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pkirwin 2021-03-02 20:09:54 -07:00
parent ce398d08bf
commit 8147ca4366
7 changed files with 193 additions and 193 deletions
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PS2/doc.aux
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Output written on d:/Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\doc.pdf (13 pages, 1844779 bytes). Output written on d:/Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\doc.pdf (12 pages, 1843727 bytes).
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@ -1,6 +1,7 @@
\documentclass{article} \documentclass{article}
\usepackage{graphicx} \usepackage{graphicx}
\usepackage{caption}
\usepackage{setspace} \usepackage{setspace}
\usepackage{listings} \usepackage{listings}
\usepackage{color} \usepackage{color}
@ -22,6 +23,8 @@
\titleformat{\subsection}[runin]{\normalfont \large \bfseries} \titleformat{\subsection}[runin]{\normalfont \large \bfseries}
{\thesubsection}{1em}{} {\thesubsection}{1em}{}
\captionsetup[figure]{width=0.75\textwidth}
\renewcommand{\thesubsection}{\indent(\alph{subsection})} \renewcommand{\thesubsection}{\indent(\alph{subsection})}
\definecolor{dkgreen}{rgb}{0,0.6,0} \definecolor{dkgreen}{rgb}{0,0.6,0}
@ -175,10 +178,10 @@ legend({'Numerical','Analytical'},'Location','northwest');
\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 The analytical solution is:
\begin{equation} \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} \label{eq:analytical}
\end{equation} \end{equation}
@ -196,41 +199,42 @@ legend({'Numerical','Analytical'},'Location','northwest');
\label{eq:integral_ident} \label{eq:integral_ident}
\end{equation} \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. to the above identity.
Evaluating the integral gives us the following relationship Evaluating the integral gives us the following relationship:
\begin{equation*} \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*} \end{equation*}
From this, we find: From this, we find:
\begin{equation*} \begin{equation*}
\boxed{A = \sqrt{\frac{2}{L}}} \boxed{A = \sqrt{\frac{2}{L}}.}
\end{equation*} \end{equation*}
\item %1bii \item %1bii
Starting with the normalization condition for the numerical case: Starting with the normalization condition for the numerical case:
\begin{align} \begin{align}
a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\ a\sum_{\ell=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|B\sin{\left(\frac{n\pi}{L}x_\ell\right)}\right|^2 &= a, \nonumber
\label{eq:numerical_norm} \label{eq:numerical_norm}
\end{align} \end{align}
recalling that \(x = al\), and allowing \(a \to 0\) while holding \(L\) recalling that \(x = a\ell\), and allowing \(a \to 0\), while holding \(L\)
constant implies that \(N \to \infty\), since \(a=\frac{L}{N}\). constant, implies that \(N \to \infty\), since \(a=\frac{L}{N}\).
An integral is defined as the limit of a Riemann sum as follows: An integral is defined as the limit of a Riemann sum as follows:
\begin{equation} \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} \label{eq:Riemann_sum}
\end{equation} \end{equation}
where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case, 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 \(f(x) = \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\). Therefore we can write
\begin{equation*} \begin{equation*}
\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx \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. = a.
\end{equation*} \end{equation*}
@ -242,7 +246,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
= \frac{a}{B^2}. = \frac{a}{B^2}.
\end{equation*} \end{equation*}
This means that B must be This means that \(B\) must be
\begin{equation*} \begin{equation*}
\boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.} \boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.}
\end{equation*} \end{equation*}
@ -254,25 +258,25 @@ legend({'Numerical','Analytical'},'Location','northwest');
\item %1ci \item %1ci
% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen) % TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} a \ell \right)$, 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 $\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
$sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$ and $\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$ and
$sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$, respectively, where $\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$, respectively, where
$a = \frac{n \pi a \ell}{L}$ and $B = \frac{n \pi a}{L}$. $a = \frac{n \pi a \ell}{L}$ and $B = \frac{n \pi a}{L}$.
Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation: Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation:
\begin{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) -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*} \end{equation*}
At this point, we notice that $\phi _ \ell = B sin \left( \frac {n \pi}{L} a \ell \right)$, At this point, we notice that $\phi _ \ell = B \sin{\left( \frac {n \pi}{L} a \ell \right)} $,
so we can factor it out. so we can factor it out.
With some minor rearranging, this leaves us with the final expression for $E$: With some minor rearranging, this leaves us with the final expression for $E$:
\begin{equation} \begin{equation}
\boxed{E = 2t_0 \left( 1 - cos \left( \frac{n \pi a}{L} \right) \right)} \boxed{E = 2t_0 \left( 1 - \cos{\left( \frac{n \pi a}{L} \right)} \right).}
\label{eq:numerical_E} \label{eq:numerical_E}
\end{equation} \end{equation}
@ -282,7 +286,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
\centering \centering
\adjustbox{valign=t}{ \adjustbox{valign=t}{
\includegraphics[width=\textwidth]{q1cii.png} \includegraphics[width=0.5\textwidth]{q1cii.png}
} }
\captionof{figure}{Analytic solution to numeric system, plotted.} \captionof{figure}{Analytic solution to numeric system, plotted.}
\end{minipage} \end{minipage}
@ -292,24 +296,23 @@ legend({'Numerical','Analytical'},'Location','northwest');
\item %1ciii \item %1ciii
Applying the approximation $cos(\theta) = 1 - \frac{\theta^2}{2}$ for small $\theta$, 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 on equation (\ref{eq:numerical_E}), we get the following expression:
\begin{equation*} \begin{equation*}
E = 2 t_0 \left( \frac{n^2 \pi^2 a^2}{2 L^2} \right) E = 2 t_0 \left( \frac{n^2 \pi^2 a^2}{2 L^2} \right).
\end{equation*} \end{equation*}
Fully substituting the known value of $t_0$, and we can get our final analytical expression for $E$: We can get our final analytical expression for $E$ by fully substituting the explicit form of $t_0$:
\begin{equation*} \begin{equation*}
\boxed{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*} \end{equation*}
\item %1civ \item %1civ
With the decreased lattice spacing, and increased number of points, we can see the numerical solution 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 more closely matches the analytical solution. As well, the \(n=50\) case is
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
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. plot in section (a), which has a low-frequency envelope around it.
\begin{minipage}[t]{\linewidth} \begin{minipage}[t]{\linewidth}
@ -319,13 +322,16 @@ legend({'Numerical','Analytical'},'Location','northwest');
\begin{subfigure}{0.5\textwidth} \begin{subfigure}{0.5\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q1civ_fig1.png} \includegraphics[width=\textwidth]{q1civ_fig1.png}
\caption{Probability densities for \(n=1\)\\and \(n=50\).} \caption{}
\end{subfigure}% \end{subfigure}%
\begin{subfigure}{0.5\textwidth} \begin{subfigure}{0.5\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q1civ_fig2.png} \includegraphics[width=\textwidth]{q1civ_fig2.png}
\caption{Comparison of first 101 numerical and analytic eigenvalues.} \caption{}
\end{subfigure} \end{subfigure}
\caption{(a) Probability densities for \(n=1\) and \(n=50\). (b) Comparison of first 101 numerical and analytic eigenvalues.}
\end{figure} \end{figure}
\end{minipage} \end{minipage}
@ -372,7 +378,7 @@ H(N, 1) = -t0;
\begin{minipage}[t]{\linewidth} \begin{minipage}[t]{\linewidth}
\centering \centering
\adjustbox{valign=t}{ \adjustbox{valign=t}{
\includegraphics[width=\textwidth]{q1div.jpg} \includegraphics[width=0.5\textwidth]{q1div.jpg}
} }
\captionof{figure}{Sketch of degenerate energy levels.} \captionof{figure}{Sketch of degenerate energy levels.}
\end{minipage} \end{minipage}
@ -380,8 +386,8 @@ H(N, 1) = -t0;
\item %1div \item %1div
Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing
by the requisite $q$), we get the energy levels of $0eV, 0.0147eV, and 0.0589eV$, for the by the requisite $q$), we get energy levels of \(0\) eV, \(0.0147\) eV, and \(0.0589\) eV for the
eigenvalue numbers $n = 0, 1, and 2$, respectively. These match very closely, to within indices \(n=0\), \(1\), and \(2\), respectively. These match very closely, to within
acceptable margin of the numerical results from part (ii). acceptable margin of the numerical results from part (ii).
@ -412,6 +418,7 @@ H(N, 1) = -t0;
clear all; clear all;
%physical constants in MKS units %physical constants in MKS units
hbar = 1.054e-34; hbar = 1.054e-34;
q = 1.602e-19; q = 1.602e-19;
m = 9.110e-31; m = 9.110e-31;
@ -475,13 +482,14 @@ axis([0 1e-9 0 0.04]);
\begin{subfigure}{0.5\textwidth} \begin{subfigure}{0.5\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2c_fig1_1s.png} \includegraphics[width=\textwidth]{q2c_fig1_1s.png}
\caption{1s probability density.} \caption{}
\end{subfigure}% \end{subfigure}%
\begin{subfigure}{0.5\textwidth} \begin{subfigure}{0.5\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2c_fig2_2s.png} \includegraphics[width=\textwidth]{q2c_fig2_2s.png}
\caption{2s probability density.} \caption{}
\end{subfigure} \end{subfigure}
\caption{(a) 1s probability density. (b) 2s probability density.}
\end{figure} \end{figure}
\end{minipage} \end{minipage}
\item %2d \item %2d
@ -515,7 +523,7 @@ 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). 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} \begin{minipage}[t]{\linewidth}
\centering \centering
\includegraphics[width=\textwidth]{q2f_fig.png} \includegraphics[width=0.5\textwidth]{q2f_fig.png}
\captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).} \captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).}
\end{minipage} \end{minipage}