commit

PS3/doc.tex

@@ -66,14 +66,14 @@
 
 \newcommand{\angstrom}{\textup{\AA}}
 
-\title{ECE 456 - Problem Set 3 (Part 1)}
+\title{ECE 456 - Problem Set 3}
 \date{2021-03-31}
 \author{David Lenfesty \\ lenfesty@ualberta.ca
     \and Phillip Kirwin \\ pkirwin@ualberta.ca}
     
 \pagestyle{fancy}
 
-\fancyhead[L]{\textbf{ECE 456} - Problem Set 3 (Part 1)}
+\fancyhead[L]{\textbf{ECE 456} - Problem Set 3}
 \fancyhead[R]{David Lenfesty and Phillip Kirwin}
 \fancyfoot[C]{Page \thepage}
 \renewcommand{\headrulewidth}{1pt}
@@ -418,10 +418,12 @@ S_u = [1, s; s, 1];
             From this, we can finally obtain an expression for \( E(\vec{k}) \):
 
             \begin{align}
-                E(\vec{k}) = E_0 \pm t_c \sqrt{1 + 4\cos(k_x a)\cos(k_y b) 4 \cos^2(k_y b)}
+                E(\vec{k}) = E_0 \pm t_c \sqrt{1 + 4\cos(k_x a)\cos(k_y b) + 4 \cos^2(k_y b)}
             \end{align}
 
         \end{enumerate}
+    
+    \newpage
 
     \section*{Problem 4}
         \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] 
@@ -430,7 +432,7 @@ S_u = [1, s; s, 1];
             We can substitute (from the assignment) equation (24) into equation (23):
 
             \begin{align}
-                \sum _k i \hbar \frac{\delta}{\delta t} c_k(t) \phi_k(x) e^{-i [E(k) / \hbar] t} &= \sum _k c_k(t) \hat{H_0} \phi(x)e^{-i [E(k) \ \hbar] t} \\
+                \sum _k i \hbar \frac{\partial}{\partial t} c_k(t) \phi_k(x) e^{-i [E(k) / \hbar] t} &= \sum _k c_k(t) \hat{H_0} \phi(x)e^{-i [E(k) / \hbar] t} \\
                  &+ \sum _k c_k(t) U_s(x, t) \phi _k (x) e^ {-i [E(k) / \hbar] t}
                  \label{eq:phi}
             \end{align}
@@ -438,7 +440,7 @@ S_u = [1, s; s, 1];
             We can partially evaluate the derivative on the left hand side:
 
             \begin{align*}
-                \frac{\delta}{\delta t} c_k(t) \phi _k(t)e^{-i [E(k) / \hbar]t} = \frac{\delta c_k(t)}{\delta t} \phi _k(t)e^{-i [E(k) / \hbar]t} - \frac{i}{\hbar} c_k(t)E(k)\phi _k (x) e^{-i [E(k) / \hbar]t}
+                \frac{\partial}{\partial t} c_k(t) \phi_k(t)e^{-i [E(k) / \hbar]t} = \frac{\partial c_k(t)}{\partial t} \phi _k(t)e^{-i [E(k) / \hbar]t} - \frac{i}{\hbar} c_k(t)E(k)\phi _k (x) e^{-i [E(k) / \hbar]t}
             \end{align*}
 
             Expanding the rightmost term out into the summation, we get the expression \( \sum _k c_k(t) E(k) \phi _k(x) e^ {-i [E(k) / \hbar]t} \).
@@ -446,7 +448,7 @@ S_u = [1, s; s, 1];
             and we get the final differential equation:
 
             \begin{align*}
-                \sum_k c_k(t) U_s(x, t) \phi _k (x) e^{-i [E(k) / \hbar]t} = \sum_k i \hbar \frac{\delta c_k(t)}{\delta t} \phi_k(x) e^{-i [E(k) / \hbar] t}
+                \sum_k c_k(t) U_s(x, t) \phi _k (x) e^{-i [E(k) / \hbar]t} = \sum_k i \hbar \frac{\partial c_k(t)}{\partial t} \phi_k(x) e^{-i [E(k) / \hbar] t}
             \end{align*}
 
             \item %4b
@@ -461,20 +463,20 @@ S_u = [1, s; s, 1];
             Starting with the initial equation
 
             \begin{align*}
-                \sum _k c_k(t) I_{k_f k} e ^ {-i [ E(k) / \hbar ] t } = i \hbar \frac{\delta c_{k_f}(t)}{\delta t} e ^ {-i [ E(k_f) / \hbar ] t}
+                \sum _k c_k(t) I_{k_f k} e ^ {-i [ E(k) / \hbar ] t } = i \hbar \frac{\partial c_{k_f}(t)}{\partial t} e ^ {-i [ E(k_f) / \hbar ] t}
             \end{align*}
 
-            Since we approximated that \( c_k = 1 \), only when \( k  = k_i \), and 0 otherwise, we can simplify
+            Since we approximated that \( c_k = 1 \) only when \( k  = k_i \) and is 0 otherwise, we can simplify
             the sum on the left hand side to a single element, and divide out the exponentials.
 
             \begin{align*}
-                I_{k_f k_i} e^{ i [- E(k_i) + E(k_f) / \hbar] t} = i \hbar \frac{\delta c_k(t)}{\delta t}
+                I_{k_f k_i} e^{ i [- E(k_i) + E(k_f) / \hbar] t} = i \hbar \frac{\partial c_k(t)}{\partial t}
             \end{align*}
 
-            Defining a new symbol \( \Lambda = [E(k_f) - E(k_i)] / \hbar \), we can simplify the equation to it's final form.
+            Defining a new symbol \( \Lambda = [E(k_f) - E(k_i)] / \hbar \), we can simplify the equation to its final form.
 
             \begin{align}
-                I_{k_f k_i} e^{i \Lambda t} = i \hbar \frac{\delta c_k(t)}{\delta t}
+                I_{k_f k_i} e^{i \Lambda t} = i \hbar \frac{\partial c_k(t)}{\partial t}
             \end{align}
 
             \item %4d
@@ -491,11 +493,41 @@ S_u = [1, s; s, 1];
 
             \item %4e
 
-            To generate the points, we used a simple MATLAB script, found in Appendix (A).
+            To generate the points, we used a simple MATLAB script:
 
+            \begin{lstlisting}
+% Parameters to change
+N_D = 4.7e15;
+N_A = 1.6e15;
+
+% Logarithmic tick marks
+T = [ 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100 200 300 ];
+
+gamma = 1.057e7 * T / sqrt(N_D - N_A);
+
+mu_n_first = 21.15e17 *  (T.^(3/2) / (N_D + N_A));
+mu_n_last = (log(1 + gamma.^2) - (gamma.^2 ./ (1 + gamma.^2)));
+
+mu_n = mu_n_first ./ mu_n_last;
+
+for n = 1 : length(T)
+   fprintf("T = %dK: %f\n", T(n), mu_n(n));
+end
+            \end{lstlisting}
+
+            Plotting on the provided graph, gives the following:
+
+            \begin{minipage}[t]{\linewidth}
+
+                \centering
+                \adjustbox{valign=t}{
+                    \includegraphics[width=0.5\textwidth]{q4_plot.png}
+                }
+                \captionof{figure}{Graph of electron mobility in GaAs.}
+            \end{minipage}
+            
+            It's clear that our calculated results match up well with the theoretical results.
 
         \end{enumerate}
 
-    \app
-
 \end{document}