567 lines
23 KiB
TeX
567 lines
23 KiB
TeX
\documentclass{article}
|
|
|
|
\usepackage{graphicx}
|
|
\usepackage{setspace}
|
|
\usepackage{listings}
|
|
\usepackage{color}
|
|
\usepackage{amsmath}
|
|
\usepackage{amssymb}
|
|
\usepackage{float}
|
|
\usepackage[justification=centering]{caption}
|
|
\usepackage{subcaption}
|
|
\usepackage[margin=0.75in]{geometry}
|
|
\usepackage[parfill]{parskip}
|
|
\usepackage{fancyhdr}
|
|
\usepackage{titlesec}
|
|
\usepackage{lipsum}
|
|
\usepackage{enumitem}
|
|
\usepackage{cancel}
|
|
\usepackage{siunitx}
|
|
\usepackage{adjustbox}
|
|
|
|
\titleformat{\subsection}[runin]{\normalfont \large \bfseries}
|
|
{\thesubsection}{1em}{}
|
|
|
|
\renewcommand{\thesubsection}{\indent(\alph{subsection})}
|
|
|
|
\definecolor{dkgreen}{rgb}{0,0.6,0}
|
|
\definecolor{gray}{rgb}{0.5,0.5,0.5}
|
|
\definecolor{mauve}{rgb}{0.58,0,0.82}
|
|
|
|
|
|
\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue
|
|
\definecolor{mylilas}{RGB}{170,55,241}
|
|
|
|
|
|
\lstset{language=Matlab,%
|
|
%basicstyle=\color{red},
|
|
breaklines=true,%
|
|
morekeywords={matlab2tikz},
|
|
keywordstyle=\color{blue},%
|
|
morekeywords=[2]{1}, keywordstyle=[2]{\color{black}},
|
|
identifierstyle=\color{black},%
|
|
stringstyle=\color{mylilas},
|
|
commentstyle=\color{mygreen},%
|
|
showstringspaces=false,%without this there will be a symbol in the places where there is a space
|
|
numbers=left,%
|
|
numberstyle={\tiny \color{black}},% size of th_sume numbers
|
|
numbersep=9pt, % this defines how far the numbers are from the text
|
|
emph=[1]{for,end,break},emphstyle=[1]\color{red}, %some words to emphasise
|
|
%emph=[2]{word1,word2}, emphstyle=[2]{style},
|
|
xleftmargin=2em
|
|
}
|
|
%\lstset{basicstyle=\small,
|
|
% keywordstyle=\color{mauve},
|
|
% identifierstyle=\color{dkgreen},
|
|
% stringstyle=\color{gray},
|
|
% numbers=left,
|
|
% xleftmargin=5em
|
|
% }
|
|
|
|
|
|
\title{ECE 456 - Problem Set 2}
|
|
\date{2021-03-01}
|
|
\author{David Lenfesty \\ lenfesty@ualberta.ca
|
|
\and Phillip Kirwin \\ pkirwin@ualberta.ca}
|
|
|
|
\pagestyle{fancy}
|
|
|
|
\fancyhead[L]{\textbf{ECE 456} - Problem Set 2}
|
|
\fancyhead[R]{David Lenfesty and Phillip Kirwin}
|
|
\fancyfoot[C]{Page \thepage}
|
|
\renewcommand{\headrulewidth}{1pt}
|
|
\renewcommand{\footrulewidth}{1pt}
|
|
|
|
\begin{document}
|
|
\doublespacing
|
|
\maketitle
|
|
\thispagestyle{empty}
|
|
\singlespacing
|
|
\newpage
|
|
|
|
\section*{Problem 1}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
\item %1a
|
|
Code:
|
|
\begin{lstlisting}[language=Matlab]
|
|
clear all;
|
|
%physical constants in MKS units
|
|
|
|
hbar = 1.054e-34;
|
|
q = 1.602e-19;
|
|
m = 9.110e-31;
|
|
|
|
%generate lattice
|
|
|
|
N = 100; %number of lattice points
|
|
n = [1:N]; %lattice points
|
|
a = 1e-10; %lattice constant
|
|
x = a * n; %x-coordinates
|
|
t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor
|
|
L = a * (N+1); %total length of consideration
|
|
|
|
%set up Hamiltonian matrix
|
|
|
|
U = 0*x; %0 potential at all x
|
|
main_diag = diag(2*t0*ones(1,N)+U,0); %create main diagonal matrix
|
|
lower_diag = diag(-t0*ones(1,N-1),-1); %create lower diagonal matrix
|
|
upper_diag = diag(-t0*ones(1,N-1),+1); %create upper diagonal matrix
|
|
|
|
H = main_diag + lower_diag + upper_diag; %sum to get Hamiltonian matrix
|
|
|
|
[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each
|
|
%column is an eigenvector
|
|
%"E_diag" is a diagonal matrix where the
|
|
%corresponding eigenvalues are on the
|
|
%diagonal.
|
|
|
|
E_col = diag(E_diag); %folds E_diag into a column vector of eigenvalues
|
|
|
|
% return eigenvectors for the 1st and 50th eigenvalues
|
|
|
|
phi_1 = eigenvectors(:,1);
|
|
phi_50 = eigenvectors(:,50);
|
|
|
|
% find the probability densities of position for 1st and 50th eigenvectors
|
|
|
|
P_1 = phi_1 .* conj(phi_1);
|
|
P_50 = phi_50 .* conj(phi_50);
|
|
|
|
% Find first N analytic eigenvalues
|
|
E_col_analytic = (1/q) * (hbar^2 * pi^2 * n.*n) / (2*m*L^2);
|
|
|
|
% Plot the probability densities for 1st and 50th eigenvectors
|
|
|
|
figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'k-');
|
|
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
|
|
xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
|
|
legend('n=1','n=50');
|
|
|
|
% Plot numerical eigenvalues
|
|
figure(2); clf; h = plot(n,E_col,'kx'); grid on;
|
|
set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
|
|
xlabel('EIGENVALUE NUMBER'); ylabel('ENERGY [eV]');
|
|
axis([0 100 0 40]);
|
|
|
|
% Add analytic eigenvalues to above plot
|
|
|
|
hold on;
|
|
plot(n,E_col_analytic,'k-');
|
|
legend({'Numerical','Analytical'},'Location','northwest');
|
|
|
|
\end{lstlisting}
|
|
%
|
|
\begin{figure}[H]
|
|
\centering
|
|
\begin{subfigure}{0.4\linewidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1a_1and50.png}
|
|
\caption{}
|
|
\label{fig:q3_ne}
|
|
\end{subfigure}%
|
|
\begin{subfigure}{0.4\linewidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1a_eigenvals.png}
|
|
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
|
|
\label{fig:q3_current}
|
|
\end{subfigure}
|
|
\caption{(a) Probability densities for \(n=1\) and \(n=50\).
|
|
(b) Comparison of first 101 numerical and analytic eigenvalues.}
|
|
\end{figure}
|
|
\item %1b
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1bi
|
|
The analytical solution is
|
|
|
|
\begin{equation}
|
|
\phi(x) = A sin \left( \frac{n \pi}{L} x \right)
|
|
\label{eq:analytical}
|
|
\end{equation}
|
|
|
|
In order to normalise this equation it must conform to the following:
|
|
|
|
\begin{equation}
|
|
\int_{0}^{L} \left| \phi(x) \right| ^2 dx = 1.
|
|
\label{eq:normalise}
|
|
\end{equation}
|
|
|
|
We use the following identity:
|
|
|
|
\begin{equation}
|
|
\int\sin^2{(a x)}\:dx = \frac{1}{2} x -\frac{1}{4a}\sin{(2a x)}.
|
|
\label{eq:integral_ident}
|
|
\end{equation}
|
|
|
|
Given that $sin$ of a real value is always real, we can disregard the norm operation, and directly relate (\ref{eq:analytical})
|
|
to the above identity.
|
|
Evaluating the integral gives us the following relationship
|
|
|
|
\begin{equation*}
|
|
\frac{1}{A^2} = \frac{1}{2} L - \cancel{\frac{L}{4n \pi} sin \left( \frac{2n \pi}{L} L \right)} - \cancel{\frac{1}{2} \cdot 0} + \cancel{\frac{L}{4n \pi} sin(0)}
|
|
\end{equation*}
|
|
|
|
From this, we find:
|
|
|
|
\begin{equation*}
|
|
\boxed{A = \sqrt{\frac{2}{L}}}
|
|
\end{equation*}
|
|
|
|
\item %1bii
|
|
|
|
Starting with the normalization condition for the numerical case:
|
|
\begin{align}
|
|
a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\
|
|
a\sum_{l=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 = a,
|
|
\label{eq:numerical_norm}
|
|
\end{align}
|
|
recalling that \(x = al\), and allowing \(a \to 0\) while holding \(L\)
|
|
constant implies that \(N \to \infty\), since \(a=\frac{L}{N}\).
|
|
An integral is defined as the limit of a Riemann sum as follows:
|
|
\begin{equation}
|
|
\int_c^df(x)\:dx \equiv \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i).
|
|
\label{eq:Riemann_sum}
|
|
\end{equation}
|
|
where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case,
|
|
\(n = N\), \(i = l\), \(c = 0\), \(d = L\), and \(\Delta x = a\), \(x_i = x_l\),
|
|
\(f(x) = \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\). Therefore we can write
|
|
\begin{equation*}
|
|
\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
|
|
= \lim_{N \to \infty}\sum_{l=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2
|
|
= a.
|
|
\end{equation*}
|
|
|
|
Using (\ref{eq:integral_ident}), we have
|
|
\begin{equation*}
|
|
\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
|
|
= \frac{1}{2}L - \cancel{\frac{L}{4n\pi}\sin{\left(\frac{2n\pi}{L}L\right)}}
|
|
- 0 + 0
|
|
= \frac{a}{B^2}.
|
|
\end{equation*}
|
|
|
|
This means that B must be
|
|
\begin{equation*}
|
|
\boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.}
|
|
\end{equation*}
|
|
|
|
\end{enumerate}
|
|
|
|
\item %1c
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1ci
|
|
|
|
% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
|
|
From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} a \ell \right)$, we can see that
|
|
$\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
|
|
$sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$ and
|
|
$sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$, respectively, where
|
|
$a = \frac{n \pi a \ell}{L}$ and $B = \frac{n \pi a}{L}$.
|
|
|
|
Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation:
|
|
|
|
\begin{equation*}
|
|
-t_0 B sin \left( \frac{n \pi a \ell}{L} \right) + 2 t_0 \phi _{\ell} - t_0 sin \left( \frac{n \pi a \ell}{L} \right)
|
|
\end{equation*}
|
|
|
|
At this point, we notice that $\phi _ \ell = B sin \left( \frac {n \pi}{L} a \ell \right)$,
|
|
so we can factor it out.
|
|
|
|
With some minor rearranging, this leaves us with the final expression for $E$:
|
|
|
|
\begin{equation}
|
|
\boxed{E = 2t_0 \left( 1 - cos \left( \frac{n \pi a}{L} \right) \right)}
|
|
\label{eq:numerical_E}
|
|
\end{equation}
|
|
|
|
\item %1cii
|
|
We can see here that the "predicted" numerical response matches nearly exactly the actual
|
|
calculated numerical solution.
|
|
\begin{figure}[H]
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1cii.png}
|
|
\caption{Analytic solution to numeric system, plotted.}
|
|
\end{figure}
|
|
|
|
|
|
|
|
\item %1ciii
|
|
|
|
Applying the approximation $cos(\theta) = 1 - \frac{\theta^2}{2}$ for small $\theta$,
|
|
on equation (\ref{eq:numerical_E}), we get the following expression
|
|
|
|
\begin{equation*}
|
|
E = 2 t_0 \left( \frac{n^2 \pi^2 a^2}{2 L^2} \right)
|
|
\end{equation*}
|
|
|
|
Fully substituting the known value of $t_0$, and we can get our final analytical expression for $E$:
|
|
|
|
\begin{equation*}
|
|
\boxed{E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }}
|
|
\end{equation*}
|
|
|
|
\item %1civ
|
|
With the decreased lattice spacing, and increased number of points, we can see the numerical solution
|
|
of eigenvalues matches the analytical solution much closer. As well, we can see that the probability
|
|
density function appears to be "squeezed" in the centre, for the $n = 50$ case. The \(n=50\) case is
|
|
now a constant-amplitude wave, which corresponds to the expected analytic result - in contrast to the
|
|
plot in section (a), which has a low-frequency envelope around it.
|
|
\begin{figure}[H]
|
|
\centering
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1civ_fig1.png}
|
|
\caption{Probability densities for \(n=1\)\\and \(n=50\).}
|
|
\end{subfigure}%
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1civ_fig2.png}
|
|
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
|
|
\end{subfigure}
|
|
\end{figure}
|
|
|
|
|
|
|
|
\end{enumerate}
|
|
\item %1d
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1di
|
|
|
|
In order to modify the computations to those for a particle in a "ring"
|
|
we simply had to add $-t_0$ elements as the "corner" elements of the
|
|
hamiltonian operator array:
|
|
\begin{lstlisting}[language=Matlab]
|
|
% Modify hamiltonian for circular boundary conditions
|
|
H(1, N) = -t0;
|
|
H(N, 1) = -t0;
|
|
\end{lstlisting}
|
|
|
|
\begin{figure}[H]
|
|
\centering
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1di_fig1.png}
|
|
\caption{Probability densities for \(n=4\)\\and \(n=5\).}
|
|
\end{subfigure}%
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q1di_fig2.png}
|
|
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
|
|
\end{subfigure}
|
|
\end{figure}
|
|
|
|
\item %1dii
|
|
|
|
The energy levels for eigenvalues number 4 and 5 are both \boxed{0.06\:\mathrm{eV}}.
|
|
These eigenstates are degenerate because they both have the same
|
|
eigenvalue/energy.
|
|
|
|
\item %1diii
|
|
|
|
\begin{minipage}[t]{\linewidth}
|
|
\centering
|
|
\adjustbox{valign=t}{
|
|
\includegraphics[width=\textwidth]{q1div.jpg}
|
|
}
|
|
\captionof{figure}{Sketch of degenerate energy levels.}
|
|
\end{minipage}
|
|
|
|
\item %1div
|
|
|
|
Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing
|
|
by the requisite $q$), we get the energy levels of $0eV, 0.0147eV, and 0.0589eV$, for the
|
|
eigenvalue numbers $n = 0, 1, and 2$, respectively. These match very closely, to within
|
|
acceptable margin of the numerical results from part (ii).
|
|
|
|
|
|
\end{enumerate}
|
|
|
|
\end{enumerate}
|
|
|
|
\newpage
|
|
\section*{Problem 2}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %2a
|
|
Since \(t_0 = \frac{\hbar^2}{2ma^2}\), and \(U_l = -\frac{q^2}{4\pi\varepsilon_0r_l}
|
|
+ \frac{l_o(l_o + 1)\hbar^2}{2mr_l^2}\), the middle diagonal elements will have values
|
|
\begin{equation*}
|
|
\boxed{\hat{H}_{ll} = \frac{\hbar^2}{ma^2}-\frac{q^2}{4\pi\epsilon_0r_l} + \frac{l_o(l_o + 1)\hbar^2}{2mr_l^2},}
|
|
\end{equation*}
|
|
and the upper and lower diagonals elements will have values
|
|
\begin{equation*}
|
|
\boxed{\hat{H}_{l(l\pm1)} = -\frac{\hbar^2}{2ma^2}.}
|
|
\end{equation*}
|
|
\item %2b
|
|
Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be \boxed{0}.
|
|
\item %2c
|
|
Code:
|
|
\begin{lstlisting}
|
|
clear all;
|
|
%physical constants in MKS units
|
|
|
|
hbar = 1.054e-34;
|
|
q = 1.602e-19;
|
|
m = 9.110e-31;
|
|
epsilon_0 = 8.854e-12;
|
|
|
|
%generate lattice
|
|
|
|
N = 100; %number of lattice points
|
|
n = [1:N]; %lattice points
|
|
a = 0.1e-10; %lattice constant
|
|
r = a * n; %x-coordinates
|
|
t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor
|
|
L = a * (N+1); %total length of consideration
|
|
|
|
%set up Hamiltonian matrix
|
|
|
|
U = -q^2./(4*pi*epsilon_0.*r) * (1/q); %potential at r in [eV]
|
|
main_diag = diag(2*t0*ones(1,N)+U,0); %create main diagonal matrix
|
|
lower_diag = diag(-t0*ones(1,N-1),-1); %create lower diagonal matrix
|
|
upper_diag = diag(-t0*ones(1,N-1),+1); %create upper diagonal matrix
|
|
|
|
H = main_diag + lower_diag + upper_diag; %sum to get Hamiltonian matrix
|
|
|
|
[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each column is an eigenvector
|
|
%"E_diag" is a diagonal matrix where the
|
|
%corresponding eigenvalues are on the
|
|
%diagonal.
|
|
|
|
E_col = diag(E_diag); %folds E_diag into a column vector of eigenvalues
|
|
|
|
% return eigenvectors for the 1st and 50th eigenvalues
|
|
|
|
phi_1 = eigenvectors(:,1);
|
|
phi_2 = eigenvectors(:,2);
|
|
|
|
% find the probability densities of position for 1st and 50th eigenvectors
|
|
|
|
P_1 = phi_1 .* conj(phi_1);
|
|
P_2 = phi_2 .* conj(phi_2);
|
|
|
|
% Plot the probability densities for 1st and 2nd eigenvectors
|
|
|
|
figure(1); clf; h = plot(r,P_1,'k-');
|
|
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
|
|
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
|
|
yticks([0.02 0.04 0.06 0.08 0.10 0.12]);
|
|
legend('n=1');
|
|
axis([0 1e-9 0 0.12]);
|
|
|
|
figure(2); clf; h = plot(r,P_2,'k-');
|
|
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
|
|
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
|
|
yticks([0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04]);
|
|
legend('n=2');
|
|
axis([0 1e-9 0 0.04]);
|
|
\end{lstlisting}
|
|
|
|
\begin{figure}[H]
|
|
\centering
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q2c_fig1_1s.png}
|
|
\caption{1s probability density.}
|
|
\end{subfigure}%
|
|
\begin{subfigure}{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q2c_fig2_2s.png}
|
|
\caption{2s probability density.}
|
|
\end{subfigure}
|
|
\end{figure}
|
|
\item %2d
|
|
For the 1s level, \boxed{E = -13.4978\:\mathrm{eV}}.
|
|
\item %2e
|
|
Beginning with equation (11) from the assignment, with \(l_o=0\):
|
|
\begin{align*}
|
|
\left[-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} - \frac{q^2}{4\pi\epsilon_0r}\right]f(r) &= Ef(r)\\
|
|
-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\left(\frac{2r}{a_0^{3/2}}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
|
|
-\frac{\hbar^2}{2m}\frac{2}{a_0^{3/2}}\frac{d}{dr}\left(e^{-r/a_0} - \frac{r}{a_0}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
|
|
-\frac{\hbar^2}{2m}\frac{2}{a_0^{3/2}}\left(-\frac{1}{a_0}e^{-r/a_0} - \frac{1}{a_0}e^{-r/a_0} + \frac{r}{a_0^2}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
|
|
-\frac{\hbar^2}{2m}\left(-\frac{2}{a_0r} + \frac{1}{a_0^2}\right)f(r) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
|
|
-\frac{\hbar^2}{2m}\left(-\frac{2}{a_0r} + \frac{1}{a_0^2}\right) - \frac{q^2}{4\pi\epsilon_0r} &= E.
|
|
\end{align*}
|
|
Recalling that \(a_0 = 4\pi\epsilon_0\hbar^2/mq^2\), we can eliminate \(r\):
|
|
\begin{align*}
|
|
\cancel{\frac{\hbar^2}{2m}}\frac{\cancel{2m}q^2}{4\pi\epsilon_0\cancel{\hbar^2}}\frac{1}{r} - \frac{\hbar^2}{2m}\frac{1}{a_0^2} - \frac{q^2}{4\pi\epsilon_0}\frac{1}{r} &= E\\
|
|
\cancel{\frac{q^2}{4\pi\epsilon_0}\frac{1}{r}} - \frac{\hbar^2}{2m}\frac{1}{a_0^2} - \cancel{\frac{q^2}{4\pi\epsilon_0}\frac{1}{r}} &= E.
|
|
\end{align*}
|
|
We can then solve for \(E\):
|
|
\begin{equation*}
|
|
E\:\mathrm{[eV]} = -\frac{1}{q}\cdot\frac{\hbar^2}{2ma_0^2} = -\frac{1}{q}\cdot\frac{(\SI{1.054e-34}{\joule\cdot\second})^2}{2(\SI{9.110e-31}{\kilogram})(\SI{0.0529}{\nm})^2}
|
|
= \boxed{- 13.6\:\mathrm{eV}.}
|
|
\end{equation*}
|
|
This is very similar to the result in (d).
|
|
|
|
\item %2f
|
|
In the figure below we can see that the numerical and analytical results agree up to scaling by \(a\). The scale difference is expected,
|
|
as discussed in Problem 1. From (d), we also expect agreement in the curve shapes because the numerical and analytical energies for the 1s
|
|
level are very similiar. We can see that the peak value of the analytic result is very slightly higher than that of the numerical result,
|
|
which corresponds to the analytical result for the energy being slightly greater in magnitude (\(-13.6\) eV versus \(-13.4978\) eV).
|
|
\begin{figure}[H]
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q2f_fig.png}
|
|
\caption{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
|
|
\end{figure}
|
|
|
|
|
|
|
|
\end{enumerate}
|
|
|
|
\newpage
|
|
\section*{Problem 3}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %3a
|
|
|
|
\begin{equation}
|
|
\theta (x') = \sqrt{\frac{2}{L}} sin \left( \frac{\pi (x' + L/2)}{L} \right) = \sqrt{\frac{2}{L}} cos \left( \frac{\pi x'}{L} \right)
|
|
\end{equation}
|
|
|
|
\item %3b
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %3bi
|
|
\lipsum[1]
|
|
\item %3bii
|
|
\lipsum[1]
|
|
\item %3biii
|
|
\lipsum[1]
|
|
\item %3biv
|
|
\lipsum[1]
|
|
\item %3bv
|
|
\lipsum[1]
|
|
\item %3bvi
|
|
\lipsum[1]
|
|
\item %3bvii
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
\item %3c
|
|
\lipsum[1]
|
|
\item %3d
|
|
\lipsum[1]
|
|
|
|
\end{enumerate}
|
|
\newpage
|
|
\section*{Problem 4}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %4a
|
|
\lipsum[1]
|
|
\item %4b
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %4bi
|
|
\lipsum[1]
|
|
\item %4bii
|
|
\lipsum[1]
|
|
\item %4biii
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
|
|
\end{enumerate}
|
|
|
|
|
|
\end{document} |