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\subsubsection*{3.1.3 Harmonic Oscillator}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/6753780059de11ed.png}
\end{center}
\begin{flashcard}[harmonic-oscillator-potential]
    \prompt{Potential for quantum harmonic oscillator?}
    \[ U(x) = \cloze{\half kx^2} \]
\end{flashcard}
$k \in \RR$ elastic constant. $\omega =
\sqrt{\frac{k}{m}}$. Classical mechanics: Newton 2
is $\ddot{x}(t) = -\omega^2 x(t)$.
\[ \implies x(t) = A \sin \omega t + B \cos\omega
t\]
with $T = \frac{2\pi}{\omega}$ period
oscillations. \\
Quantum mechanics:
\[ -\frac{\hbar^2}{2m} \chi''(x) + \half m\omega^2
x^2 \chi(x) = E \chi(x) \tag{1} \]
We know:
\begin{itemize}
    \item Discrete eigenvalues
    \item even / odd eigenfunctions
\end{itemize}
Change of variables:
\[ \xi^2 \equiv \frac{m\omega}{\hbar}x^2 \]
\[ \eps \equiv \frac{2E}{\hbar\omega} \]
Plug into (1)
\[ -\dfrac[2]{\chi}{\xi}(\xi) + \xi^2 \chi(\xi) =
\eps \chi(\xi) \tag{2} \]
Solve it by starting from a particular solution
\[ \eps = 1 \quad \left( E_0 =
\frac{\hbar\omega}{2} \right) \]
ansatz:
\[ \chi_0(\xi) = e^{-\xi^2 / 2} \tag{3} \]
Plug (3) into (2) with $\eps = 1$ works. We found
one eigenvalues $E_0 = \frac{\hbar\omega}{2}$,
$\chi_0(x) = A e^{-\frac{m\omega}{2\hbar} x^2}$ To
find other eigenfunction of $\hat{H}$ take general
form
\[ \xi(\xi) = f(\xi) e^{-\xi^2 / x} \tag{4} \]
Plug (4) into (2)
\[ -\dfrac[2]{f}{\xi} + 2\xi \dfrac{f}{\xi} + (1 -
\eps) f = 0 \tag{5} \]
Use power series method ($\xi = 0$ regular point)
\[ f(\xi) = \sum_{n = 0}^\infty a_n \xi^n \tag{6} \]
$a_n \in \RR$. Clearly
\[ \xi \dfrac{f}{\xi} = \sum_{n = 0}^\infty na_n
\xi^n \tag{7} \]
\[ \dfrac[2]{f}{\xi} = \sum_{n = 0}^\infty n(n -
1)a_n \xi^{n - 2} = \sum_{n = 0}^\infty (n + 1)(n
+ 2) a_{n + 2} \xi^n \]
Plug (6)-(8) into (5):
\begin{align*}
    \sum_{n = 0}^\infty [(n + 1)(n + 2)a_{n + 2} -
    2na_n + (\eps - 1) a_n]\xi^n
    &= 0 \\
    \implies a_{n + 2}
    &= \frac{(2n - \eps + 1)}{(n + 1)(n + 2)}a_n
\end{align*}
Because of parity of eigenfunction:
\begin{itemize}
    \item Either $a_n = 0$ for odd $n$ ($f(\xi) =
        f(-\xi)$) even eigenfunction
    \item or $a_n = 0$ for even $n$, ($f(\xi) =
        -f(-\xi)$) odd eigenfunction.
\end{itemize}

\begin{proposition*}
    If series (6) does \emph{not} terminate then
    eigenfunction of $\hat{H}$ would \emph{not} be
    normalisable.
\end{proposition*}

\begin{proof}
    Suppose that the series in (6) does \emph{not}
    terminate. Hence can look at asymptotic
    behaviour of series. Take (0)
    \[ \frac{a_{n + 2}}{a_n} \to \frac{2}{n} \]
    as $n \to \infty$. This is same asymptotic
    behaviour as
    \[ g(\xi) = e^{\xi^2} = \sum_{m = 0}^\infty
    \frac{\xi^{2m}}{m!} = \sum_{m = 0}^\infty b_m
    \xi^m \]
    where
    \[ b_m = \begin{cases}
        \frac{1}{m!} & \text{for $m$ even} \\
        0 & \text{for $m$ odd}
    \end{cases} \]
    asymptotic behaviour of $g(\xi)$
    \[ \frac{b_{n + 2}}{b_n} = \frac{\left(
    \frac{m}{2} \right)!}{ \left( \frac{m}{2} + 1
    \right)!} = \frac{2}{m + 2} \to \frac{2}{m} \]
    as $m \to \infty$. So if $e^{\xi^2/2}$ and
    $f(\xi)$ have same asymptotic behaviour
    \[ \chi(\xi) \sim e^{\xi^2} e^{-\xi^2/2} =
    e^{\xi^2/2} \to \infty \]
\end{proof}

\myskip
Given that the series (6) terminates then there
exists $N $such that
\[ a_{N + 2} = 0 \tag{10} \]
with $a_N \neq 0$. Plug (10) into (9)
\[ a_{N + 2} = \frac{(2N - \eps + 1)}{(N + 1)(N +
2)}a_N = 0 \]
\[ \implies 2N - \eps + 1 = 0 \]
Plugging in definition of $\eps$
\[ \implies E_N = \left( N + \half \right)
\hbar\omega \]
eigenvalues $N = 0$, $E_0 = \frac{\hbar\omega}{2}$
\[ E_{N + 1} - E_n = \hbar \omega \]
eigenfunction $\chi_N(\xi) = f_N(\xi)
e^{-\xi^2/2}$
\[ \chi_N(-\xi) = (-1)^N \chi_N(\xi) \]
Hermite polynomials are defined with recursive
relation
\[ f_N(\xi) = (-1)^N e^{\xi^2} \dfrac[N]{}{\xi}
(e^{-\xi^2}) \]
\begin{center}
    \begin{tabular}{c|c|c}
        $N$ & $E_N$ & $f_N(\xi)$ \\
        \hline
        $0$ & $\frac{\hbar\omega}{2}$ & $1$ \\
        $1$ & $\frac{3\hbar\omega}{2}$ & $\xi$ \\
        $2$ & $\frac{5\hbar\omega}{2}$ & $(1 - 2\xi^2)$ \\
        $3$ & $\frac{7\hbar\omega}{2}$ & $(\xi - \frac{2}{3}\xi^3)$
    \end{tabular}
\end{center}