rangle_\psi)^2 \\
    &+ \langle \hat{A} \rangle_\psi^2 - (\langle
    \hat{A} \rangle_\psi)^2
\end{align*}

\begin{flashcard}[uncertainty-lemma]
\begin{lemma}
    $(\Delta_\psi A)^2 \ge 0$ and $(\Delta_\psi A)
    = 0 \iff \psi$ is eigenfunction of $\hat{A}$.
\end{lemma}
\end{flashcard}

\begin{proof}
    \begin{align*}
        (\Delta_\psi A)^2
        &= \langle (\hat{A} - \langle \hat{A}
        \rangle_\psi \hat{I})^2 \rangle_\psi \\
        &= (\psi, (\hat{A} - \langle \hat{A}
        \rangle_\psi \hat{I})^2 \psi) \\
        &= ((\hat{A} - \langle \hat{A}
        \rangle_\psi \hat{I}) \psi, (\hat{A} -
        \langle \hat{A} \rangle_\psi \hat{I})\psi)
        \\
        &= \langle \phi, \phi) \\
        &\ge 0
    \end{align*}
    (Call $\phi = (\hat{A} - \langle \hat{A}
    \rangle_\psi \hat{I})\psi$)
    Now prove that $(\Delta_\psi A)^2 = 0 \iff
    \phi = 0$.
    \begin{enumerate}
        \item[$\Rightarrow$] $(\Delta_\psi A)^2 =
            (\phi, \phi) = 0$ if $\phi = 0$
            implies
            \[ \hat{A} \psi = \langle \hat{A}
            \rangle_\psi \psi \]
            i.e. $\psi$ eigenfunction of
            $\hat{A}$.
        \item If $\psi$ is eigenfunction of
            $\hat{A}$ with eigenvalue $a \in \RR$
            then
            \[ \langle \hat{A} \rangle_\psi =
            (\psi, \hat{A} \psi) = a(\psi, \psi) =
            a \]
            \[ \langle \hat{A} \rangle_\psi =
            (\psi, \hat{A}^2 \psi) = a^2 (\psi,
            \psi) = a^2 \]
            using second definition,
            \[ (\Delta_\psi A)^2 = \langle
            \hat{A}^2 \rangle_\psi - (\langle
            \hat{A} \rangle_\psi)^2 = a^2 -  a^2 =
            0 \qedhere \]
    \end{enumerate}
\end{proof}

\begin{lemma}
    If $\psi, \phi \in \mathcal{H}$, then
    \[ |(\phi, \psi)|^2 \le (\phi, \phi) (\psi,
    \psi) \]
    and $|(\phi, \psi)|^2 = (\phi, \phi)(\psi,
    \psi)$ if and only if $\phi = a \psi$ for $a
    \in \CC$.
\end{lemma}

\noindent
(proof comes from Schwarz inequality
and is available in Maria Ubiali's notes).

\begin{flashcard}[generalised-uncertainty-theorem]
\begin{theorem}[Generalised uncertainty theorem]
    If $A$ and $B$ observables and $\psi \in
    \mathcal{H}$ then
    \[ (\Delta_\psi A)(\Delta_\psi B) \ge \cloze{\half
    |(\psi, [\hat{A}, \hat{B}]\psi)|} \]
\end{theorem}
\end{flashcard}

\begin{proof}
    \begin{align*}
        (\Delta_\psi A)^2
        &= ((\hat{A} - \langle \hat{A}
        \rangle_\psi \hat{I}) \psi, (\hat{A} -
        \langle \hat{A} \rangle_\psi \hat{I})\psi)
    \end{align*}
    Define
    \[ \hat{A}' = \hat{A} - \langle \hat{A}
    \rangle_\psi \hat{I} \]
    \[ \hat{B}' = \hat{B} - \langle \hat{B}
    \rangle_\psi \hat{I} \]
    Hence
    \[ (\Delta_\psi A)^2 = (\hat{A}'\psi,
    \hat{A}'\psi) \]
    \[ (\Delta_\psi B)^2 = (\hat{B}' \psi,
    \hat{B}' \psi) \]
    Using lemma 4.3:
    \[ (\Delta_\psi A)^2 (\Delta_\psi B)^2 \ge
    |(\hat{A}' \psi, \hat{B}' \psi)|^2 \tag{1} \]
    and RHS is equal to $|(\psi, \hat{A}'\hat{B}'
    \psi)|^2$ because $\hat{A}'$ is Hermitian.
    Define
    \[ [\hat{A}', \hat{B}'] = \hat{A}'\hat{B}' -
    \hat{B}'\hat{A}' \tag{2} \]
    \[ \{\hat{A}', \hat{B}'\} = \hat{A}'\hat{B}' +
    \hat{B}'\hat{A}' \tag{3} \]
    if $\hat{A}', \hat{B}'$ Hermitian
    \[ [\hat{A}', \hat{B}']^\dag = -[\hat{A}',
    \hat{B}'] \tag{4} \]
    Now writing
    \[ \hat{A}' \hat{B}' = \half ([\hat{A}',
    \hat{B}'] + \{\hat{A}', \hat{B}'\}) \tag{5} \]
    Plug (5) into (1)
    \[ (\Delta_\psi A)^2(\Delta_\psi B)^2 \ge
    \frac{1}{4} |(\psi, [\hat{A}', \hat{B}']\psi)
    + (\psi, \{A', B'\}\psi)|^2 \]
    Given that:
    \begin{itemize}
        \item $(\psi, \{\hat{A}', \hat{B}'\} \psi)
            \in \RR$
        \item $(\psi, [\hat{A}', \hat{B}'] \psi) =
            ir$ with $r \in \RR$
    \end{itemize}
    then
    \[ (\Delta_\psi A)^2 (\Delta_\psi B)^2 \ge
    \frac{1}{4} |(\psi, [\hat{A}', \hat{B}']
    \psi)|^2 + \frac{1}{4} |\psi, \{\hat{A}',
    \hat{B}'\} \psi)|^2 \]
    \[ \implies (\Delta_\psi A)(\Delta \psi B) \ge
    \half |(\psi, [\hat{A}, \hat{B}]\psi)|
    \qedhere \]
\end{proof}
