318 lines
9.7 KiB
TeX
318 lines
9.7 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}
|
|
|
|
\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
|
|
hello
|
|
\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{probability densities for \(n=1\)\\and \(n=50\).}
|
|
\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}
|
|
\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}
|
|
|
|
Using the following identity:
|
|
|
|
\begin{equation}
|
|
foo
|
|
\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 to the above identity.
|
|
Evaluating the integral gives us the following relationship
|
|
|
|
\begin{equation*}
|
|
\frac{1}{A^2} = \frac{1}{2} L - \frac{L}{4n \pi} sin \left( \frac{2n \pi}{L} L \right) - \frac{1}{2} \cdot 0 + \frac{L}{4n \pi} sin(0)
|
|
\end{equation*}
|
|
|
|
From this, we find:
|
|
|
|
\begin{equation*}
|
|
A = \sqrt{\frac{2}{L}}
|
|
\end{equation*}
|
|
|
|
\item %1bii
|
|
|
|
{\em I really don't like this explanation}.
|
|
|
|
The output of the numerical sum will change, as the value of $\alpha$ changes, by a factor if $\alpha$.
|
|
To correct for this, we need to factor the result of the summation by $\alpha$, which corresponds to
|
|
|
|
% Idk how to explain this at all, I need to chew on it internally and come back later.
|
|
\begin{equation*}
|
|
foo
|
|
\end{equation*}
|
|
|
|
This means that B must be
|
|
\begin{equation*}
|
|
B = \sqrt{\frac{2 \alpha}{L}}
|
|
\end{equation*}
|
|
|
|
\end{enumerate}
|
|
|
|
\item %1c
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1ci
|
|
\lipsum[1]
|
|
\item %1cii
|
|
\lipsum[1]
|
|
\item %1ciii
|
|
\lipsum[1]
|
|
\item %1civ
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
\item %1d
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1di
|
|
\lipsum[1]
|
|
\item %1dii
|
|
\lipsum[1]
|
|
\item %1diii
|
|
\lipsum[1]
|
|
\item %1div
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
|
|
\end{enumerate}
|
|
|
|
|
|
\section*{Problem 2}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %2a
|
|
\lipsum[1]
|
|
\item %2b
|
|
\lipsum[1]
|
|
\item %1c
|
|
\lipsum[1]
|
|
\item %1d
|
|
\lipsum[1]
|
|
\item %1e
|
|
\lipsum[1]
|
|
\item %1f
|
|
\lipsum[1]
|
|
|
|
\end{enumerate}
|
|
|
|
\section*{Problem 3}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %2a
|
|
\lipsum[1]
|
|
\item %2b
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1di
|
|
\lipsum[1]
|
|
\item %1dii
|
|
\lipsum[1]
|
|
\item %1diii
|
|
\lipsum[1]
|
|
\item %1div
|
|
\lipsum[1]
|
|
\item %1dv
|
|
\lipsum[1]
|
|
\item %1dvi
|
|
\lipsum[1]
|
|
\item %1dvii
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
\item %1c
|
|
\lipsum[1]
|
|
\item %1d
|
|
\lipsum[1]
|
|
|
|
\end{enumerate}
|
|
|
|
\section*{Problem 4}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %2a
|
|
\lipsum[1]
|
|
\item %2b
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %1di
|
|
\lipsum[1]
|
|
\item %1dii
|
|
\lipsum[1]
|
|
\item %1diii
|
|
\lipsum[1]
|
|
\end{enumerate}
|
|
|
|
\end{enumerate}
|
|
|
|
|
|
\end{document} |