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@ -67,7 +67,7 @@
\newcommand{\angstrom}{\textup{\AA}}
\title{ECE 456 - Problem Set 2}
\date{2021-03-01}
\date{2021-03-08}
\author{David Lenfesty \\ lenfesty@ualberta.ca
\and Phillip Kirwin \\ pkirwin@ualberta.ca}
@ -292,7 +292,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1cii.png}
}
\captionof{figure}{Analytic solution to numeric system, plotted.}
\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
@ -309,9 +309,9 @@ legend({'Numerical','Analytical'},'Location','northwest');
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*}
\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
@ -319,7 +319,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
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{minipage}[b]{\linewidth}
\begin{figure}[H]
\centering
@ -385,15 +385,16 @@ H(N, 1) = -t0;
\adjustbox{valign=t}{
\includegraphics[width=0.5\textwidth]{q1div.jpg}
}
\captionof{figure}{Sketch of degenerate energy levels.}
\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 very closely, to within
acceptable margin of the numerical results from part (ii).
indices \(n=0\), \(1\), and \(2\), respectively. These match within an
acceptable margin to the numerical results from part (ii).
\end{enumerate}
@ -528,7 +529,7 @@ axis([0 1e-9 0 0.04]);
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}
\includegraphics[width=0.6\textwidth]{q2f_fig.png}
\captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
\end{minipage}
@ -542,10 +543,15 @@ axis([0 1e-9 0 0.04]);
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %3a
Recalling the identity
\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)
\cos{u} = \sin{\left(\frac{\pi}{2} + u\right)},
\end{equation}
we can write
\begin{align}
\phi (x') &= \sqrt{\frac{2}{L}} \sin{ \left( \frac{\pi (x' + L/2)}{L} \right) } \nonumber \\
\phi (x') &= \boxed{\sqrt{\frac{2}{L}} \cos{\left( \frac{\pi x'}{L} \right)}.}
\end{align}
\item %3b
@ -553,13 +559,18 @@ axis([0 1e-9 0 0.04]);
\item %3bi
Mapping the provided Fourier identities from \( t \) and \( \omega \) onto
\(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\):
\(x'\) 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*}
\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\}.
\mathcal{F}\left[\rect{\left(\frac{x'}{L}\right)} \right] &= \frac{L}{\sqrt{2 \pi}} \sinc{ \left( \frac{k' L}{2 \pi} \right) }\\
\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]
\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
@ -568,8 +579,9 @@ axis([0 1e-9 0 0.04]);
\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\}.
\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
@ -580,23 +592,28 @@ axis([0 1e-9 0 0.04]);
\item %3biv
\( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral.
As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \)
and \( B = \frac{L}{2 \pi \hbar} \).
\( sinc \) 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{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*}
\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*}
Given property (26) of the \( sinc \) function, we can simplify the left and right terms.
Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central term.
Given property (26) of the \( sinc \) function in the assignment, we can evaluate the left and right terms to be \(1/B\).
Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central cross term:
\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] \\
&= \frac{1}{2} \left[ sinc(2p_1) + 2 \right]
\end{align}
\begin{align*}
\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
@ -621,7 +638,7 @@ axis([0 1e-9 0 0.04]);
\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
\(v = \pm \SI{3.6e4}{m/s}\).
\(\boxed{v = \pm \SI{3.6e4}{m/s}}\).
\begin{minipage}[t]{\linewidth}
@ -646,7 +663,7 @@ axis([0 1e-9 0 0.04]);
\end{enumerate}
\item %3c
Because the probability density is even about \(p' = 0\), we can surmise that \(\left\langle p'\right\rangle = 0\).
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
@ -659,13 +676,13 @@ axis([0 1e-9 0 0.04]);
\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,
\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 wavefunction \( \theta (x') = e^{ik'x'} \) is sharp,
and the corresponding value is \( p' = \hbar k' \).
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*}
\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'}
@ -692,12 +709,12 @@ axis([0 1e-9 0 0.04]);
\includegraphics[width=\textwidth]{q4a_e2.png}
\caption{}
\end{subfigure}
\caption{Probability Density Plots for first two energy levels.}
\caption{Probability densities versus position for first two energy levels.}
\end{figure}
\end{minipage}
An \( a \) value of \( 0.53 \angstrom \) was chosen in order to provide an adequately
shaped graph without sacrificing too much computation time and ensure that the first two
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 \(\SI{0.43140}{\electronvolt} \),
@ -710,29 +727,29 @@ axis([0 1e-9 0 0.04]);
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %4bi
The energies were used were \( 0.14395eV \) and \( 0.43185eV \) for the first and second energy levels,
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=\textwidth]{q4bi_I-V.png}
\captionof{figure}{Current vs. Voltage.}
\includegraphics[width=0.5\textwidth]{q4bi_I-V.png}
\captionof{figure}{Current-voltage characteristic of a 2-level molecule.}
\end{minipage}
\item %4bii
Between \( 0V \) and \( 0.25V \), only the first energy level is carrying any current.
This current drops to 0 above \( 0.25V\) because the coupling between the contacts and that
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 \(0.4V\) and \(0.65V\), only the second energy level is carrying current.
This energy level stops conducting current because it's shifted energy drops below
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 Drain Resistance (NDR) is present in this design from a \(V_D\) of
approximately \(0.27V\) to \(0.45V\), as well as from \(0.65V\) to \(0.8V\)
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}