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\subsubsection*{Consequences of generalised
uncertainty theorem}

\begin{itemize}
    \item $[\hat{A}, \hat{B}] = 0$ if and only if
        there exists joint set of eigenstates
        which form a complete basis of
        $\mathcal{H}$ which happens if and only if
        $A, B$ can be measured simultaneously with
        arbitrary precision on a given state.
    \item Take $\hat{A} = \hat{x}$, $\hat{B} =
        \hat{p}$. Given that $[\hat{x}, \hat{p}] =
        i\hbar \hat{I}$
        \[ \implies (\Delta_\psi x)(\Delta_\psi p)
        \ge \frac{\hbar}{2} \]
        (Heisenberg's uncertainty principle).
\end{itemize}
We had shown explicitly that, if $\psi =
\psi_{\text{GP}}$ then
\[ (\Delta_{\psi_{\text{GP}}} x)
(\Delta_{\psi_{\text{GP}}} p) = \frac{\hbar}{2} \]
at $t = 0$. (this is the minimum uncertainty). The
reason for this lies in two lemmas:
\begin{enumerate}[(i)]
    \item Lemma 4.5: $\psi$ is a state of minimum
        uncertainty
        \[ \iff \hat{x} \psi = ia \hat{p} \psi
        \quad a \in \RR \]
    \item Lemma 4.6: The condition for 4.5 to hold
        is
        \[ \psi(x) = C e^{-bx^2} \quad c \in \CC,
        b \in \RR^+ \]
\end{enumerate}
Exercise: Verify that $\psi_k(x, t) = e^{ikx}
e^{-E_k t/\hbar}$ does not satisfy equation of
Lemma 4.5.

\subsection{Ehrenfest theorem}

Time evolution of operators.

\begin{theorem}
    The expectation value of an Hermitian operator
    $\hat{A}$ evolves according to
    \[ \dfrac{}{t} \langle \hat{A} \rangle_\psi =
    \frac{i}{\hbar} \langle [\hat{H}, \hat{A}]
    \rangle_\psi + \left\langle \pfrac{\hat{A}}{t}
    \right\rangle_\psi \]
\end{theorem}

\begin{proof}
    \begin{align*}
        \dfrac{}{t} \langle \hat{A} \rangle_\psi
        &= \dfrac{}{t} \int_{-\infty}^\infty
        \psi^*(x, t) \hat{A} \psi(x, t) \dd x \\
        &= \int_{-\infty}^\infty \pfrac{}{t}
        (\psi^* \hat{A} \psi) \dd x \\
        &= \int_{-\infty}^\infty \left(
        \pfrac{\psi^*}{t} \hat{A} \psi + \psi^*
        \pfrac{\hat{A}}{t} \psi + \psi^* \hat{A}
        \pfrac{\psi}{t} \right) \dd x \\
        &= \frac{i}{\hbar} \int_{-\infty}^\infty
        \psi^* (\hat{H} \hat{A} - \hat{A} \hat{H})
        \psi \dd x + \left\langle \pfrac{\hat{A}}{t}
        \right\rangle_\psi \\
        &= \frac{i}{\hbar} \int_{-\infty}^\infty
        \psi^* [\hat{H}, \hat{A}] \psi \dd x +
        \left\langle \pfrac{\hat{A}}{t}
        \right\rangle_\psi \\
        &= \frac{i}{\hbar} \langle [\hat{H},
        \hat{A}] \rangle_\psi + \left\langle
        \pfrac{\hat{A}}{t} \right\rangle_\psi \qedhere
    \end{align*}
\end{proof}

\subsubsection*{Examples}
\begin{enumerate}[(1)]
    \item Take $\hat{A} = \hat{H}$
        \[ \implies \dfrac{\langle \hat{H}
        \rangle_\psi}{t} = 0 \]
        ($\dfrac{E}{t} = 0$)
    \item Take $\hat{A} = \hat{p}$.
        \begin{align*}
            [\hat{H}, \hat{p}] \psi
            &= \left[ \frac{\hat{p}^2}{2m} +
            U(\hat{x}), \hat{p} \right] \psi \\
            &= [U(\hat{x}), \hat{p}] \psi \\
            &= U(x) \left( -i\hbar \pfrac{}{x}
            \right) \psi(x, t) - \left(
            -i\hbar \pfrac{}{x} \right) [U(x) \psi(x,
            t) \\
            &= \cancel{i\hbar U(x)
            \pfrac{\psi}{x}(x, t)} +
            \cancel{i\hbar U(x) \pfrac{\psi}{x}(x,
            t)} + i\hbar \pfrac{U}{x}(x) \psi(x,
            t) \\
            \implies \dfrac{\langle \hat{p}
            \rangle_\psi}{t}
            &= \frac{i}{\hbar} \langle [\hat{H},
            \hat{p}] \rangle_\psi \\
            &= - \left\langle \pfrac{U}{x}
            \right\rangle_\psi
        \end{align*}
    \item $\hat{A} = \hat{x}$
        \begin{align*}
            [\hat{H}, \hat{x}]
            &= \left[ \frac{\hat{p}^2}{2m} +
            U(\hat{x}), \hat{x} \right] \\
            &= \frac{1}{2m} [\hat{p}^2, \hat{x}^2]
            \\
            &= \frac{1}{2m} (\hat{p} \ub{[\hat{p},
            \hat{x}]}_{-i\hbar \hat{I}} +
            \ub{[\hat{p}, \hat{x}]}_{i\hbar \hat{I}}
            \hat{p}) \\
            &= -\frac{i\hbar}{m} \hat{p} \\
            \dfrac{\langle \hat{x} \rangle_\psi}{t}
            &= \frac{i}{\hbar} \langle [\hat{H},
            \hat{x}] \rangle_\psi \\
            &= \frac{\langle \hat{p}
            \rangle_\psi}{m}
        \end{align*}
        (matches the classical $\dot{x} =
        \frac{p}{m}$)
\end{enumerate}

\subsection{Harmonic oscillator revisited
(non-examinable)}

\[ \hat{H} = \frac{\hat{p}^2}{2m} + \half
m\omega^2 \hat{x}^2 \]
($k = m\omega^2$, elastic constant). Eigenvalues,
eigenfunctions of $\hat{H}$. Rewrite:
\begin{align*}
    \hat{H}
    &= \frac{1}{2m} (\hat{p} + im\omega
    \hat{x})(\hat{p} - im\omega \hat{x}) +
    \frac{i\omega}{2} \ub{[\hat{p}, \hat{x}]}_{-i\hbar
    \hat{I}} \\
    &= \frac{1}{2m}(\hat{p} + im\omega
    \hat{x})(\hat{p} - im\omega \hat{x}) +
    \frac{\hbar\omega}{2} \hat{I} \tag{1}
\end{align*}

\begin{definition*}
    Ladder operators
    \[ \hat{a} =\frac{1}{\sqrt{2m}} (\hat{p} -
    im\omega \hat{x}) \tag{2} \]
    \[ \hat{a}^\dag = \frac{1}{\sqrt{2m}} (\hat{p}
    + im\omega \hat{x}) \]
    \[ \implies \boxed{\hat{H} = \hat{a}^\dag \hat{a} +
    \frac{\hbar\omega}{2} \hat{I}} \tag{4} \]
\end{definition*}

\noindent
Compute
\begin{align*}
    [\hat{a}, \hat{a}^\dag]
    &= \frac{1}{2m} [\hat{p} - im \omega \hat{x},
    \hat{p} + im\omega \hat{x}] \\
    &= -\frac{im\omega}{2m} [\hat{x}, \hat{p}] +
    \frac{im\omega}{2m} [\hat{p}, \hat{x}] \\
    &= \hbar \omega \hat{I} \tag{5} \\
    [\hat{H}, \hat{a}]
    &= [\hat{a}^\dag, \hat{a}, \hat{a}] \\
    &= -\hbar \omega \hat{a} \tag{6} \\
    [\hat{H}, \hat{a}^\dag] = \hbar \omega
    \hat{a}^\dag \tag{7}
\end{align*}
Suppose $\chi$ eigenfunction of $\hat{H}$ with
eigenvalue $E$,
\[ \hat{H} \chi = E \chi \]
Take $(\hat{a} \chi)$. What is its energy?
\begin{align*}
    \hat{H}(\hat{a}, \chi)
    &= [\hat{H}, \hat{a}] \chi + \hat{a} \hat{H}
    \chi \\
    &= -\hbar \omega \hat \chi + E \hat{a} \chi \\
    &= (E - \hbar \omega) \hat{a} \chi
\end{align*}
$\hat{a} \chi)$ is eigenfunction of $\hat{H}$ with
eigenvalue $(E - \hbar \omega)$ and $\hat{a}^\dag
\chi)$ is eigenfunction of $\hat{H}$ with
eigenvalue $(E + \hbar\omega)$. Prove by
induction:
\[ (\hat{a}^n \chi) \to \text{eigenfunction with
eigenvalue $E - n \hbar\omega$} \]
\[ (\hat{a}^{\dag n} \chi) \to \text{eigenfunction with
eigenvalue $E + n \hbar\omega$} \]
Using the fact that
\[ \langle \hat{H} \rangle_\psi \ge 0 \]
then $\exists$ eigenfunction $\chi_0$ such that
\[ \hat{a} \chi_0 = 0 \]
Find $\chi_0$
\[ \frac{1}{\sqrt{2m}}(\hat{p} - im\omega \hat{x})
\chi_0) = 0 \]
\[  -i\hbar \pfrac{\chi_0}{x} - im\omega x\chi_0 =
0 \]
\[ \implies \chi_0(x0 = ce^{-m\omega x^2 /2\hbar} \]
\[ \hat{H} \chi_0 = \hat{a}^\dag \hat{a} \chi_0 +
\frac{\hbar \omega}{2} \hat{I} \chi_0 =
\frac{\hbar\omega}{2} \chi_0 \]
The excited states with $E > E_0$
\begin{align*}
    \chi_n
    &= (a^\dag)^n \chi_0 \\
    &= \frac{1}{(\sqrt{2m})^2} (\hat{p} + im\omega
    \hat{x})^n \chi_0 \\
    &= \frac{c}{(\sqrt{2m})^n} \left( -i\hbar
    \pfrac{}{x} + im\omega x \right)^n e^{-m\omega
    x^2 /2\hbar}
\end{align*}
Eigenvalues
\[ E_n = \frac{\hbar\omega}{2} + n\hbar\omega =
\left( n + \half \right) \hbar\omega \]