commit

PS2/doc.aux

@@ -7,7 +7,8 @@
 \newlabel{eq:analytical}{{1}{3}}
 \newlabel{eq:normalise}{{2}{3}}
 \newlabel{eq:integral_ident}{{3}{3}}
-\newlabel{eq:numerical_E}{{4}{4}}
+\newlabel{eq:numerical_norm}{{5}{4}}
-\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Analytic solution to numeric system, plotted.\relax }}{4}{}\protected@file@percent }
+\newlabel{eq:Riemann_sum}{{6}{4}}
-\newlabel{fig:q1b_electrons}{{2}{4}}
+\newlabel{eq:numerical_E}{{7}{4}}
+\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Analytic solution to numeric system, plotted.\relax }}{5}{}\protected@file@percent }
 \gdef \@abspage@last{9}

PS2/doc.fdb_latexmk

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PS2/doc.fls

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 INPUT d:\Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\q1a_eigenvals.png
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PS2/doc.log

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PS2/doc.tex

@@ -155,7 +155,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
                 \begin{subfigure}{0.4\linewidth}
                     \centering
                     \includegraphics[width=\textwidth]{q1a_1and50.png}
-                    \caption{probability densities for \(n=1\)\\and \(n=50\).}
+                    \caption{Probability densities for \(n=1\)\\and \(n=50\).}
                     \label{fig:q3_ne}
                 \end{subfigure}%
                 \begin{subfigure}{0.4\linewidth}
@@ -206,29 +206,38 @@ legend({'Numerical','Analytical'},'Location','northwest');
                 \item %1bii
 
                 Starting with the normalization condition for the numerical case:
-                \begin{align*}
+                \begin{align}
-                    a\sum_{l=1}^N|\phi_l|^2 = a \\
+                    a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\
-                    a\sum_{l=1}^N|B\sin{(\frac{n\pi}{L}x_l)}|^2 = a,
+                    a\sum_{l=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 = a,
-                \end{align*}
+                    \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_a^bf(x)\:dx = \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i).
+                    \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
-                The output of the numerical sum will change, as the value of $\alpha$ changes, by a factor of $\alpha$.
-                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
+                    \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*}
-                    B = \sqrt{\frac{2 \alpha}{L}}
+                    B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.
                 \end{equation*}
 
             \end{enumerate}
@@ -266,7 +275,6 @@ legend({'Numerical','Analytical'},'Location','northwest');
                     \centering
                     \includegraphics[width=\textwidth]{q1cii.png}
                     \caption{Analytic solution to numeric system, plotted.}
-                    \label{fig:q1b_electrons}
                 \end{figure}
 
                 We can see here that the "predicted" numerical response matches nearly exactly the actual
@@ -288,18 +296,46 @@ legend({'Numerical','Analytical'},'Location','northwest');
                 \end{equation*}
 
                 \item %1civ
-                \lipsum[1]
+
+                \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}
+
+                % TODO idk if this is enough
+                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.
+
             \end{enumerate}
             \item %1d
             \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
                 \item %1di
-                \lipsum[1]
+                
+                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.
+
+                % TODO do we need a listing here
+
                 \item %1dii
-                \lipsum[1]
+
+
                 \item %1diii
-                \lipsum[1]
+
+
                 \item %1div
-                \lipsum[1]
+
+
             \end{enumerate}
 
         \end{enumerate}

PS2/q1d.m

@@ -0,0 +1,60 @@
+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
+
+% Modify hamiltonian for circular boundary conditions
+H(1, N) = -t0;
+H(N, 1) = -t0;
+
+[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_4  = eigenvectors(:,4);
+phi_5 = eigenvectors(:,5);
+
+% find the probability densities of position for 1st and 50th eigenvectors
+
+P_4 = phi_4 .* conj(phi_4);
+P_5 = phi_5 .* conj(phi_5);
+
+% 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_4,'kx',x,P_5,'k-');
+grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
+xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
+legend('n=4','n=5');
+
+% 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]');