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\newpage
\section{1 dimensional solutions of Schr\"odinger
equation}

TISE (Time independent Schr\"odinger equation):
\[ \hat{H} \chi(X) = E\chi(x) \]
\[ -\frac{\hbar^2}{2m} \chi''(x) + U(x) \chi(x) =
E \chi(x) \]
with $E \in \RR$. We will solve TISE in 3 cases:
\begin{enumerate}
    \item[3.1] Bound states
    \item[3.2] Free particle
    \item[3.3] scattering states.
\end{enumerate}

\subsection{Bound states}

\subsubsection*{3.1.1 Infinite potential well}

\[ U(x) = \begin{cases}
    0 & |x| \le a \\
    +\infty & |x| > 0
\end{cases} \]
$a \in \RR^+$.
\begin{itemize}
    \item for $|x| > 0$, $\chi(x) = 0$ otherwise
        $U \cdot \chi = \infty$ so boundary
        condition $\chi(\pm a) = 0$.
    \item for $|x| \le a$ we look for solutions of
        \[ -\frac{\hbar^2}{2m}\chi''(x) = E
        \chi(x) \]
        \[ \implies \chi''(x) + k^2 \chi(x) = 0 \]
        with $k = \sqrt{\frac{2mE}{\hbar^2}}$ and
        we also have $\chi(\pm a) = 0$. Solution:
        \[ \chi(x) = A \sin (kx) + B \cos(kx) \]
        $\chi(a) = 0$, $\chi(-a) = 0$ implies
        \[ A\sin (ka) = 0, \quad B\cos (ka) = 0 \]
        so two options:
        \begin{enumerate}[(i)]
            \item $A = 0$, $\cos (ka) = 0$ then
                $k_n = \frac{n\pi}{2a}$, $n$ an
                odd integer.
                \[ \chi_n(x) = B \cos(k_n x) \]
                the \emph{even solutions}.
            \item $B = 0$, $\sin(ka) = 0$ then
                $k_n = \frac{n\pi}{2a}$, $n$ even
                integer.
                \[ \chi_n(x) = A \sin(k_n x) \]
                the \emph{odd solutions}.
        \end{enumerate}
        Determine $A$, $B$ by requiring
        normalisation of eigenfunction
        \[ \int_{-a}^a |\chi_n(x)|^2 \dd x = 0
        \implies A = B = \sqrt{\frac{1}{Q}} \]
        Solution: eigenvalues of $\hat{H}$ are
        \[ E_n = \frac{\hbar^2}{2m} k_n^2 =
        \hbar^2 \frac{\pi^2}{8ma^2}n^2 \]
        eigenfunction of $\hat{H}$
        \[ \chi_n(x) = \sqrt{\frac{1}{Q}} \begin{cases}
            \cos \left( \frac{n\pi x}{2a} \right)
            & n = 1, 3, \dots \\
            \sin \left( \frac{n\pi x}{2a} \right)
            & n = 2, 4, \dots
        \end{cases} \]
        .image
        \begin{enumerate}[(i)]
            \item Ground state has  $E \neq 0$.
                Note (contrarily to classical
                mechanics)
            \item $n \to \infty$, $|\chi_n(x)|^2
                \to \text{const}$ (Classical
                mechanics limits)
        \end{enumerate}
\end{itemize}
In classical mechanics
\[ P(x) \propto \frac{1}{\mathcal{N}(x)} \quad
P(x) = \frac{A}{\mathcal{N}(x)} \]
In this case particle free inside the wall
\[ \implies \mathcal{N} \text{constant} \implies P
\text{constant} \]

\begin{proposition*}
    If quantum system has non-degenerate
    eigenstates ($E_i \neq E_j$ for $i \neq j$)
    then, if $U(x) = U(-x)$ the eigenfunction of
    $\hat{H}$ have to be either odd or even.
\end{proposition*}

\begin{proof}
    If $U(x) = U(-x)$ then TISE invariant under $x
    \to -x$. If $\chi(x)$ is a solution with
    eigenvalue $E$, then also $\chi(-x)$ solution
    and $\chi(-x) = \alpha \chi(x)$ solutions must
    be the same up to a normalisation
    factor $\alpha$. Then
    \[ \chi(x) = \chi(-(-x)) = \alpha \chi(-x) =
    \alpha^2 \chi (x) \]
    \[ \implies \alpha^2 = 1 \implies \alpha = \pm
    1 \]
    \[ \implies \chi(x) = \pm \chi(-x) \qedhere \]
\end{proof}

\subsubsection*{3.1.2 Finite potential well}

\[ U(x) = \begin{cases}
    0 & |x| \le a \\
    U_0 & |x| > a
\end{cases} \]
Consider $E > 0$ ($E < 0$ does not exist in this
case) and $E < U_0$ (bound state) We look for odd
/ even eigenfunction
\begin{enumerate}[(i)]
    \item even parity bound states
        \[ \chi(-x) = \chi(x) \]
        solve
        \[ -\frac{\hbar^2}{2m} \chi''(x) = E
        \chi(x) \qquad |x| \le a \tag{I} \]
        \[ -\frac{\hbar^2}{2m} \chi''(x) = (E -
        U_0) \chi(x) \qquad |x| > a \tag{II} \]
        \begin{enumerate}[(I)]
            \item $\chi''(x) + k^2 \chi(x) = 0$
                with $k =
                \sqrt{\frac{2mE}{\hbar^2}}$
                \[ \chi(x) = A \sin(kx) + B
                \cos(kx) \]
                but $A = 0$ (even parity)
                \[ \chi(x) = B\cos(kx) \]
            \item $\chi''(x) - \ol{k}^2 \chi(x) =
                0$ with $k = \sqrt{\frac{2m(U_0 -
                E)}{\hbar^2}}$
                \[ \chi(x) = ce^{+\ol{k} x} + D
                e^{-\ol{k} x} \]
                but impose normalisability implies
                $x > a$, $c = 0$, $x < -a$, $D =
                0$. Impose even parity $C = D$.
        \end{enumerate}
        To summarise:
        \[ \chi(x) = \begin{cases}
            C e^{\ol{k} x} & x < -a \\
            B \cos(kx) & |x| \le a \\
            C e^{-\ol{k} x} & x > a
        \end{cases} \]
        Impose continuity of $\chi(x)$ at $x = \pm
        a$, $\chi'(x)$ at $x = \pm a$. Then
        \[ \chi(a) \to C e^{-\ol{k} a} = B
        \cos(ka) \]
        \[ \chi'(a) \to -\ol{k} C e^{-\ol{k} a} =
        -kB \sin(ka) \]
        if take ratio from definition
        \[ k\tan(ka) = \ol{k} \]
        \[ k^2 + \ol{k}^2 = \frac{2m U_0}{\hbar^2} \]
        Define rescaled variables $\xi = ka$, $\eta
        = \ol{k} a$.
        \[ \xi \tan \xi = \eta \]
        \[ \xi^2 + \eta^2 = r_0^2 \]
        \[ r_)^2 = \frac{2m U_0}{\hbar^2}a^2 \]
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/fdbe470859dd11ed.png}
        \end{center}
        eigenvalues of $\hat{H}$ corespond to
        points of intersection
        \[ E_n = \frac{\hbar^2}{2ma^2} \xi_n^2
        \quad n = 1, \dots, p \]
        \begin{note*}
            $U_0 \to \infty \implies r_0 \to
            \infty \implies E_n =
            \frac{\hbar^2}{8ma^2} (2n - 1)^2
            \pi^2$, $\chi \to \chi_n$ of infinite
            well.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/19c8ce0059de11ed.png}
            \end{center}
        \end{note*}
\end{enumerate}

\noindent
Exercise:
\begin{enumerate}[(1)]
    \item Use the unused condition in system to
        write $C$ in terms of $B$
    \item Impose normalisation to 1 to find $B$.
\end{enumerate}
