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PS2/doc.fls

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PS2/doc.log

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PS2/doc.tex

@@ -46,7 +46,7 @@
         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=5em
+        xleftmargin=2em
     }
     %\lstset{basicstyle=\small,
     %        keywordstyle=\color{mauve},
@@ -81,18 +81,144 @@
 
         \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] 
             \item %1a
-            \lipsum[1]
+            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 (\frac{n \pi}{L} x)
+                    \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
-                \lipsum[1]
+
+                {\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

PS2/q1a.asv

@@ -0,0 +1,57 @@
+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) / (
+% 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]);
+

PS2/q1a.m

@@ -1,4 +1,4 @@
-
+clear all;
 %physical constants in MKS units
 
 hbar = 1.054e-34;
@@ -7,7 +7,59 @@ m = 9.110e-31;
 
 %generate lattice
 
-N = 100;    %number of lattice points
+N  = 100;                   %number of lattice points
-n = [1:N];  %lattice points
+n  = [1:N];                 %lattice points
-a = 1e-10;  %lattice constant
+a  = 1e-10;                 %lattice constant
-x = a * [1:1:N]; 
+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');
+
+