%%%%		-*- 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:   Model of the Hydrogen Atom
%%%%

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\title{Model of the Hydrogen Atom}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
% - Use the \inst{?} command only if the authors have different
%   affiliation.

\institute[Millersville University of Pennsylvania]
% (optional, but mostly needed)
{
  Department of Mathematics
}
% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

\date{ % (optional)
Fall
%Spring
%Summer
\the\year}

\subject{Talks}
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% out. 



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% the beginning of each subsection:
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\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\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 model the simplest atom,
\item solve the Schr\"{o}dinger equation for the hydrogen atom, and
\item discuss some of the properties of the solution.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Hydrogen Atom}

\begin{itemize}
\item Mass of a proton $m_{p} = 1.67262 \times
10^{-27}$ kg.
\item Mass of an electron is
$m_{e} = 9.10938 \times 10^{-31}$ kg.
\item The proton's mass is about $1836$ times that of the
electron, thus the center of mass of a hydrogen atom is located nearly
at the position of the proton in the nucleus.
\item The proton and electron possess charges of $e_{0} = 1.60218 \times
10^{-19}$ Coulomb and $-e_{0}$ respectively.
\item The potential for this attractive force between electron and
  proton is $V(\rho) = -e_{0}^{2}/(4 \pi \epsilon_{0} \rho)$ where $\rho$ is the distance
separating the proton and electron.
\item $\epsilon_{0}$ is known as the permittivity of free
space and has value $8.854 \times 10^{-12}$ Farad/meter.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Schr\"{o}dinger Equation}

\[
%\label{eq:apps:hydrogenpde}
i \hbar \frac{\partial \Psi}{\partial t}
	= -\frac{\hbar^{2}}{2 \mu} \Delta \Psi -
        \frac{e_{0}^{2}}{4 \pi \epsilon_{0} \rho} \Psi	.
\]
The origin of the spherical coordinate system is assumed
to lie at the center of the proton.

\vspace*{0.10in}

$\mu$ represents the reduced mass of the electron,
\[
\mu
	= \frac{m_{e} m_{p}}{m_{e} + m_{p}} \approx m_{e}	.
\]
The expression $\Psi(\rho, \varphi, \theta, t)$ is the wave function
of the electron.
Solutions should have the property that
\[
%\label{eq:apps:hydrogennormal}
\int_{0}^{\infty} \int_{0}^{\pi} \int_{-\pi}^{\pi}
\Psi(\rho, \varphi, \theta, t)
\overline{\Psi(\rho, \varphi, \theta, t)}
\rho^{2} \sin \varphi \, d\theta \, d\varphi \, d\rho
	= 1	.
\]
This condition is sometimes referred to as the
solution being \textbf{normalizable}.
The wave function should also be $2 \pi$-periodic in $\theta$, should
vanish as $\rho \to \infty$, and should be bounded as $\rho \to
0^{+}$.

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (1 of 5)}

Assume a product solution of the
form $\Psi(\rho, \varphi, \theta, t) = T(t) R(\rho) F(\varphi,
\theta)$.
\[
%\label{eq:apps:hydrogenseparated}
i \hbar \frac{T'(t)}{T(t)}
	= -\frac{\hbar^{2}}{2 \mu} \frac{\Delta \left[ R(\rho) F(\varphi,
    \theta) \right]}{R(\rho) F(\varphi, \theta)} -
\frac{e_{0}^{2}}{4 \pi \epsilon_{0} \rho}
\]
The left-hand side of this equation depends on $t$ while the
right-hand side depends on the spherical coordinate system variables,
thus each side is equal to a constant denoted as $E$.
Expanding
the Laplacian in spherical coordinates result in
\begin{align*}
%\label{eq:apps:hydrogentimeindep}
E
	= {} & -\frac{\hbar^{2}}{2 \mu} \frac{\Delta \left[ R(\rho) F(\varphi,
    \theta) \right]}{R(\rho) F(\varphi, \theta)} -
              \frac{e_{0}^{2}}{4 \pi \epsilon_{0} \rho}	\\
\left( E + \frac{e_{0}^{2}}{4 \pi \epsilon_{0} \rho} \right) R F
	= {} & -\frac{\hbar^{2}}{2 \mu}
\left( R'' F + \frac{2}{\rho} R' F \right)	\\
	&- \frac{\hbar^{2}}{2 \mu \rho^{2}}
\left( R F_{\varphi\varphi} + \cot \varphi R F_{\varphi} +
\csc^{2} \varphi R F_{\theta\theta}
\right) 	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (2 of 5)}

Multiply both sides by
$2 \mu \rho^{2}/(\hbar^{2} R F)$ and
move terms depending on $\rho$ to the left-hand side.
\[
\rho^{2} \left(
\frac{R''}{R} + \frac{2}{\rho} \frac{R'}{R} + \frac{2 \mu E}{\hbar^{2}}
\right) +
\frac{\mu e_{0}^{2}}{2 \pi \epsilon_{0} \hbar^{2}} \rho
	= -\left( \frac{F_{\varphi\varphi}}{F} + \cot \varphi
  \frac{F_{\varphi}}{F} + \csc^{2} \varphi \frac{F_{\theta\theta}}{F} \right)
\]
The radial variable $\rho$ is separated from the
angular variables $(\varphi, \theta)$.
If both sides of the equation are equal to a constant $\lambda$ then
the partial differential equation for the angular variables becomes
\[
F_{\varphi\varphi} + \cot \varphi F_{\varphi} + \csc^{2} \varphi
F_{\theta\theta} + \lambda F
	= 0	.
\]
If $\lambda = l (l+1)$ for $l = 0, 1, 2, \ldots$, the solutions are
the spherical harmonic functions $Y_{l}^{m}(\varphi, \theta)$.

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (3 of 5)}

Setting the $\rho$-dependent side of the equation equal to $l (l+1)$
produces
\[
%\label{eq:apps:hydrogenradial}
\rho^{2} R''(\rho) + 2 \rho R'(\rho) + \left(
\frac{2 \mu E}{\hbar^{2}} \rho^{2} + \frac{\mu
  e_{0}^{2}}{2 \pi \epsilon_{0} \hbar^{2}} \rho - l (l+1)
\right) R(\rho)
	= 0
\]
This can be re-written as
\[
%\label{eq:apps:genlaguerre}
x y''(x) + ((2 l + 1) + 1 - x) y'(x) +
\left( e_{0}^{2} \sqrt{\frac{-\mu}{32 \pi^{2} \epsilon_{0}^{2} E
      \hbar^{2}}} - l - 1 \right) y(x)
	= 0	.
\]
If the coefficient of $y(x)$ is a
nonnegative integer, this is an example of
the associated Laguerre differential
equation.

\vspace*{0.10in}

Let $n = e_{0}^{2} \sqrt{-\mu/(32 \pi^{2} \epsilon_{0}^{2} E
  \hbar^{2})}$, and thus when $n$ is a positive
integer such that $n \ge l + 1$, there is a polynomial solution
$y(x) = L_{n-l-1}^{2l+1}(x)$.

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (4 of 5)}

Define $\alpha = \frac{2}{\hbar} \sqrt{-2
  \mu E}$ and the radial factor of the solution
can be written as
\[
%\label{eq:apps:hydrogenradialsoln}
R_{n, l}(\rho)
	= e^{-\frac{\alpha \rho}{2}} (\alpha \rho)^{l} L_{n-l-1}^{2l+1}(\alpha \rho)
\]
for $l = 0, 1, 2, \ldots$ and $n$ an integer for which $n \ge l+1$.

\vspace*{0.10in}

The time-independent solution to the boundary value problem
\[
R(\rho) F(\varphi, \theta)
	= e^{-\frac{\alpha \rho}{2}} (\alpha \rho)^{l} L_{n-l-1}^{2l+1}(\alpha \rho)
Y_{l}^{m}(\varphi, \theta)	,
\]
where $l$, $m$, and $n$ are integers satisfying the inequality $|m|
\le l < n$ and $n \in \mathbb{N}$.
To aid with notation, define $u_{l, m, n}(\rho, \varphi, \theta)$ to be
the expression on the right-hand side of the last equation.

\vspace*{0.10in}

The assumption that $n$ is a positive integer implies the possible
values of $E$, the energy of the electron, are given by
\[
%\label{eq:apps:hydrogenelectronenergy}
E
	\equiv E_{n}
	= -\frac{\mu e_{0}^{4}}{32 \pi^{2} \epsilon_{0}^{2} \hbar^{2} n^{2}}	.
\]
These energies are called the energies of the bound (or stationary)
states of the hydrogen atom.

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (5 of 5)}
\fontsize{9pt}{10.8}\selectfont

The time-dependent
factor $T(t)$ of the product solution is
\[
T'(t) = \frac{-i E_{n}}{\hbar} T(t)	,
\]
and thus $T_{n}(t) = e^{-i E_{n} t/\hbar}$ for $n \in \mathbb{N}$.

\vspace*{0.10in}

Since the magnitude of $T_{n}(t)$ is one, the wave functions are
normalized when $u_{l, m, n}(\rho, \varphi, \theta)$ is normalized.
\begin{align*}
\lefteqn{\int_{0}^{\infty} \int_{0}^{\pi} \int_{-\pi}^{\pi}
u_{l, m, n}(\rho, \varphi, \theta) \overline{u_{l, m, n}(\rho, \varphi, \theta)}
\rho^{2} \sin \varphi \, d\theta \, d\varphi \, d\rho}	\\
	&= \int_{0}^{\infty} e^{-\alpha \rho} (\alpha \rho)^{2 l}
   \left( L_{n-l-1}^{2l+1}(\alpha \rho) \right)^{2} \rho^{2} \, d\rho
\int_{0}^{\pi} \int_{-\pi}^{\pi} Y_{l}^{m}(\varphi, \theta)
   \overline{Y_{l}^{m}(\varphi, \theta)} \sin \varphi \, d\theta \, d\varphi	\\
	&= \int_{0}^{\infty} e^{-\alpha \rho} (\alpha \rho)^{2 l}
   \left( L_{n-l-1}^{2l+1}(\alpha \rho) \right)^{2} \rho^{2} \, d\rho	\\
	&= \frac{1}{\alpha^{3}} \int_{0}^{\infty} 
r^{2 l+2} \left( L_{n-l-1}^{2l+1}(r) \right)^{2} e^{-r} \, dr	\\
	&= \frac{(2n) \Gamma(n+l+1)}{\alpha^{3} (n-l-1)!}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}

\[
%\label{eq:apps:hydrogennormwave}
\Psi_{l, m, n}(\rho,\varphi,\theta, t)
	= \sqrt{\frac{\alpha^{3} (n-l-1)!}{(2n) \Gamma(n+l+1)}}
e^{-\frac{i E_{n}}{\hbar} t} e^{-\frac{\alpha}{2} \rho} (\alpha \rho)^{l}
L_{n-l-1}^{2l+1}(\alpha \rho) Y_{l}^{m}(\varphi, \theta)	,
\]
for $|m| \le l < n$ and $n \in \mathbb{N}$.

\begin{itemize}
\item Index $l$ is often called the
\textbf{orbital} or \textbf{angular momentum quantum
number}.
\item Index $m$ is the
\textbf{magnetic quantum number}.
\item Index $n$ is the \textbf{principal} or \textbf{total quantum
  number}.
\end{itemize}

When the electron of the hydrogen atom is in energy state $E_{n}$, the wave
function of the electron will be one of the $n^{2}$ functions given above.

\end{frame}
%%

\begin{frame}\frametitle{Probability Density Function}
\fontsize{9pt}{10.8}\selectfont

The probability density function of the electron is
\begin{align*}
P_{l, m, n}(\rho,\varphi,\theta)
%\label{eq:apps:hydrogenpdf}
	&= \Psi_{l, m, n}(\rho,\varphi,\theta, t)
\overline{\Psi_{l, m, n}(\rho,\varphi,\theta, t)}	\\
	&= \frac{\alpha^{3} (n-l-1)!}{(2n) \Gamma(n+l+1)}
e^{-\alpha \rho} (\alpha \rho)^{2l}
\left( L_{n-l-1}^{2l+1}(\alpha \rho) \right)^{2}
Y_{l}^{m}(\varphi, \theta) \overline{Y_{l}^{m}(\varphi, \theta)}              .
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Example: $P_{2, 0, 3}(\rho, \varphi, \theta)$}

\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{hydrogenl2m0n3}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

%%%
\end{document}

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