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pkirwin 2021-04-09 16:29:46 -06:00
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@ -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}