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@ -66,14 +66,14 @@
\newcommand{\angstrom}{\textup{\AA}} \newcommand{\angstrom}{\textup{\AA}}
\title{ECE 456 - Problem Set 2} \title{ECE 456 - Problem Set 3 (Part 1)}
\date{2021-03-08} \date{2021-03-31}
\author{David Lenfesty \\ lenfesty@ualberta.ca \author{David Lenfesty \\ lenfesty@ualberta.ca
\and Phillip Kirwin \\ pkirwin@ualberta.ca} \and Phillip Kirwin \\ pkirwin@ualberta.ca}
\pagestyle{fancy} \pagestyle{fancy}
\fancyhead[L]{\textbf{ECE 456} - Problem Set 2} \fancyhead[L]{\textbf{ECE 456} - Problem Set 3 (Part 1)}
\fancyhead[R]{David Lenfesty and Phillip Kirwin} \fancyhead[R]{David Lenfesty and Phillip Kirwin}
\fancyfoot[C]{Page \thepage} \fancyfoot[C]{Page \thepage}
\renewcommand{\headrulewidth}{1pt} \renewcommand{\headrulewidth}{1pt}
@ -90,670 +90,225 @@
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %1a \item %1a
Code: \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\begin{lstlisting}[language=Matlab] \item %1bi
clear all; The matrix equation is
%physical constants in MKS units \begin{equation*}
[\hat{H}]\left\{\phi\right\} = E \left\{\phi\right\},
\end{equation*}
where \([\hat{H}]\) is an \(N\)-by-\(N\) matrix with \([\hat{H}]_{nm} = 0\) except for the following elements:
\begin{align*}
&[\hat{H}]_{nn} = 2t_0 + U_n \\
&[\hat{H}]_{n,n \pm 1} = -t_0 \\
&[\hat{H}]_{0,N} = [\hat{H}]_{N,0} = -t_0,
\end{align*}
with \(t_0 = \hbar^2/(2ma^2)\) and \(U_n = U(na)\). The \(N\)-vector \(\left\{\phi\right\}\) has elements \(\phi_n\) which each represent
the value of the eigenvector at the point \(na = x_n\).
\item %1bii
hbar = 1.054e-34; The expression of the wave function \( \phi (x) \) as a sum of basis functions is as below:
q = 1.602e-19;
m = 9.110e-31;
%generate lattice \begin{equation*}
\phi (x) = \sum _{n = 1} ^{N} \phi _n u_n(x)
\end{equation*}
N = 100; %number of lattice points The derived matrix equation:
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 \begin{equation*}
[\hat{H}] _u \{ \phi \} = [S]_u \{ \phi \}
\end{equation*}
U = 0*x; %0 potential at all x Where \( [\hat{H}]_u \) is a matrix with the elements:
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 \begin{equation*}
H_{nm} = \int u _n ^* (x) \hat{H} u_m(x) dx
\end{equation*}
[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each and \( [S]_u \) is a matrix with the elements:
%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 \begin{equation*}
S_{nm} = \int u _n ^* (x) u_m(x) dx
\end{equation*}
% return eigenvectors for the 1st and 50th eigenvalues \([\hat{H}]_u\) and \([S]_u\) are both of size \(N\)-by-\(N\). The elements
of \(\left\{\phi\right\}\), \(\phi_n\), are the expansion coefficients of \(\phi(x)\).
\end{enumerate}
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{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\linewidth}
\centering
\includegraphics[width=\textwidth]{q1a_1and50.png}
\caption{}
\label{fig:q3_ne}
\end{subfigure}%
\begin{subfigure}{0.5\linewidth}
\centering
\includegraphics[width=\textwidth]{q1a_eigenvals.png}
\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 \item %1b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1bi \item %1bi
The analytical solution is: code:
\begin{equation}
\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}
We use the following identity:
\begin{equation}
\int\sin^2{(a x)}\:dx = \frac{1}{2} x -\frac{1}{4a}\sin{(2a x)}.
\label{eq:integral_ident}
\end{equation}
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:
\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)}}.
\end{equation*}
From this, we find:
\begin{equation*}
\boxed{A = \sqrt{\frac{2}{L}}.}
\end{equation*}
\item %1bii
Starting with the normalization condition for the numerical case:
\begin{align}
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 \(x = a\ell\), 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),
\label{eq:Riemann_sum}
\end{equation}
where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case,
\(n = N\), \(i = \ell\), \(c = 0\), \(d = L\), 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_{\ell=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2
= a.
\end{equation*}
Using (\ref{eq:integral_ident}), we have
\begin{equation*}
\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
= \frac{1}{2}L - \cancel{\frac{L}{4n\pi}\sin{\left(\frac{2n\pi}{L}L\right)}}
- 0 + 0
= \frac{a}{B^2}.
\end{equation*}
This means that \(B\) must be
\begin{equation*}
\boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.}
\end{equation*}
\end{enumerate}
\item %1c
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\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 = 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
$\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$ 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 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} a \ell \right)} $,
so we can factor it out.
With some minor rearranging, this leaves us with the final expression for $E$:
\begin{equation}
\boxed{E = 2t_0 \left( 1 - \cos{\left( \frac{n \pi a}{L} \right)} \right).}
\label{eq:numerical_E}
\end{equation}
\item %1cii
\begin{minipage}[t]{\linewidth}
\centering
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1cii.png}
}
\captionof{figure}{Comparison between analytical result and numerical result. Above \(n=50\), the results diverge substantially.}
\end{minipage}
We can see here that the "predicted" numerical response matches nearly exactly the actual
calculated numerical solution.
\item %1ciii
Applying the approximation $\cos{(\theta)} \approx 1 - \frac{\theta^2}{2}$ 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 a^2}{2 L^2} \right).
\end{equation*}
We can get our final analytical expression for $E$ by fully substituting the explicit form of $t_0$:
\begin{equation}
\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}[b]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1civ_fig1.png}
\caption{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1civ_fig2.png}
\caption{}
\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}
\end{enumerate}
\item %1d
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1di
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:
\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{}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q1di_fig2.png}
\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 \boxed{0.06\:\mathrm{eV}}.
These eigenstates are degenerate because they both have the same
eigenvalue/energy.
\item %1diii
\begin{minipage}[t]{\linewidth}
\centering
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1div.jpg}
}
\captionof{figure}{Sketch of degenerate energy levels. In the sketch, closely-spaced
levels are in fact degenerate.}
\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 within an
acceptable margin to 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
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
Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be \boxed{0}.
\item %2c
Code:
\begin{lstlisting} \begin{lstlisting}
clear all; %constants
%physical constants in MKS units E1 = -13.6;
R = 0.074;
a0 = 0.0529;
%matrix elements
R0 = R/a0;
hbar = 1.054e-34; a = 2*E1*(1-(1+R0)*exp(-2*R0))/R0;
q = 1.602e-19; b = 2*E1*(1+R0)*exp(-R0);
m = 9.110e-31; s = exp(-R0)*(1+R0+(R0^2/3));
epsilon_0 = 8.854e-12;
%generate lattice %matrices
H_u = [E1 + a, E1*s+b; E1*s+b, E1 + a];
S_u = [1, s; s, 1];
N = 100; %number of lattice points %find eigenvalues and eigenvectors
n = [1:N]; %lattice points [vectors,energies] = eig(inv(S_u)*H_u);
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} \end{lstlisting}
The bonding and antibonding eigenenergies are \(\boxed{\SI{-32.2567}{eV}}\) and \(\boxed{\SI{-15.5978}{eV}}\) respectively.
\item %1bii
Neglecting normalization, we have the following expressions for \(\phi_B(z)\) and \(\phi_A(z)\):
\begin{align*}
\phi_B(z) = u_L(z) + u_R(z) \\
\phi_A(z) = u_L(z) - u_R(z).
\end{align*}
We obtain the following plot:
\begin{minipage}[t]{\linewidth} \begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering \centering
\begin{subfigure}{0.5\textwidth} \adjustbox{valign=t}{
\centering \includegraphics[width=0.5\textwidth]{q1bii.png}
\includegraphics[width=\textwidth]{q2c_fig1_1s.png} }
\caption{} \captionof{figure}{non-normalized probability densities for bonding and antibonding solutions.}
\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} \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=0.6\textwidth]{q2f_fig.png}
\captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
\end{minipage}
\end{enumerate} \end{enumerate}
\end{enumerate}
\newpage \newpage
\section*{Problem 3}
\section*{Problem 2}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %2ai
\begin{minipage}[t]{\linewidth}
\centering
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q2ai.png}
}
% TODO still don't like this caption
\captionof{figure}{Energy vs. wave vector relationship (note: not discretized to account for N)}
\end{minipage}
\item %2aii
Energy values from \( -2 \) eV to \(2\) eV are allowed.
\item %2aiii
The vector \( \{ \phi \} \), which is of length \( N \), and has elements \( n \) is:
\begin{equation*}
\{ \phi \} = C e ^{ i k \cdot n a }
\end{equation*}
The corresponding wave function is:
\begin{equation*}
\phi (x) = \sum _ {n = 1} ^ N C e ^ { i k \cdot n a } u_n (x)
\end{equation*}
There is one wave function and thus one energy level for each value of \( k \).
This means that there is one electronic state per \( k \).
\end{enumerate}
\item %2b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %2bi
\begin{align*}
\left\{\phi\right\} = \begin{bmatrix}
C_Ae^{ika} \\
C_Be^{ika} \\
C_Ae^{ik2a} \\
C_Be^{ik2a} \\
\vdots \\
C_Ae^{ikNa} \\
C_Ae^{ikNa}
\end{bmatrix}
\end{align*}
\item %2bii
\begin{equation*}
\phi(x) = \sum_{n=1}^{N} C_Ae^{ikna} u_{nA}(x) + C_Be^{ikna} u_{nB}(x)
\end{equation*}
\item %2biii
\([h(k)]\) is of size 2-by-2. Thus there will be two values of \(E(k)\) for a fixed \(k\).
This also means there are two \(\phi(x)\) for each \(k\).
\end{enumerate}
\item %2c
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %2ci
\begin{align*}
\phi _2 + 2 \phi _3 + \phi _4 &= E \phi_1 \\
\phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\
2 \phi _1 + \phi _2 + \phi _4 &= E \phi_3 \\
\phi _1 + 2 \phi _2 + \phi _3 &= E \phi_4 \\
\end{align*}
\item %2cii
\begin{align*}
\phi _2 + 2 \phi _3 + \phi _0 &= E \phi_1 \\
\phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\
2 \phi _5 + \phi _2 + \phi _4 &= E \phi_3 \\
\phi _5 + 2 \phi _6 + \phi _3 &= E \phi_4 \\
\end{align*}
\item %2ciii
A generalized form of the $n$th equation is:
\begin{equation*}
E \phi_n = \phi_{n-1} + \phi_{n+1} + 2 \phi_{n+2}
\end{equation*}
\item %2civ
\item %3a
Recalling the identity
\begin{equation}
\cos{u} = \sin{\left(\frac{\pi}{2} + u\right)},
\end{equation}
we can write
\begin{align} \begin{align}
\phi (x') &= \sqrt{\frac{2}{L}} \sin{ \left( \frac{\pi (x' + L/2)}{L} \right) } \nonumber \\ E C e ^{ikna} &= Ce ^ {ik (n + 1)a} + Ce^{ik (n - 1)a} + 2 Ce^{ik (n + 2)a} \nonumber \\
\phi (x') &= \boxed{\sqrt{\frac{2}{L}} \cos{\left( \frac{\pi x'}{L} \right)}.} E &= e^{ika} + e^{-ika} + 2 e^{2ika} \nonumber \\
E(k) &= 2 e^{2ika} + 2\cos(ka) \label{eq:e_k}
\end{align} \end{align}
\item %3b \item %2cv
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] Imposing the repeating boundary conditions \( \phi _{n + 4} = \phi _{n} \), We obtain the following relationship:
\item %3bi
Mapping the provided Fourier identities from \( t \) and \( \omega \) onto
\(x'\) and \(k'\), we can evaluate the Fourier transform of \(\phi(x') = \sqrt{\frac{2}{L}}\cos{\left(\frac{\pi}{L} x' \right) }\times\rect{\left(\frac{x'}{L}\right)}\), denoted \(A(k')\),
using the following:
\begin{align*} \begin{align*}
\mathcal{F}\left[\rect{\left(\frac{x'}{L}\right)} \right] &= \frac{L}{\sqrt{2 \pi}} \sinc{ \left( \frac{k' L}{2 \pi} \right) }\\ Ce^{ikna} &= Ce^{ik(n+4)a} \\
\mathcal{F}\left[f(x')\cos{\left(\frac{\pi}{L}x'\right)}\right] &= \frac{1}{2} \left[ F(k' + k_1) + F(k' - k_1) \right] 1 &= e^{i4ka}
\end{align*}
Letting \(f(x') = \sqrt{\frac{2}{L}}\rect{\left(\frac{x'}{L}\right)}\), we can obtain
\begin{align*}
A(k') &= \boxed{\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*}
where \(k_1 = \pi/L\).
\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*}
\boxed{ \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*}
where \(p_1 = \hbar\pi/L\).
\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 taking the norm of the integrand.
As well, to simplify the intermediate equations we will define the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \)
and \( B = \frac{L}{2 \pi \hbar} \). Then we have
\begin{align*}
\int _{-\infty} ^{\infty} \Phi (p')^2 \:dp' = \int _{-\infty} ^{\infty} A^2 &\left[ \sinc{\left( B\left(p'+p_1\right) \right)}^2 \right.\\
&\;\left. + 2 \sinc{ \left( B(p'+p_1) \right)} \sinc{ \left( B(p'-p_1) \right) } + \sinc \left( B(p'-p_1) \right)^2 \right] dp'.
\end{align*} \end{align*}
Given property (26) of the \( sinc \) function in the assignment, we can evaluate the left and right terms to be \(1/B\). For this to hold, \(4ka\) must be some multiple of \(2 \pi \), and this mean \( k = \frac{\pi}{2a} \cdot integer \).
Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central cross term:
\begin{align*} \item %2cvi
\int _{-\infty} ^{\infty} \Phi (p')^2 \:dp &= A^2 \frac{2}{B} - \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'' \\
&= \frac{2A^2}{B} - A^2\sinc(2Bp_1) \\
&= 1 + A^2\sinc(1) \\
&= \boxed{1.}
\end{align*}
Since we obtained \(\Phi(p')\) from a normalized
position wave function and we have reasoned that it should have the same properties, but with respect to momentum rather than
position, it makes sense that this normalization integral should be 1, just as it would be for the associated position wave function.
\item %3bv Using the E-k relationship from equation \ref{eq:e_k},
we know that \( k \) must always be real, so the \( 2\cos(ka) \) portion of the E-k relationship must be
real. As well, if we substitute in the equation for \( k \) we obtained in part \nolinebreak (v), we get the following (partial) expression:
\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 wave function 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
\(\boxed{v = \pm \SI{3.6e4}{m/s}}\).
\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 %3c
Because the probability density is even about \(p' = 0\), we can surmise that \(\boxed{\left\langle p'\right\rangle = 0}\).
\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) = \boxed{0},
\end{align*}
which verifies our above inference.
\item %3d
The momentum associated with the wave function \( \theta (x') = e^{ik'x'} \) is \boxed{\mathrm{sharp}},
and the corresponding value is \(\boxed{ p' = \hbar k' } \).
\begin{equation*} \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'} 2e^{i2ka} = 2 e^{i\pi \cdot integer}
\end{equation*} \end{equation*}
Which we know will always be real (with a value of \( \pm 2 \)).
\item %2cvii
Since \(e^{2\pi n} = 1\), we have:
\begin{align*}
\phi_n(k+\frac{2\pi}{a}) &= Ce^{i(k+\frac{2\pi}{a} \cdot nA} = Ce^{i(k \cdot nA} \\
\phi_n(k+\frac{2\pi}{a}) &= \phi_n(k).
\end{align*}
Therefore wavefunctions for which \(k\) is separated by \(\frac{2\pi}{a}\) are equivalent, and we
only need consider the range \(k \in [-\frac{\pi}{a}, \frac{\pi}{a}]\).
\end{enumerate} \end{enumerate}
\newpage
\section*{Problem 4}
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %4a
\begin{minipage}[t]{\linewidth}
\begin{figure}[H]
\centering
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q4a_e1.png}
\caption{}
\end{subfigure}%
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{q4a_e2.png}
\caption{}
\end{subfigure}
\caption{Probability densities versus position for first two energy levels.}
\end{figure}
\end{minipage}
An \( a \) value of \(\boxed{ 0.53 \angstrom } \) was chosen in order to provide an adequately
shaped graph without sacrificing too much computation time and to ensure that the first two
numerical energies correspond to the given experimental results. The experimental results are
\(\SI{0.14395}{\electronvolt}\) and \(\SI{0.43185}{\electronvolt} \) for the first and second energy levels
respectively, and the numerical results with our chosen \(a\) are \(\SI{0.14386}{\electronvolt}\) and \(\SI{0.43140}{\electronvolt} \),
which are in agreement.
\item %4b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %4bi
The energies used were \( \SI{0.14395}{eV} \) and \( 0.43185 - 0.1\:\si{eV} \) for the first and second energy levels,
respectively.
\begin{minipage}[H]{\linewidth}
\centering
\includegraphics[width=0.5\textwidth]{q4bi_I-V.png}
\captionof{figure}{Current-voltage characteristic of a 2-level molecule.}
\end{minipage}
\item %4bii
Between \( \SI{0}{V} \) and \( \SI{0.25}{V} \), only the first energy level is carrying any current.
This current drops to 0 above \( \SI{0.25}{V}\) because the coupling between the contacts and that
energy level drops to 0, meaning no electrons can transfer.
Between \(\SI{0.4}{V}\) and \(\SI{0.65}{V}\), only the second energy level is carrying current.
This energy level stops conducting current above \(\SI{0.65}{V}\) because its shifted energy drops below
the threshold where the contacts have any coupling with it.
\item %4biii
Negative differential resistance is present in this design from a \(V_D\) of
approximately \(\SI{0.27}{V}\) to \(\SI{0.45}{V}\), as well as from \(\SI{0.65}{V}\) to \(\SI{0.8}{V}\).
\end{enumerate} \end{enumerate}
\end{enumerate}
\end{document} \end{document}

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%constants
E1 = -13.6;
R = 0.074;
a0 = 0.0529;
%matrix elements
R0 = R/a0;
a = 2*E1*(1-(1+R0)*exp(-2*R0))/R0;
b = 2*E1*(1+R0)*exp(-R0);
s = exp(-R0)*(1+R0+(R0^2/3));
%matrices
H_u = [E1 + a, E1*s+b; E1*s+b, E1 + a];
S_u = [1, s; s, 1];
%find eigenvalues and eigenvectors
[vectors,energies] = eig(inv(S_u)*H_u);
z = linspace(-2,2);
z_L = -0.37;
z_R = 0.37;
a0 = 0.529;
u_L = exp(-abs(z - z_L)./a0)./sqrt(pi*a0^3);
u_R = exp(-abs(z - z_R)./a0)./sqrt(pi*a0^3);
phi_B = u_L + u_R;
phi_A = u_L - u_R;
figure(1);
plot(z, phi_B.^2, 'k-');
hold on
plot(z, phi_A.^2, 'b-');
grid on;
xlabel('z [Angstroms]');
ylabel('probability [arbitrary scaling]');
legend('Bonding', 'Antibonding');

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% Atomic spacing
a = 5e-10;
%
t = -1;
E_0 = 0;
N = 100;
k = linspace(- pi / a, pi / a, N);
e_k = E_0 + 2* t * cos(k*a);
figure(1);
plot(k, e_k, 'k-');
grid on;
xlabel('k');
ylabel('E[k] (eV)');

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