%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%%
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      03/11/2025
%%%%
%%%% Purpose:   The Quantum Harmonic Oscillator
%%%%

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\title{The Quantum Harmonic Oscillator}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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Fall
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\the\year}

\subject{Talks}
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\begin{frame}
  \titlepage
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%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
  % You might wish to add the option [pausesections]
%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item review the classical harmonic oscillator and its properties,
\item introduce the wave function and some axioms of quantum mechanics,
\item present the Schr\"{o}dinger equation, and
\item solve the Schr\"{o}dinger equation for the quantum analogue of
  the classical harmonic oscillator.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Classical Harmonic Oscillator}

Ignoring damping, the initial value problem for the classical harmonic
oscillator resembles,
\begin{align*}
m \, u'' + k \, u
	&= 0	\\
u(0)
	&= u_{0}	\\
u'(0)
	&= v_{0}	.
\end{align*}
The displacement of the mass can be described by the
function,
\[
%\label{eq:apps:classicalsho}
u(t)
	= u_{0} \cos \left( \sqrt{\frac{k}{m}} t \right) +
v_{0} \sqrt{\frac{m}{k}} \sin \left( \sqrt{\frac{k}{m}} t \right)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Properties of the Classical Harmonic Oscillator}

\begin{itemize}
\item The period of the
oscillation is $T = 2 \pi/\omega$ and the frequency of the oscillation
is $f = \omega/(2 \pi) = 1/T$.
\item The amplitude of the oscillation is
$A = \sqrt{u_{0}^{2} + v_{0}^{2}/\omega^{2}}$.
\item The total energy of the spring-mass system is the sum
of the kinetic and potential energies:
\[
%\label{eq:apps:classicaltotalenergy}
E
	= \frac{1}{2} m \left( u'(t) \right)^{2} + \frac{1}{2} k \left( u(t) \right)^{2}	.
\]
\item Since there are no other forces present in the spring-mass system the
total energy is constant, 
and thus
\[
%\label{eq:apps:classicaltotalenergy2}
E
	= \frac{1}{2} k \left( u_{0}^{2} + \frac{v_{0}^{2}}{\omega^{2}} \right)	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Location of the Mass}

Suppose an observer examines a subinterval of $[-A, A]$ of length
$\Delta u$.
What is the probability of observing the mass in this subinterval?

\begin{itemize}
\item Let the probability distribution for the displacement $u$ defined for
$u \in [-A, A]$ be denoted as $P(u)$.
\item For any fixed $u \in (-A, A)$, if $\Delta u$ is small,
the probability of observing the
mass in the subinterval $[u, u + \Delta u]$ is $P(u) \Delta
u$.
\item Let $\Delta t$ be the time required for the mass to travel
distance $\Delta u$.
\item During a complete period of oscillation the mass will traverse the
subinterval of length $\Delta u$ twice,
\[
P(u) \Delta u
	= \frac{2 \Delta t}{T}
\implies
P(u) \frac{\Delta u}{\Delta t}
	= \frac{2}{T}	.
\]
As $\Delta u$ and $\Delta t$ approach zero,
$P(u) = 2/(| u'(t) | T)$
where $|u'(t)| = \omega \sqrt{A^{2} - \left( u(t) \right)^{2}}$ is the
speed of the mass.
\item The probability
density function for the displacement of the mass is
\[
%\label{eq:apps:classicalpdf}
P(u)
	= \frac{2}{\omega \sqrt{A^{2} - \left( u(t) \right)^{2}} \frac{2
    \pi}{\omega}}
	= \frac{1}{\pi \sqrt{A^{2} - u^{2}}}	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Background}

\begin{itemize}
\item In 1926, Louis de Broglie stated that all matter has wave-like properties.
This assertion has since become known as the
\textbf{de Broglie hypothesis}.
\item The fundamental property of an object is its
\textbf{wave function} $\Psi(x, t)$, from which
quantities such as the position and velocity of the object can be derived.
The wave function is a complex-valued function.
\item Max Born developed the statistical interpretation of the wave
function which implies the probability of finding an object in the
interval $[x, x + dx]$ at time $t$ is $P(x, t) \, dx = \Psi(x, t)
\overline{\Psi(x, t)} \, dx$ where $\overline{\Psi(x, t)}$ is the
complex conjugate of the wave function.
\item The wave function of an object solves the linear partial differential
equation known as the linear
\textbf{Schr\"{o}dinger equation}.
The one-dimensional version of the linear Schr\"{o}dinger equation can be
written as
\[
i \hbar \frac{\partial \Psi}{\partial t}
	= -\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2} \Psi}{\partial x^{2}} +
V(x) \Psi(x, t)	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Solving Schr\"{o}dinger's Equation}

Assume $V(x) =
k x^{2}/2$ (the same as for the classical harmonic oscillator),
Schr\"{o}dinger's equation can be written as
\[
%\label{eq:apps:1dSchrodinger}
i \hbar \frac{\partial \Psi}{\partial t}
	= -\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2} \Psi}{\partial x^{2}} +
\frac{1}{2} k x^{2} \Psi(x, t)	.
\]
Assume this equation is defined for $-\infty < x < \infty$ and the
solutions asymptotically approach $0$ as $|x| \to \infty$.
Let the initial wave function be $\Psi(x,0) = f(x)$.

\vspace*{0.10in}

Use separation of variables assuming $\Psi(x, t) = X(x) T(t)$.
\begin{align*}
i \hbar X(x) T'(t)
	&= -\frac{\hbar^{2}}{2 \mu} X''(x) T(t) + \frac{1}{2} k x^{2} X(x) T(t)	\\
i \hbar \frac{T'(t)}{T(t)}
	&= -\frac{\hbar^{2}}{2 \mu} \frac{X''(x)}{X(x)} + \frac{1}{2} k x^{2}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Energy}

\[
i \hbar \frac{T'(t)}{T(t)}
	= -\frac{\hbar^{2}}{2 \mu} \frac{X''(x)}{X(x)} + \frac{1}{2} k x^{2}
	= E
\]
The induced ordinary differential equation involving $x$ is
\[
%\label{eq:apps:xoscillator}
-\frac{\hbar^{2}}{2 \mu} X''(x) + \left( \frac{1}{2} k x^{2} - E
\right) X(x)
	= 0	.
\]
Let $\omega$ be a constant such that $k = \mu \omega^{2}$ and
make the substitution $x = \xi \sqrt{\hbar/(\mu \omega)}$.
\[
%\label{eq:apps:quantumHermite}
\frac{d^{2} X}{d \xi^{2}} + \left( \frac{2 E}{\hbar \omega} - \xi^{2} \right) X
	= 0
\]

\end{frame}
%%

\begin{frame}\frametitle{Hermite's Equation}

\[
%\label{eq:spc:hermite}
y'' - 2 x y' + \lambda y
	= 0
\mbox{ for $-\infty < x < \infty$,}
\]
Let $\phi(x) = e^{-x^{2}/2} y(x)$ where $y(x)$ is a solution to
Hermite's ordinary differential equation then
\begin{align*}
0
	&= \left[ e^{x^{2}/2} \phi(x) \right]'' -
2 x \left[ e^{x^{2}/2} \phi(x) \right]' +
\lambda e^{x^{2}/2} \phi(x)	\\
%\label{eq:apps:hermiteode}
0
	&= \left( \phi''(x) + (\lambda + 1 - x^{2}) \phi(x) \right)
   e^{x^{2}/2}	.
\end{align*}
Dividing both sides of by
$e^{x^{2}/2}$ and note that this is of the form of
the previous equation if $2 E/(\hbar \omega) = \lambda + 1$.
If $\lambda = 2 n$ where $n = 0, 1, 2, \ldots$, then $y(x) = H_{n}(x)$,
the $n$th Hermite polynomial.
Thus for $E \equiv E_{n} = \hbar \omega (n + 1/2)$ the function
$X_{n}(\xi) = e^{-\xi^{2}/2} H_{n}(\xi)$.

\end{frame}
%%

\begin{frame}\frametitle{Hermite Functions}

\[
%\label{eq:apps:hermitefunction}
X_{n}(x)
	= e^{-\frac{\mu \omega x^{2}}{2 \hbar}} H_{n}\left( \sqrt{\frac{\mu
    \omega}{\hbar}} x \right)
\mbox{ for $n = 0, 1, 2, \ldots$.}
\]
Functions of this form are
non-normalized \textbf{Hermite functions}.
\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{hermitefuncs}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Product Solutions}

Product solutions to the one-dimensional linear
Schr\"{o}dinger equation
for the quantum harmonic oscillator have the form,
\[
\Psi_{n}(x, t)
	= e^{-i \omega (n + \frac{1}{2}) t} e^{-\frac{\mu \omega x^{2}}{2 \hbar}}
H_{n}\left( \sqrt{\frac{\mu \omega}{\hbar}} x \right)
\mbox{ for $n = 0, 1, 2, \ldots$.}
\]
The symbol $E_{n}$ was chosen for the eigenvalues, since these are
typically called \textbf{energy eigenvalues}.
The implication is that energies of the product solutions
for the quantum harmonic oscillator
take on values from an infinite, but discrete set of values.
The energy levels are separated by the finite positive quantity $\hbar
\omega$.
The least energy the quantum harmonic oscillator can possess is $E_{0}
= \hbar \omega/2 > 0$ which is sometimes called the \textbf{zero point
  energy}.

\end{frame}
%%

\begin{frame}\frametitle{Series Solution}

\[
%\label{eq:apps:quantumsoln}
\Psi(x, t)
	= e^{-\frac{\mu \omega x^{2}}{2 \hbar}} \sum_{n=0}^{\infty} A_{n}
        e^{-i \omega (n + 1/2) t} H_{n}\left( \sqrt{\frac{\mu
              \omega}{\hbar}} x \right)	,
\]
where the expressions $A_{n}$ are constants.
If the initial form of the wave function is $f(x)$, then
\[
f(x)
	= e^{-\frac{\mu \omega x^{2}}{2 \hbar}} \sum_{n=0}^{\infty} A_{n}
        H_{n}\left( \sqrt{\frac{\mu \omega}{\hbar}} x \right)	,
\]
where
\[
%\label{eq:apps:quantumcoeff}
A_{n}
	= \frac{1}{2^{n} n!} \sqrt{\frac{\mu \omega}{\hbar \pi}}
\int_{-\infty}^{\infty} f(x) H_{n} \left( \sqrt{\frac{\mu \omega}{\hbar}} x \right)
e^{-\frac{\mu \omega x^{2}}{2 \hbar}} \, dx	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Example}

Suppose the probability distribution function for
the position of the mass is initially Gaussian in shape with
$f(x) \overline{f(x)} = \frac{1}{\sqrt{2 \pi}} e^{-x^{2}/2}$.
This implies $f(x) = \frac{1}{\sqrt[4]{2 \pi}} e^{-x^{2}/4}$.
The coefficients of the infinite series solution are then
\[
A_{n}
	= \frac{1}{2^{n} n!} \sqrt{\frac{\mu \omega}{\hbar \pi}}
\int_{-\infty}^{\infty} \frac{1}{\sqrt[4]{2 \pi}} e^{-\frac{x^{2}}{4}} H_{n}
              \left( \sqrt{\frac{\mu \omega}{\hbar}} x \right) e^{-\frac{\mu
              \omega x^{2}}{2 \hbar}} \, dx	.
\]
Since $H_{n}(x)$ is an odd function when $n$ is odd, $A_{n} = 0$ for
$n$ odd.
The calculation of the remaining coefficients is made easier if it is
assumed that $\mu \omega/\hbar = 1$.
In this case,
\[
A_{2n}
	= \frac{1}{2^{2n+1/4} \pi^{3/4} (2n)!}
\int_{-\infty}^{\infty} e^{-3 x^{2}/4} H_{2n}(x) \, dx
	= \frac{2}{12^{n} (n!) \sqrt[4]{18 \pi}}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Comparison Between Classical and Quantum
    Solutions}

As $n$ increases the energy $E_{n}$ increases and the relative change
in energy vanishes,
\[
\lim_{n \to \infty} \frac{\hbar \omega}{E_{n}}
	= \lim_{n \to \infty} \frac{\hbar \omega}{\left( n + \frac{1}{2} \right)
  \hbar \omega}
	= 0	.
\]
Thus for energetic oscillators the discrete differences between energy
levels is less discrete and more continuous.

\vspace*{0.10in}

Let the amplitude of the classical harmonic oscillator be $A =
\sqrt{2 E/k}$.
If the classical harmonic oscillator has energy equal to one of the
discrete energy levels of the quantum harmonic oscillator, then
\[
A
	= \sqrt{\frac{2 (n + 1/2) \hbar \omega}{k}}
	= \sqrt{\frac{(2n+1) \hbar}{\mu \omega}}	,
\]
for some $n \in \mathbb{N}$.

\end{frame}
%%

\begin{frame}\frametitle{Probability $n= 5$}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{hopdfsn5}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Probability $n= 10$}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{hopdfsn10}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

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