diff --git a/PS3/doc.pdf b/PS3/doc.pdf index 69fbf70..cf254b8 100644 Binary files a/PS3/doc.pdf and b/PS3/doc.pdf differ diff --git a/PS3/doc.tex b/PS3/doc.tex index 5536614..81042dc 100644 --- a/PS3/doc.tex +++ b/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} \ No newline at end of file