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pkirwin 2021-03-01 20:21:04 -07:00
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8 changed files with 95 additions and 68 deletions

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\newlabel{sub@fig:q3_current}{{b}{3}}
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@ -12,4 +13,5 @@
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@ -155,7 +155,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
\begin{subfigure}{0.4\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}
@ -164,6 +164,8 @@ legend({'Numerical','Analytical'},'Location','northwest');
\caption{Comparison of first 101 numerical and analytic eigenvalues.}
\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}
\item %1b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
@ -185,7 +187,7 @@ 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}
@ -247,25 +249,25 @@ 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 $\phi _\ell = Bsin \left( \frac {n \pi}{L} a \ell \right)$, 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)
E = 2t_0 \left( 1 - cos \left( \frac{n \pi a}{L} \right) \right)
\label{eq:numerical_E}
\end{equation}
@ -286,7 +288,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
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 $E$:
@ -388,7 +390,7 @@ H(N, 1) = -t0;
\end{equation*}
\item %2b
Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be 0.
\item %1c
\item %2c
Code:
\begin{lstlisting}
clear all;
@ -464,12 +466,22 @@ axis([0 1e-9 0 0.04]);
\caption{2s probability density.}
\end{subfigure}
\end{figure}
\item %1d
\item %2d
For the 1s level, \(E = -13.4978\) eV.
\item %1e
\lipsum[1]
\item %1f
\item %2e
\lipsum[1]
\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{figure}[H]
\centering
\includegraphics[width=\textwidth]{q2f_fig.png}
\caption{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
\end{figure}
\end{enumerate}
@ -477,28 +489,28 @@ axis([0 1e-9 0 0.04]);
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\item %3a
\lipsum[1]
\item %2b
\item %3b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1di
\item %3bi
\lipsum[1]
\item %1dii
\item %3bii
\lipsum[1]
\item %1diii
\item %3biii
\lipsum[1]
\item %1div
\item %3biv
\lipsum[1]
\item %1dv
\item %3bv
\lipsum[1]
\item %1dvi
\item %3bvi
\lipsum[1]
\item %1dvii
\item %3bvii
\lipsum[1]
\end{enumerate}
\item %1c
\item %3c
\lipsum[1]
\item %1d
\item %3d
\lipsum[1]
\end{enumerate}
@ -507,15 +519,15 @@ axis([0 1e-9 0 0.04]);
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
\item %2a
\item %4a
\lipsum[1]
\item %2b
\item %4b
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1di
\item %4bi
\lipsum[1]
\item %1dii
\item %4bii
\lipsum[1]
\item %1diii
\item %4biii
\lipsum[1]
\end{enumerate}

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