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Merge branch 'master' of github.com:davidlenfesty/ECE_456

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David Lenfesty 2021-03-05 15:40:46 -07:00
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\relax
\providecommand*\caption@xref[2]{\@setref\relax\@undefined{#1}}
\newlabel{fig:q3_ne}{{1a}{3}}
\newlabel{sub@fig:q3_ne}{{a}{3}}
\newlabel{fig:q3_current}{{1b}{3}}
\newlabel{sub@fig:q3_current}{{b}{3}}
\newlabel{eq:analytical}{{1}{3}}
\newlabel{eq:normalise}{{2}{3}}
\newlabel{eq:integral_ident}{{3}{3}}
\newlabel{eq:numerical_norm}{{5}{4}}
\newlabel{eq:Riemann_sum}{{6}{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{10}

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@ -1,6 +1,7 @@
\documentclass{article}
\usepackage{graphicx}
\usepackage{caption}
\usepackage{setspace}
\usepackage{listings}
\usepackage{color}
@ -16,10 +17,16 @@
\usepackage{lipsum}
\usepackage{enumitem}
\usepackage{cancel}
\usepackage{siunitx}
\usepackage{adjustbox}
\titleformat{\subsection}[runin]{\normalfont \large \bfseries}
{\thesubsection}{1em}{}
\captionsetup[figure]{width=0.75\textwidth,labelfont=normalfont,font=it,labelsep=period}
\DeclareMathOperator*{\sinc}{sinc}
\DeclareMathOperator*{\rect}{rect}
\renewcommand{\thesubsection}{\indent(\alph{subsection})}
\definecolor{dkgreen}{rgb}{0,0.6,0}
@ -57,6 +64,8 @@
% }
\newcommand{\angstrom}{\textup{\AA}}
\title{ECE 456 - Problem Set 2}
\date{2021-03-01}
\author{David Lenfesty \\ lenfesty@ualberta.ca
@ -76,7 +85,7 @@
\thispagestyle{empty}
\singlespacing
\newpage
hello
\section*{Problem 1}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
@ -150,28 +159,33 @@ legend({'Numerical','Analytical'},'Location','northwest');
\end{lstlisting}
%
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.4\linewidth}
\begin{subfigure}{0.5\linewidth}
\centering
\includegraphics[width=\textwidth]{q1a_1and50.png}
\caption{Probability densities for \(n=1\)\\and \(n=50\).}
\caption{}
\label{fig:q3_ne}
\end{subfigure}%
\begin{subfigure}{0.4\linewidth}
\begin{subfigure}{0.5\linewidth}
\centering
\includegraphics[width=\textwidth]{q1a_eigenvals.png}
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
\caption{}
\label{fig:q3_current}
\end{subfigure}
\caption{(a) Probability densities for \(n=1\) and \(n=50\).
(b) Comparison of first 101 numerical and analytic eigenvalues.}
\end{figure}
\end{minipage}
\item %1b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1bi
The analytical solution is
The analytical solution is:
\begin{equation}
\phi(x) = A sin \left( \frac{n \pi}{L} x \right)
\phi(x) = A \sin{\left( \frac{n \pi}{L} x \right)}.
\label{eq:analytical}
\end{equation}
@ -185,45 +199,46 @@ legend({'Numerical','Analytical'},'Location','northwest');
We use the following identity:
\begin{equation}
\int\sin^2{(\alpha x)}\:dx = \frac{1}{2} x -\frac{1}{4\alpha}\sin{(2\alpha x)}.
\int\sin^2{(a x)}\:dx = \frac{1}{2} x -\frac{1}{4a}\sin{(2a x)}.
\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 (\ref{eq:analytical})
Given that the sine of a real value is always real, we can disregard the norm operation, and directly relate (\ref{eq:analytical})
to the above identity.
Evaluating the integral gives us the following relationship
Evaluating the integral gives us the following relationship:
\begin{equation*}
\frac{1}{A^2} = \frac{1}{2} L - \cancel{\frac{L}{4n \pi} sin \left( \frac{2n \pi}{L} L \right)} - \cancel{\frac{1}{2} \cdot 0} + \cancel{\frac{L}{4n \pi} sin(0)}
\frac{1}{A^2} = \frac{1}{2} L - \cancel{\frac{L}{4n \pi} \sin{\left( \frac{2n \pi}{L} L \right)} }
- \cancel{\frac{1}{2} \cdot 0} + \cancel{\frac{L}{4n \pi} \sin{(0)}}.
\end{equation*}
From this, we find:
\begin{equation*}
A = \sqrt{\frac{2}{L}}
\boxed{A = \sqrt{\frac{2}{L}}.}
\end{equation*}
\item %1bii
Starting with the normalization condition for the numerical case:
\begin{align}
a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\
a\sum_{l=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 = a,
a\sum_{\ell=1}^N\left|\phi_l\right|^2 &= a \\
a\sum_{\ell=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_\ell\right)}\right|^2 &= a, \nonumber
\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 \(a=\frac{L}{N}\).
recalling that \(x = a\ell\), 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).
\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 \(\Delta x = a\), \(x_i = x_l\),
\(n = N\), \(i = \ell\), \(c = 0\), \(d = L\), and \(\Delta x = a\), \(x_i = x_\ell\),
\(f(x) = \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\). Therefore we can write
\begin{equation*}
\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
= \lim_{N \to \infty}\sum_{\ell=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2
= a.
\end{equation*}
@ -235,9 +250,9 @@ legend({'Numerical','Analytical'},'Location','northwest');
= \frac{a}{B^2}.
\end{equation*}
This means that B must be
This means that \(B\) must be
\begin{equation*}
B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.
\boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.}
\end{equation*}
\end{enumerate}
@ -247,76 +262,83 @@ legend({'Numerical','Analytical'},'Location','northwest');
\item %1ci
% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} \alpha \ell \right)$, we can see that
From the base form of $\phi _\ell = B\sin{\left( \frac {n \pi}{L} a \ell \right)} $, we can see that
$\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$ and
$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$, respectively, where
$\alpha = \frac{n \pi \alpha \ell}{L}$ and $B = \frac{n \pi \alpha}{L}$.
$\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$ and
$\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$, respectively, where
$a = \frac{n \pi a \ell}{L}$ and $B = \frac{n \pi a}{L}$.
Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation:
\begin{equation*}
-t_0 B sin \left( \frac{n \pi \alpha \ell}{L} \right) + 2 t_0 \phi _{\ell} - t_0 sin \left( \frac{n \pi \alpha \ell}{L} \right)
-t_0 B \sin{\left( \frac{n \pi a \ell}{L} \right)} + 2 t_0 \phi _{\ell} - t_0 \sin{\left( \frac{n \pi a \ell}{L} \right)}.
\end{equation*}
At this point, we notice that $\phi _ \ell = B sin \left( \frac {n \pi}{L} \alpha \ell \right)$,
At this point, we notice that $\phi _ \ell = B \sin{\left( \frac {n \pi}{L} a \ell \right)} $,
so we can factor it out.
With some minor rearranging, this leaves us with the final expression for $E$:
\begin{equation}
E = 2t_0 \left( 1 - cos \left( \frac{n \pi \alpha}{L} \right) \right)
\boxed{E = 2t_0 \left( 1 - \cos{\left( \frac{n \pi a}{L} \right)} \right).}
\label{eq:numerical_E}
\end{equation}
\item %1cii
\begin{figure}[H]
\begin{minipage}[t]{\linewidth}
\centering
\includegraphics[width=\textwidth]{q1cii.png}
\caption{Analytic solution to numeric system, plotted.}
\end{figure}
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1cii.png}
}
\captionof{figure}{Analytic solution to numeric system, plotted.}
\end{minipage}
We can see here that the "predicted" numerical response matches nearly exactly the actual
calculated numerical solution.
\item %1ciii
Applying the approximation $cos(\Theta) = 1 - \frac{\Theta^2}{2}$ for small $\Theta$,
on equation (\ref{eq:numerical_E}), we get the following expression
Applying the approximation $\cos{(\theta)} \approx 1 - \frac{\theta^2}{2}$ for small $\theta$
on equation (\ref{eq:numerical_E}), we get the following expression:
\begin{equation*}
E = 2 t_0 \left( \frac{n^2 \pi^2 \alpha^2}{2 L^2} \right)
E = 2 t_0 \left( \frac{n^2 \pi^2 a^2}{2 L^2} \right).
\end{equation*}
Fully substituting the known value of $t_0$, and we can get our final expression for $E$:
We can get our final analytical expression for $E$ by fully substituting the explicit form of $t_0$:
\begin{equation*}
E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }
\boxed{E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }}
\end{equation*}
\item %1civ
With the decreased lattice spacing and increased number of points we can see the numerical solution
more closely matches the analytical solution. As well, the \(n=50\) case is
now a constant-amplitude wave, which corresponds to the expected analytic result, in contrast to the
plot in section (a), which has a low-frequency envelope around it.
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1civ_fig1.png}
\caption{Probability densities for \(n=1\)\\and \(n=50\).}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1civ_fig2.png}
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
\caption{}
\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. The \(n=50\) case is
now a constant-amplitude wave, which corresponds to the expected analytic result - in contrast to the
plot in section (a), which has a low-frequency envelope around it.
\caption{(a) Probability densities for \(n=1\) and \(n=50\). (b) Comparison of first 101 numerical and analytic eigenvalues.}
\end{figure}
\end{minipage}
\end{enumerate}
\item %1d
@ -325,104 +347,359 @@ legend({'Numerical','Analytical'},'Location','northwest');
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.
hamiltonian operator array:
\begin{lstlisting}[language=Matlab]
% Modify hamiltonian for circular boundary conditions
H(1, N) = -t0;
H(N, 1) = -t0;
\end{lstlisting}
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1di_fig1.png}
\caption{Probability densities for \(n=4\)\\and \(n=5\).}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1di_fig2.png}
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
\caption{}
\end{subfigure}
\caption{(a) Probability densities for \(n=4\) and \(n=5\). (b) Comparison of first 101 numerical and analytic eigenvalues.}
\end{figure}
\end{minipage}
\item %1dii
The energy levels for eigenvalues number 4 and 5 are both $0.06eV$.
Because these eigenstates both have the same eigenvalue (0.06).
The energy levels for eigenvalues number 4 and 5 are both \boxed{0.06\:\mathrm{eV}}.
These eigenstates are degenerate because they both have the same
eigenvalue/energy.
\item %1diii
% TODO add figure for sketch
\begin{minipage}[t]{\linewidth}
\centering
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1div.jpg}
}
\captionof{figure}{Sketch of degenerate energy levels.}
\end{minipage}
\item %1div
Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing
by the requisite $q$), we get energy levels of \(0\) eV, \(0.0147\) eV, and \(0.0589\) eV for the
indices \(n=0\), \(1\), and \(2\), respectively. These match very closely, to within
acceptable margin of the numerical results from part (ii).
\end{enumerate}
\end{enumerate}
\newpage
\section*{Problem 2}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\lipsum[1]
Since \(t_0 = \frac{\hbar^2}{2ma^2}\), and \(U_l = -\frac{q^2}{4\pi\varepsilon_0r_l}
+ \frac{l_o(l_o + 1)\hbar^2}{2mr_l^2}\), the middle diagonal elements will have values
\begin{equation*}
\boxed{\hat{H}_{ll} = \frac{\hbar^2}{ma^2}-\frac{q^2}{4\pi\epsilon_0r_l} + \frac{l_o(l_o + 1)\hbar^2}{2mr_l^2},}
\end{equation*}
and the upper and lower diagonals elements will have values
\begin{equation*}
\boxed{\hat{H}_{l(l\pm1)} = -\frac{\hbar^2}{2ma^2}.}
\end{equation*}
\item %2b
\lipsum[1]
\item %1c
\lipsum[1]
\item %1d
\lipsum[1]
\item %1e
\lipsum[1]
\item %1f
\lipsum[1]
Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be \boxed{0}.
\item %2c
Code:
\begin{lstlisting}
clear all;
%physical constants in MKS units
hbar = 1.054e-34;
q = 1.602e-19;
m = 9.110e-31;
epsilon_0 = 8.854e-12;
%generate lattice
N = 100; %number of lattice points
n = [1:N]; %lattice points
a = 0.1e-10; %lattice constant
r = 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 = -q^2./(4*pi*epsilon_0.*r) * (1/q); %potential at r in [eV]
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_2 = eigenvectors(:,2);
% find the probability densities of position for 1st and 50th eigenvectors
P_1 = phi_1 .* conj(phi_1);
P_2 = phi_2 .* conj(phi_2);
% Plot the probability densities for 1st and 2nd eigenvectors
figure(1); clf; h = plot(r,P_1,'k-');
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
yticks([0.02 0.04 0.06 0.08 0.10 0.12]);
legend('n=1');
axis([0 1e-9 0 0.12]);
figure(2); clf; h = plot(r,P_2,'k-');
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
yticks([0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04]);
legend('n=2');
axis([0 1e-9 0 0.04]);
\end{lstlisting}
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q2c_fig1_1s.png}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q2c_fig2_2s.png}
\caption{}
\end{subfigure}
\caption{(a) 1s probability density. (b) 2s probability density.}
\end{figure}
\end{minipage}
\item %2d
For the 1s level, \boxed{E = -13.4978\:\mathrm{eV}}.
\item %2e
Beginning with equation (11) from the assignment, with \(l_o=0\):
\begin{align*}
\left[-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} - \frac{q^2}{4\pi\epsilon_0r}\right]f(r) &= Ef(r)\\
-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\left(\frac{2r}{a_0^{3/2}}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
-\frac{\hbar^2}{2m}\frac{2}{a_0^{3/2}}\frac{d}{dr}\left(e^{-r/a_0} - \frac{r}{a_0}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
-\frac{\hbar^2}{2m}\frac{2}{a_0^{3/2}}\left(-\frac{1}{a_0}e^{-r/a_0} - \frac{1}{a_0}e^{-r/a_0} + \frac{r}{a_0^2}e^{-r/a_0}\right) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
-\frac{\hbar^2}{2m}\left(-\frac{2}{a_0r} + \frac{1}{a_0^2}\right)f(r) - \frac{q^2}{4\pi\epsilon_0r}f(r) &= Ef(r)\\
-\frac{\hbar^2}{2m}\left(-\frac{2}{a_0r} + \frac{1}{a_0^2}\right) - \frac{q^2}{4\pi\epsilon_0r} &= E.
\end{align*}
Recalling that \(a_0 = 4\pi\epsilon_0\hbar^2/mq^2\), we can eliminate \(r\):
\begin{align*}
\cancel{\frac{\hbar^2}{2m}}\frac{\cancel{2m}q^2}{4\pi\epsilon_0\cancel{\hbar^2}}\frac{1}{r} - \frac{\hbar^2}{2m}\frac{1}{a_0^2} - \frac{q^2}{4\pi\epsilon_0}\frac{1}{r} &= E\\
\cancel{\frac{q^2}{4\pi\epsilon_0}\frac{1}{r}} - \frac{\hbar^2}{2m}\frac{1}{a_0^2} - \cancel{\frac{q^2}{4\pi\epsilon_0}\frac{1}{r}} &= E.
\end{align*}
We can then solve for \(E\):
\begin{equation*}
E\:\mathrm{[eV]} = -\frac{1}{q}\cdot\frac{\hbar^2}{2ma_0^2} = -\frac{1}{q}\cdot\frac{(\SI{1.054e-34}{\joule\cdot\second})^2}{2(\SI{9.110e-31}{\kilogram})(\SI{0.0529}{\nm})^2}
= \boxed{- 13.6\:\mathrm{eV}.}
\end{equation*}
This is very similar to the result in (d).
\item %2f
In the figure below we can see that the numerical and analytical results agree up to scaling by \(a\). The scale difference is expected,
as discussed in Problem 1. From (d), we also expect agreement in the curve shapes because the numerical and analytical energies for the 1s
level are very similiar. We can see that the peak value of the analytic result is very slightly higher than that of the numerical result,
which corresponds to the analytical result for the energy being slightly greater in magnitude (\(-13.6\) eV versus \(-13.4978\) eV).
\begin{minipage}[t]{\linewidth}
\centering
\includegraphics[width=0.5\textwidth]{q2f_fig.png}
\captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
\end{minipage}
\end{enumerate}
\newpage
\section*{Problem 3}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\lipsum[1]
\item %2b
\item %3a
\begin{equation}
\theta (x') = \sqrt{\frac{2}{L}} sin \left( \frac{\pi (x' + L/2)}{L} \right) = \sqrt{\frac{2}{L}} cos \left( \frac{\pi x'}{L} \right)
\end{equation}
\item %3b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1di
\lipsum[1]
\item %1dii
\lipsum[1]
\item %1diii
\lipsum[1]
\item %1div
\lipsum[1]
\item %1dv
\lipsum[1]
\item %1dvi
\lipsum[1]
\item %1dvii
\lipsum[1]
\item %3bi
Mapping the provided Fourier identities from \( t \) and \( \omega \) onto
\(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\):
\begin{align*}
\mathcal{F}\left[rect(\frac{x'}{L}) \right] &= \frac{L}{\sqrt{2 \pi}} sinc \left( \frac{k' L}{2 \pi} \right) \\
\mathcal{F}\left[f(x')cos(\frac{\pi}{L}x')\right] &= \frac{1}{2} \left[ F(k' + k_1) + F(k' - k_1) \right] \\
A(k') &= \frac{1}{2} \sqrt{\frac{L}{\pi}} \left\{ \sinc \left( \frac{L}{2\pi} (k' + k_1) \right) + \sinc \left( \frac{L}{2\pi} (k' - k_1) \right) \right\}.
\end{align*}
\item %3bii
Beginning with the result for \(A(k')\) above, and writing
\begin{equation*}
\Phi(p') \equiv \frac{1}{\sqrt{\hbar}} A \left(\frac{p'}{\hbar}\right),
\end{equation*}
we can obtain
\begin{equation*}
\Phi(p') = \frac{1}{2} \sqrt{\frac{L}{\pi\hbar}} \left\{ \sinc \left( \frac{L}{2\pi\hbar} (p' + p_1) \right) + \sinc \left( \frac{L}{2\pi\hbar} (p' - p_1) \right) \right\}.
\end{equation*}
\item %3biii
\(\left|\Phi(p')\right|^2\) has units of [\si{\s.kg^{-1}.m^{-1}}], which are those of inverse momentum. Thus, multiplication
(or integration) by a differential of momentum results in a unitless probability, as we should expect. This holds in the 1D case
and can easily be generalized to higher dimensions.
\item %3biv
\( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral.
% TODO just replace these in the formulas themselves
As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \)
and \( B = \frac{L}{2 \pi \hbar} \).
\begin{equation*}
\int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[ sinc\left( B(p'+p_1) \right)^2 + 2 sinc \left( B(p'+p_1) \right) sinc \left( B(p'-p_1) \right) + sinc \left( B(p'-p_1) \right) \right] dp
\end{equation*}
Given property (26) of the \( sinc \) function, we can simplify the left and right elements. % TODO better word than elements
Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central element.
\begin{align}
\int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[2 sinc \left( B(2p_1 - p'') \right) sinc \left( B(-p'') \right) \right] dp + A^2 \frac{2}{B} \\
&= \int _{-\infty} ^{\infty} A^2 \left[ 2 sinc \left(B(2p_1 - p'') \right) sinc(B p'') \right] dp + A^2 \frac{2}{B} \\
&= \frac{A^2}{B} \left[ sinc(2p_1) + 2 \right] \\
\end{align}
\item %3bv
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q3bv_fig1.png}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q3bv_fig2.png}
\caption{}
\end{subfigure}
\caption{(a) momentum wavefunction versus normalized momentum. (b) Probability density versus normalized momentum.}
\end{figure}
\end{minipage}
\item %3bvi
The points of classical momentum are given by \( p_1 = \pm \sqrt{2mE}\). On the normalized plots, these occur at \(\pm 1\) on the \(p/p_1\) axis.
Given that \(L = \SI{101}{\angstrom}\), we can find the velocity of the electron by taking \(v = \frac{p_1}{m_e}\). We find that
\(v = \pm \SI{3.6e4}{m/s}\).
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q3bvi_fig1.png}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q3bvi_fig2.png}
\caption{}
\end{subfigure}
\caption{(a),(b) Previous plots but with the classical momentum marked in red.}
\end{figure}
\end{minipage}
\item %3bvii
From the plot of the probability density, we can clearly see that the particle can take a continuum of momentum values.
Thus the statement is false.
\end{enumerate}
\item %1c
\lipsum[1]
\item %1d
\lipsum[1]
\item %3c
Because the probability density is even about \(p' = 0\), we can surmise that \(\left\langle p'\right\rangle = 0\).
\newline
To verify this, we find \(\left\langle p'\right\rangle\)
from \(\phi(x') = \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\times\rect{\left(\frac{x'}{L}\right)}\) according to
\begin{align*}
\left\langle p'\right\rangle\ &= \int_{-\infty}^\infty \phi^*(x')\:\hat{p}\:\phi(x')\:dx'\\
\left\langle p'\right\rangle\ &= -i\hbar\int_{-L/2}^{L/2} \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\:\frac{d}{dx'}\left[\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi}{L}x'\right)}\right]\:dx'\\
\left\langle p'\right\rangle\ &= \frac{-2i\pi\hbar}{L^2}\int_{-L/2}^{L/2} \sin{\left(\frac{\pi}{L}x'\right)}\:\cos{\left(\frac{\pi}{L}x'\right)}\:dx'.
\end{align*}
Using Equation (31) in the assignment we can write
\begin{align*}
\left\langle p'\right\rangle\ &= \left.\frac{-i\hbar}{L}\sin^2{\left(\frac{\pi}{L}x'\right)}\right|_{-L/2}^{L/2}\\
\left\langle p'\right\rangle\ &= \frac{-i\hbar}{L}\left(1 - 1\right) = 0,
\end{align*}
which verifies our above inference.
\item %3d
The momentum associated with the wavefunction \( \theta (x') = e^{ik'x'} \) is sharp,
and the corresponding value is \( p' = \hbar k' \).
\begin{equation*}
\hat{p} = -i \hbar \frac{d}{dx'} e ^{i k' x'} = -i^2 \hbar k' e ^{i k' x'} = \hbar k' e^{i k' x'}
\end{equation*}
\end{enumerate}
\newpage
\section*{Problem 4}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\lipsum[1]
\item %2b
\item %4a
% TODO I really don't like this explanation...
An \( a \) value of \( 0.62 \angstrom \) was chosen, in order to provide an adequately
shaped graph without sacrificing too much computation time.
% maybe talk about normalization?
\item %4b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1di
\lipsum[1]
\item %1dii
\lipsum[1]
\item %1diii
\item %4bi
\lipsum[1]
\item %4bii
\item %4biii
Negative Drain Resistance (NDR) is present in this design from a \(V_D\) of
approximately \(0.27V\) to \(0.45V\), with a mostly linear region from
aroung \(0.28V\) to \(0.3V\).
\end{enumerate}
\end{enumerate}

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% A simple MATLAB code to plot the momentum
% wave function for a particle in an escape-proof "box"
clear all
hbar = 1.054e-34;
L = 101e-10;
% Set up k' and p' axes
k_prime = linspace(-2*pi/L,2*pi/L,100);
dk_prime = k_prime(2) - k_prime(1);
p_prime = hbar * k_prime;
dp_prime = p_prime(2) - p_prime(1);
% k1 and p1
k1 = (pi/L); p1 = hbar * k1;
% Fourier transform A(k')
A1 = 0.5*sqrt(L/pi)*( sinc((k_prime+k1)*L/(2*pi)) );
A2 = 0.5*sqrt(L/pi)*( sinc((k_prime-k1)*L/(2*pi)) );
A = A1 + A2;
A_sum = sum(abs(A).^2)*dk_prime;
% Momentum wave function Phi(p')
Phi1 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime+p1)*L/(2*pi*hbar)) );
Phi2 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime-p1)*L/(2*pi*hbar)) );
Phi = Phi1 + Phi2;
Phi_sum = sum(abs(Phi).^2)*dp_prime;
% Normalized p' axis for plotting purposes; the points of classical
% momenta occur when the variable pp_N is plus or minus unity
pp_N = p_prime / p1;
figure(1); clf;
h = plot(pp_N,Phi1,'kx',pp_N,Phi2,'ko',pp_N,Phi,'k--');
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]');
ylabel('WAVE FUNCTION [sqrt(s / kg m)]');
legend('Phi1','Phi2','Phi');
figure(2); clf; h = plot(pp_N,abs(Phi).^2,'k-');
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]');
ylabel('PROBABILITY DENSITY [s / kg m]');

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@ -4,19 +4,20 @@ clear all;
hbar = 1.054e-34;
q = 1.602e-19;
m = 9.110e-31;
epsilon_0 = 8.854e-12;
%generate lattice
N = 100; %number of lattice points
n = [1:N]; %lattice points
a = 1e-10; %lattice constant
x = a * n; %x-coordinates
a = 0.1e-10; %lattice constant
r = 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
U = -q^2./(4*pi*epsilon_0.*r) * (1/q); %potential at r in [eV]
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
@ -33,25 +34,25 @@ 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);
phi_2 = eigenvectors(:,2);
% 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);
P_2 = phi_2 .* conj(phi_2);
% 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
% Plot the probability densities for 1st and 2nd eigenvectors
figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'k-');
figure(1); clf; h = plot(r,P_1,'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]);
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
yticks([0.02 0.04 0.06 0.08 0.10 0.12]);
legend('n=1');
axis([0 1e-9 0 0.12]);
figure(2); clf; h = plot(r,P_2,'k-');
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('RADIAL POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
yticks([0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04]);
legend('n=2');
axis([0 1e-9 0 0.04]);

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clear all;
% Physical constants in MKS units, including free-electron mass m
hbar = 1.054e-34;
q = 1.602e-19;
m = 9.110e-31;
% SHO frequency
w0 = (0.2879*q/hbar);
% Lattice data: lattice constant "a"; SHO distance parameter beta;
% a discretized spatial grid x; total number of lattice points N;
% Hamiltonian parameter t0, as introduced in class, but now
% divided by q so that it has units of eV; total length of the lattice,
% L = a*(N+1)
% YOU MUST ENTER AN APPROPRIATE VALUE OF "a"
a = 12 / sqrt(m*w0/hbar) / 100;
beta = sqrt(m*w0/hbar);
x = [-6/beta:a:6/beta];
N = length(x);
t0 = (hbar^2)/(2*m*a^2)/q;
L = a*(N+1);
% Set up the Hamiltonian matrix, in pieces, by constructing square
% matrices containing information for the main
% diagonal, lower diagonal, and upper diagonal, and then adding them
% together; you may need to look up the diag command using help
% to see how this works; U is the potential energy
U = (1/q)*(0.5*m*w0^2*x.^2);
main_D = diag(2*t0*ones(1,N)+U,0);
lower_D = diag(-t0*ones(1,N-1),-1);
upper_D = diag(-t0*ones(1,N-1),+1);
H = main_D + lower_D + upper_D;
% Find the eigenvectors and eigenvalues of H; V will contain the
% eigenvectors in its columns, and D will contain the corresponding
% eigenvalues on its main diagonal
[V,D] = eig(H);
% Find the numerical eigenvalues
E_numerical = diag(D,0);
% Find the eigenvectors corresponding to sorted eigenvalues 1 and 50;
% we can get these from the matrix V, provided we reference the
% appropriate UNSORTED eigenvalue number
phi_numerical_1 = V(:,1);
phi_numerical_2 = V(:,2);
% Find the position-probability densities
P_numerical_1 = phi_numerical_1 .* conj(phi_numerical_1);
P_numerical_2 = phi_numerical_2 .* conj(phi_numerical_2);
% Plot out the relevant results
figure(1); h = plot(x,P_numerical_1,'k-'); grid on;
set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
legend('Eigenvector 1');
figure(2); h = plot(x,P_numerical_2,'k-'); grid on;
set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
legend('Eigenvector 2');

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clear all;
% Physical constants in MKS units
hbar = 1.054e-34;
q = 1.602e-19;
% Energy parameters in eV; included are the single-electron charging
% energy U0; the kBT product; the equilibrium Fermi level mu; and the
% energy levels cal_E1 and cal_E2, where "cal_E" is short-form for "calligraphic E"
% Please especially note that ALL ENERGY VARIABLES IN THIS CODE
% ARE IN eV (NOT joules); the equations from class must be adjusted
% accordingly, multiplying or dividing appropriate terms by a factor of q
% YOU MUST ENTER THE APPROPRIATE VALUES OF cal_E1 and cal_E2
U0 = 0.025;
kBT = 0.025;
mu = 0;
cal_E1 = 0.2879 / 2;
cal_E2 = 0.2879 * 3 / 2;
% Capacitance parameters
alpha_G = 0.5;
alpha_D = 0.5;
alpha_S = 1 - alpha_G - alpha_D;
% Energy grid in eV, from -1 eV to 1 eV
NE = 501;
E = linspace(-1,1,NE);
dE = E(2) - E(1);
% Coupling coefficients, which are now grids over E, with
% gamma_1 equal to 0.005 eV for E > 0, and 0 for E <= 0, and
% with gamma_2 equal to 0.005 eV for all E; the 1e-6
% term is included to avoid divide-by-zero errors
gamma_1 = 0.005*(E + abs(E)) ./ (E + E + 1e-6);
gamma_2 = 0.005*ones(1,NE);
gamma = gamma_1 + gamma_2;
% Reference number of electrons in the channel, assumed to be zero in
% this code
N0 = 0;
% Voltage values to consider for the final plots
NV = 121;
VV = linspace(-0.4,0.8,NV);
dV = VV(2) - VV(1);
% Loop over voltage values and compute number of electrons and current
% for each voltage value in a self-consistent manner
for count = 1:NV
% Set terminal voltages
VG = 0;
VD = VV(count);
VS = 0;
% Values of mu1 and mu2; notice that the usual factor of q multiplying
% the voltages is omitted, because in this code, energy is in eV
mu1 = mu - VS;
mu2 = mu - VD;
% Compute source and drain Fermi functions
f1 = 1./(1+exp((E - mu1)./kBT));
f2 = 1./(1+exp((E - mu2)./kBT));
% Value of Laplace potential in eV
UL = - (alpha_G*VG) - (alpha_D*VD) - (alpha_S*VS);
% Initial value of Poisson part in eV
UP = 0;
% Iterate until self-consistent potential is achieved by monitoring
% the Poisson part (the Laplace part does not change)
dUP = 1;
while dUP > 1e-6
% Lorentzian SHIFTED density of states for levels 1 and 2,
% each normalized so that its integral is unity
D1 = (0.01/(2*pi))./((E - (UL + UP) - cal_E1).^2+(0.01/2)^2);
D1 = D1./(dE*sum(D1));
D2 = (0.01/(2*pi))./((E - (UL + UP) - cal_E2).^2+(0.01/2)^2);
D2 = D2./(dE*sum(D2));
% Total density of states
D = D1 + D2;
% Compute number of channel electrons
N(count) = dE*sum( ((gamma_1./gamma).*f1 + (gamma_2./gamma).*f2).*D );
% Newly calculated Poisson part of self-consistent potential
UPnew = U0*( N(count) - N0 );
% Change in Poisson part between iterations
dUP = abs(UP - UPnew);
% New guess for next iteration, found by adding a fraction of the
% difference between iterations to the old guess
UP = UP + 0.1*(UPnew - UP);
end
% Compute the current in A after the self-consistent potential
% has been achieved; notice the extra factor of q preceding the
% equation, which is needed since the gammas are in eV
I(count) = q*(q/hbar)...
*dE*sum((f1-f2).*D.*gamma_1.*gamma_2./gamma);
run q4b_subplots;
end
% Plotting commands, including lines to modify the linewidth
% and Fontsize, just to make the plots look nicer; you don't
% need to worry about how these work
figure(1); h = plot(VV,N,'k'); grid on;
set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('DRAIN VOLTAGE [V]'); ylabel('NUMBER OF ELECTRONS');
figure(2); h = plot(VV,I,'k'); grid on;
set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
xlabel('DRAIN VOLTAGE [V]'); ylabel('CURRENT [A]');

113
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% Code to create the required subplots for Problem Set 2, Problem 5(b)
% If you've used the same variable names as in the sample code, then you
% should be able to simply insert this into the appropriate spot in your
% own code; otherwise, you'll have to modify this accordingly, which
% should be easy to do---if disaster strikes and it doesn't work, then
% please ask for help
% The "if" statement is used to choose VD values closest to the required
% values of 0.0, 0.2, 0.3, ..., 0.8 V, and you don't need to
% worry about how this works
if (abs(VD-0.0) <= dV/2)
figure(3); title('VD = 0.0 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V');
elseif (abs(VD-0.2) <= dV/2)
figure(4); title('VD = 0.2 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V');
elseif (abs(VD-0.25) <= dV/2)
figure(5); title('VD = 0.25 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V');
elseif (abs(VD-0.3) <= dV/2)
figure(6); title('VD = 0.3 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V');
elseif (abs(VD-0.4) <= dV/2)
figure(7); title('VD = 0.4 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V');
elseif (abs(VD-0.5) <= dV/2)
figure(8); title('VD = 0.5 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V');
elseif (abs(VD-0.6) <= dV/2)
figure(9); title('VD = 0.6 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V');
elseif (abs(VD-0.65) <= dV/2)
figure(10); title('VD = 0.65 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V');
elseif (abs(VD-0.7) <= dV/2)
figure(11); title('VD = 0.7 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V');
elseif (abs(VD-0.8) <= dV/2)
figure(12); title('VD = 0.8 V');
subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]);
xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V');
end

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function out = sinc(x)
% Deal with the removable singularity at 0 explicitly.
out = sin(pi*x)./(pi*x);
out(x == 0) = 1;
end