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\subsection{The free particle}

TISE ($U(x) = 0$):
\[ -\frac{\hbar^2}{2m} \chi''(X) = E \chi(x) \]
\[ \chi''(x) + \frac{2mE}{\hbar^2} \chi(x) = 0 \]
$k = \sqrt{\frac{2mE}{\hbar^2}}$
\[ \chi(x) = e^{ikx} \]
\[ E_k = \frac{\hbar^2 k^2}{2m} \to \chi_k(x) = e^{ikx} \]
\[ \psi_k(x, t) = \chi_k(x) e^{-iE_k t/\hbar} = e^{i(kx - \hbar k^2/2m)} \]
This wave function is \emph{not} square-integrable:
\[ \int_{-\infty}^\infty |\psi_k(x, t)|^2 \dd x =
\int_{-\infty}^\infty = \infty  \]
This is a consequence of
\[ \int_{-\infty}^\infty |\psi(x, t)|^2 \dd x =
\mathcal{N} < \infty \implies \lim_{R \to \infty}
\int_{|x| > R} \dd x |\psi(x, t)|^2 = 0 \]
How do we deal with unbound states?
\begin{enumerate}[Option 1]
    \item[Option 1] Build a linear superposition of
        not-normalisable states that is
        normalisable (section 3.2.1)
    \item[Option 2] We ignore the problem but change
        interpretation (section 3.2.2)
\end{enumerate}

\subsubsection*{3.2.1 Gaussian Wave Packet}

\[ \psi(x, t) = \int_{-\infty}^\infty A(k)
\psi_k(x, t) \dd k \]
($A(k)$ is a continuous coefficient of linear
combination) A possible option is Gaussian wave
packet:
\[ A(k) = A_{\text{GP}}(k) = \exp \left[
-\frac{\sigma}{2} (k - k_0)^2 \right] \quad \sigma
\in \RR^+, k_0 \in \RR \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/253f832c5d0411ed.png}
\end{center}
\[ \psi_{\text{GP}} (x, t) = \int_{-\infty}^\infty
A_{\text{GP}}(k) \psi_k(x, t) \dd k \]
\[ \psi_{\text{GP}} (x, t) = \int_{-\infty}^\infty
\exp[F(k)] \dd k \]
where
\begin{align*}
    F(k)
    &= -\frac{\sigma}{2} (k - k_0)^2 + ikx
    - \frac{i\hbar k^2}{2m}t \\
    &= -\half \left( \sigma + \frac{i\hbar t}{m}
    \right) k^2 + (k_0 \sigma + ix) k
\end{align*}
\[ \alpha \equiv \sigma + \frac{i\hbar t}{m} \]
\[ \beta \equiv k_0 \sigma + ix \]
\[ \delta = -\frac{\sigma}{2} k_0^2 \]
Complete the square:
\[ F(k) = -\frac{\alpha}{2} \left( k -
\frac{\beta}{\alpha} \right)^2 +
\frac{\beta^2}{2\alpha} + \delta \]
\[ \implies Y_{\text{GP}}(x, t) = \exp \left[
\frac{\beta^2}{2\alpha} + \delta \right]
\int_{-\infty}^\infty \exp \left[
-\frac{\alpha}{2} \left( k - \frac{\beta}{\alpha}
\right)^2 \right] \dd k \]
Shift contour $\tilde{k} = k -
\frac{\beta}{\alpha}$. Let $\nu = \Im \left(
\frac{\beta}{\alpha} \right)$.
\[ \psi_{\text{GP}} (x, t) = \exp \left[
\frac{\beta^2}{2\alpha} + \delta \right]
\int_{-\infty - i\nu}^{\infty - i\nu} \exp \left(
-\frac{\alpha}{2} \tilde{k}^2 \right) \dd
\tilde{k} \]
Using standard Gaussian integral
\[ I(\alpha) = \int_{-\infty}^\infty \exp(-ay^2)
\dd y = \sqrt{\frac{\pi}{a}} \]
We get
\[ \psi_{\text{GP}}(x, t) =
\sqrt{\frac{2\pi}{\alpha}} \exp \left[
\frac{\beta^2}{2\alpha} + \delta \right] \]
Exercise: Write $\psi_{\text{GP}}(x, t)$ by
substituting $\beta, \alpha, \delta$ and normalise
it to 1.
\[ \beta = k_0 \sigma + ix \quad \beta^2 = k_0^2
\sigma^2 - k^2 + 2ix k_0\sigma \]
The $-x^2$ in $\beta^2$ implies that
$\psi_{\text{GP}}$ is normalisable. Once
$\psi_{\text{GP}}$ is normalised,
$\ol{\psi}_{\text{GP}}$ cen define
\[ \rho_{\text{GP}}(x, t) =
|\ol{\psi}_{\text{GP}}(x, t)|^2 =
\sqrt{\frac{\sigma}{\pi \left( \sigma^2 +
\frac{\hbar^2 t^2}{m^2} \right)}} \exp \left[
\frac{-\pi \left( x - \frac{\hbar k_0 t}{m}
\right)^2}{(\sigma^2 + \frac{\hbar^2 t^2}{m^2})} \right] \]
at $t$ fixed:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/ea2575b45d0611ed.png}
\end{center}
width of distance
\[ \sqrt{\half \left( \sigma + \frac{\hbar^2
t^2}{m^2 \sigma} \right)} \]
The centre of the distribution is $\langle x
\rangle_{\psi_{\text{GP}}}$:
\begin{align*}
    \langle x \rangle_{\psi_{\text{GP}}}
    &= \int_{-\infty}^\infty
    \ol{\psi}^*_{\text{GP}}(x, t) x
    \ol{\psi}_{\text{GP}} (x, t) \dd x \\
    &= \int_{-\infty}^\infty x \rho_{\text{GP}}(x,
    t) \\
    &= \frac{\hbar k_0}{m}t
\end{align*}
Error on position of particle:
\[ \Delta x = \sqrt{\langle x^2
\rangle_{\psi_{\text{GP}}} - \langle x
\rangle_{\psi_{\text{GP}}}^2} = \sqrt{\half \left(
\sigma + \frac{\hbar^2 t^2}{m^2\sigma} \right)} \]
$\Delta x = \sqrt{\frac{\pi}{2}}$ at $t = 0$.
$\Delta x$ increases as $t$ increases. Given
$\psi_{\text{GP}}$ it is interesting to compute
$\langle p \rangle$, $\Delta p$
\begin{align*}
    \langle p \rangle_{\psi_{\text{GP}}}
    &= \int_{-\infty}^\infty
    \ol{\psi}^*_{\text{GP}}(x, t) \left( -i\hbar
    \frac{}{x} \ol{\psi}_{\text{GP}} (x, t)
    \right) \dd x \\
    &= \hbar k_0
\end{align*}
\[ \Delta p = \sqrt{\langle p^2
\rangle_{\psi_{\text{GP}}} - \langle p
\rangle_{\psi_{\text{GP}}}^2} \]
To calculate $\Delta p$ on $\psi_{\text{GP}}$ we
have
\[ \langle p \rangle_{\psi_{\text{GP}}}^2 =
\hbar^2 k_0^2 \]
we need
\[ \langle p^2 \rangle_{\psi_{\text{GP}}} =
\int_{-\infty}^\infty \ol{\psi}^*_{\text{GP}}(x,
t) \left( -\hbar^2 \dfrac[2]{}{x}
\ol{\psi}_{\text{GP}} (x, t) \right) \dd x \]
If you compute it and plug it into $\Delta p$
THE FOLLOWING SECTION IS ALL WRONG, IGNORE UNTIL
TOLD TO STOP IGNORING.
\[ \Delta p = \frac{\hbar}{\sqrt{2 \left(
\sigma + \frac{\hbar^2 t^2}{m\sigma} \right)}} \]
at $t = 0$, $\Delta p = \hbar
\sqrt{\frac{2}{\sigma}}$, as $t \to \infty$,
$\Delta p$ decreases as $\frac{1}{\sqrt{a + t^2}}$
What we learnt is
\[ \Delta x \to \infty, \Delta p \to \infty \quad
\text{as $t \to \infty$} \]
\[ \Delta x \Delta p = \frac{\hbar}{2} \]
STOP IGNORING. \\
At time $t = 0$, $\Delta x \Delta p =
\frac{\hbar}{2}$. \myskip
The GP is a state of minimum uncertainty. Other
$A(k)$ would give you a normalisable state but if
you compute $\Delta x \Delta p$ you would find
something $> \frac{\hbar}{2}$. \\
Exercise: Compare what you find for $\psi_k(x, t)$
\[ \Delta x = \infty, \Delta p = 0 \]
\[ \langle x \rangle_{\psi_k} = 0 , \langle x^2
\rangle_{\psi_k} = \infty \]