443 lines
16 KiB
TeX
443 lines
16 KiB
TeX
\documentclass{article}
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\usepackage{graphicx}
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\usepackage{lipsum}
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\usepackage{enumitem}
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\titleformat{\subsection}[runin]{\normalfont \large \bfseries}
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\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue
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\definecolor{mylilas}{RGB}{170,55,241}
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\lstset{language=Matlab,%
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%basicstyle=\color{red},
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breaklines=true,%
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morekeywords={matlab2tikz},
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keywordstyle=\color{blue},%
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identifierstyle=\color{black},%
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commentstyle=\color{mygreen},%
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showstringspaces=false,%without this there will be a symbol in the places where there is a space
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numbers=left,%
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numberstyle={\tiny \color{black}},% size of th_sume numbers
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numbersep=9pt, % this defines how far the numbers are from the text
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emph=[1]{for,end,break},emphstyle=[1]\color{red}, %some words to emphasise
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%emph=[2]{word1,word2}, emphstyle=[2]{style},
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xleftmargin=2em
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}
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%\lstset{basicstyle=\small,
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% stringstyle=\color{gray},
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% numbers=left,
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% xleftmargin=5em
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% }
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\title{ECE 456 - Problem Set 2}
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\date{2021-03-01}
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\author{David Lenfesty \\ lenfesty@ualberta.ca
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\and Phillip Kirwin \\ pkirwin@ualberta.ca}
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\pagestyle{fancy}
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\fancyhead[L]{\textbf{ECE 456} - Problem Set 2}
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\fancyhead[R]{David Lenfesty and Phillip Kirwin}
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\fancyfoot[C]{Page \thepage}
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\renewcommand{\headrulewidth}{1pt}
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\renewcommand{\footrulewidth}{1pt}
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\begin{document}
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\doublespacing
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\maketitle
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\thispagestyle{empty}
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\singlespacing
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\newpage
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hello
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\section*{Problem 1}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %1a
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Code:
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\begin{lstlisting}[language=Matlab]
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clear all;
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%physical constants in MKS units
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hbar = 1.054e-34;
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q = 1.602e-19;
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m = 9.110e-31;
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%generate lattice
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N = 100; %number of lattice points
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n = [1:N]; %lattice points
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a = 1e-10; %lattice constant
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x = a * n; %x-coordinates
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t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor
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L = a * (N+1); %total length of consideration
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%set up Hamiltonian matrix
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U = 0*x; %0 potential at all x
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main_diag = diag(2*t0*ones(1,N)+U,0); %create main diagonal matrix
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lower_diag = diag(-t0*ones(1,N-1),-1); %create lower diagonal matrix
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upper_diag = diag(-t0*ones(1,N-1),+1); %create upper diagonal matrix
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H = main_diag + lower_diag + upper_diag; %sum to get Hamiltonian matrix
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[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each
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%column is an eigenvector
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%"E_diag" is a diagonal matrix where the
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%corresponding eigenvalues are on the
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%diagonal.
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E_col = diag(E_diag); %folds E_diag into a column vector of eigenvalues
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% return eigenvectors for the 1st and 50th eigenvalues
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phi_1 = eigenvectors(:,1);
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phi_50 = eigenvectors(:,50);
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% find the probability densities of position for 1st and 50th eigenvectors
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P_1 = phi_1 .* conj(phi_1);
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P_50 = phi_50 .* conj(phi_50);
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% Find first N analytic eigenvalues
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E_col_analytic = (1/q) * (hbar^2 * pi^2 * n.*n) / (2*m*L^2);
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% Plot the probability densities for 1st and 50th eigenvectors
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figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'k-');
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grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
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xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
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legend('n=1','n=50');
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% Plot numerical eigenvalues
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figure(2); clf; h = plot(n,E_col,'kx'); grid on;
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set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
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xlabel('EIGENVALUE NUMBER'); ylabel('ENERGY [eV]');
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axis([0 100 0 40]);
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% Add analytic eigenvalues to above plot
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hold on;
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plot(n,E_col_analytic,'k-');
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legend({'Numerical','Analytical'},'Location','northwest');
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\end{lstlisting}
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%
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.4\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q1a_1and50.png}
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\caption{Probability densities for \(n=1\)\\and \(n=50\).}
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\label{fig:q3_ne}
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\end{subfigure}%
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\begin{subfigure}{0.4\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q1a_eigenvals.png}
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\caption{Comparison of first 101 numerical and analytic eigenvalues.}
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\label{fig:q3_current}
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\end{subfigure}
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\end{figure}
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\item %1b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1bi
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The analytical solution is
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\begin{equation}
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\phi(x) = A sin \left( \frac{n \pi}{L} x \right)
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\label{eq:analytical}
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\end{equation}
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In order to normalise this equation it must conform to the following:
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\begin{equation}
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\int_{0}^{L} \left| \phi(x) \right| ^2 dx = 1.
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\label{eq:normalise}
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\end{equation}
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We use the following identity:
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\begin{equation}
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\int\sin^2{(\alpha x)}\:dx = \frac{1}{2} x -\frac{1}{4\alpha}\sin{(2\alpha x)}.
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\label{eq:integral_ident}
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\end{equation}
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Given that $sin$ of a real value is always real, we can disregard the norm operation, and directly relate (\ref{eq:analytical})
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to the above identity.
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Evaluating the integral gives us the following relationship
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\begin{equation*}
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\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)}
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\end{equation*}
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From this, we find:
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\begin{equation*}
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A = \sqrt{\frac{2}{L}}
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\end{equation*}
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\item %1bii
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Starting with the normalization condition for the numerical case:
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\begin{align}
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a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\
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a\sum_{l=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 = a,
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\label{eq:numerical_norm}
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\end{align}
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recalling that \(x = al\), and allowing \(a \to 0\) while holding \(L\)
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constant implies that \(N \to \infty\), since \(a=\frac{L}{N}\).
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An integral is defined as the limit of a Riemann sum as follows:
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\begin{equation}
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\int_c^df(x)\:dx \equiv \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i).
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\label{eq:Riemann_sum}
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\end{equation}
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where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case,
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\(n = N\), \(i = l\), \(c = 0\), \(d = L\), and \(\Delta x = a\), \(x_i = x_l\),
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\(f(x) = \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\). Therefore we can write
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\begin{equation*}
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\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
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= \lim_{N \to \infty}\sum_{l=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2
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= a.
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\end{equation*}
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Using (\ref{eq:integral_ident}), we have
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\begin{equation*}
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\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
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= \frac{1}{2}L - \cancel{\frac{L}{4n\pi}\sin{\left(\frac{2n\pi}{L}L\right)}}
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- 0 + 0
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= \frac{a}{B^2}.
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\end{equation*}
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This means that B must be
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\begin{equation*}
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B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.
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\end{equation*}
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\end{enumerate}
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\item %1c
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1ci
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% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
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From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} \alpha \ell \right)$, we can see that
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$\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
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$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$ and
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$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$, respectively, where
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$\alpha = \frac{n \pi \alpha \ell}{L}$ and $B = \frac{n \pi \alpha}{L}$.
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Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation:
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\begin{equation*}
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-t_0 B sin \left( \frac{n \pi \alpha \ell}{L} \right) + 2 t_0 \phi _{\ell} - t_0 sin \left( \frac{n \pi \alpha \ell}{L} \right)
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\end{equation*}
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At this point, we notice that $\phi _ \ell = B sin \left( \frac {n \pi}{L} \alpha \ell \right)$,
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so we can factor it out.
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With some minor rearranging, this leaves us with the final expression for $E$:
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\begin{equation}
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E = 2t_0 \left( 1 - cos \left( \frac{n \pi \alpha}{L} \right) \right)
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\label{eq:numerical_E}
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\end{equation}
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\item %1cii
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\begin{figure}[H]
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\centering
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\includegraphics[width=\textwidth]{q1cii.png}
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\caption{Analytic solution to numeric system, plotted.}
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\end{figure}
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We can see here that the "predicted" numerical response matches nearly exactly the actual
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calculated numerical solution.
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\item %1ciii
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Applying the approximation $cos(\Theta) = 1 - \frac{\Theta^2}{2}$ for small $\Theta$,
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on equation (\ref{eq:numerical_E}), we get the following expression
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\begin{equation*}
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E = 2 t_0 \left( \frac{n^2 \pi^2 \alpha^2}{2 L^2} \right)
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\end{equation*}
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Fully substituting the known value of $t_0$, and we can get our final expression for $E$:
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\begin{equation*}
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E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }
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\end{equation*}
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\item %1civ
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1civ_fig1.png}
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\caption{Probability densities for \(n=1\)\\and \(n=50\).}
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1civ_fig2.png}
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\caption{Comparison of first 101 numerical and analytic eigenvalues.}
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\end{subfigure}
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\end{figure}
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With the decreased lattice spacing, and increased number of points, we can see the numerical solution
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of eigenvalues matches the analytical solution much closer. As well, we can see that the probability
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density function appears to be "squeezed" in the centre, for the $n = 50$ case. The \(n=50\) case is
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now a constant-amplitude wave, which corresponds to the expected analytic result - in contrast to the
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plot in section (a), which has a low-frequency envelope around it.
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\end{enumerate}
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\item %1d
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1di
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In order to modify the computations to those for a particle in a "ring"
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we simply had to add $-t_0$ elements as the "corner" elements of the
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hamiltonian operator array:
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\begin{lstlisting}[language=Matlab]
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% Modify hamiltonian for circular boundary conditions
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H(1, N) = -t0;
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H(N, 1) = -t0;
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1di_fig1.png}
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\caption{Probability densities for \(n=4\)\\and \(n=5\).}
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1di_fig2.png}
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\caption{Comparison of first 101 numerical and analytic eigenvalues.}
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\end{subfigure}
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\end{figure}
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\item %1dii
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The energy levels for eigenvalues number 4 and 5 are both $0.06eV$.
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These eigenstates are degenerate because they both have the same
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eigenvalue/energy (0.06).
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\item %1diii
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\begin{figure}[H]
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\centering
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\includegraphics[width=\textwidth, angle=-90]{q1div.jpg}
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\caption{Sketch of appropriate energy levels}
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\end{figure}
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\item %1div
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Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing
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by the requisite $q$), we get the energy levels of $0eV, 0.0147eV, and 0.0589eV$, for the
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eigenvalue numbers $n = 0, 1, and 2$, respectively. These match very closely, to within
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acceptable margin of the numerical results from part (ii).
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\end{enumerate}
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\end{enumerate}
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\section*{Problem 2}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %2a
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\lipsum[1]
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\item %2b
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\lipsum[1]
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\item %1c
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\lipsum[1]
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\item %1d
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\lipsum[1]
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\item %1e
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\lipsum[1]
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\item %1f
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\lipsum[1]
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\end{enumerate}
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\section*{Problem 3}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %2a
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\lipsum[1]
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\item %2b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1di
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\lipsum[1]
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\item %1dii
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\lipsum[1]
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\item %1diii
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\lipsum[1]
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\item %1div
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\lipsum[1]
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\item %1dv
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\lipsum[1]
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\item %1dvi
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\lipsum[1]
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\item %1dvii
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\lipsum[1]
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\end{enumerate}
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\item %1c
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\lipsum[1]
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\item %1d
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\lipsum[1]
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\end{enumerate}
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\section*{Problem 4}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %2a
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\lipsum[1]
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\item %2b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1di
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\lipsum[1]
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\item %1dii
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\lipsum[1]
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\item %1diii
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\lipsum[1]
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\end{enumerate}
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\end{enumerate}
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\end{document} |