diff --git a/PS2/doc.aux b/PS2/doc.aux index cc31c13..8f8388d 100644 --- a/PS2/doc.aux +++ b/PS2/doc.aux @@ -9,10 +9,11 @@ \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}} +\newlabel{eq:Riemann_sum}{{5}{4}} +\newlabel{eq:numerical_E}{{6}{4}} \@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Analytic solution to numeric system, plotted.\relax }}{5}{}\protected@file@percent } -\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Sketch of degenerate energy levels.\relax }}{7}{}\protected@file@percent } -\@writefile{lof}{\contentsline {figure}{\numberline {7}{\ignorespaces Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).\relax }}{10}{}\protected@file@percent } -\gdef \@abspage@last{13} +\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces (a) Probability densities for \(n=1\) and \(n=50\). 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PDF statistics: - 134 PDF objects out of 1000 (max. 8388607) + 130 PDF objects out of 1000 (max. 8388607) 0 named destinations out of 1000 (max. 500000) 56 words of extra memory for PDF output out of 10000 (max. 10000000) diff --git a/PS2/doc.pdf b/PS2/doc.pdf index e173e18..8bf7e6d 100644 Binary files a/PS2/doc.pdf and b/PS2/doc.pdf differ diff --git a/PS2/doc.synctex.gz b/PS2/doc.synctex.gz index abf72d6..4629e9c 100644 Binary files a/PS2/doc.synctex.gz and b/PS2/doc.synctex.gz differ diff --git a/PS2/doc.tex b/PS2/doc.tex index 246abdd..9192714 100644 --- a/PS2/doc.tex +++ b/PS2/doc.tex @@ -1,6 +1,7 @@ \documentclass{article} \usepackage{graphicx} +\usepackage{caption} \usepackage{setspace} \usepackage{listings} \usepackage{color} @@ -22,6 +23,8 @@ \titleformat{\subsection}[runin]{\normalfont \large \bfseries} {\thesubsection}{1em}{} +\captionsetup[figure]{width=0.75\textwidth} + \renewcommand{\thesubsection}{\indent(\alph{subsection})} \definecolor{dkgreen}{rgb}{0,0.6,0} @@ -175,10 +178,10 @@ legend({'Numerical','Analytical'},'Location','northwest'); \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} @@ -196,41 +199,42 @@ legend({'Numerical','Analytical'},'Location','northwest'); \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*} - \boxed{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*} @@ -242,7 +246,7 @@ 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*} \boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.} \end{equation*} @@ -254,25 +258,25 @@ 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 $\phi _\ell = Bsin \left( \frac {n \pi}{L} a \ell \right)$, 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 - $sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$ and - $sin(a + B) = sin(a)cos(B) + cos(a)sin(B)$, respectively, where + $\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$ 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) + -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 $\phi _ \ell = B sin \left( \frac {n \pi}{L} a \ell \right)$, + 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)} + \boxed{E = 2t_0 \left( 1 - \cos{\left( \frac{n \pi a}{L} \right)} \right).} \label{eq:numerical_E} \end{equation} @@ -282,7 +286,7 @@ legend({'Numerical','Analytical'},'Location','northwest'); \centering \adjustbox{valign=t}{ - \includegraphics[width=\textwidth]{q1cii.png} + \includegraphics[width=0.5\textwidth]{q1cii.png} } \captionof{figure}{Analytic solution to numeric system, plotted.} \end{minipage} @@ -292,24 +296,23 @@ legend({'Numerical','Analytical'},'Location','northwest'); \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 a^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 analytical expression for $E$: + 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 - 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 + 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} @@ -319,13 +322,16 @@ legend({'Numerical','Analytical'},'Location','northwest'); \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} + + \caption{(a) Probability densities for \(n=1\) and \(n=50\). (b) Comparison of first 101 numerical and analytic eigenvalues.} + \end{figure} \end{minipage} @@ -372,7 +378,7 @@ H(N, 1) = -t0; \begin{minipage}[t]{\linewidth} \centering \adjustbox{valign=t}{ - \includegraphics[width=\textwidth]{q1div.jpg} + \includegraphics[width=0.5\textwidth]{q1div.jpg} } \captionof{figure}{Sketch of degenerate energy levels.} \end{minipage} @@ -380,8 +386,8 @@ H(N, 1) = -t0; \item %1div 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 - eigenvalue numbers $n = 0, 1, and 2$, respectively. These match very closely, to within + 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). @@ -412,6 +418,7 @@ H(N, 1) = -t0; clear all; %physical constants in MKS units + hbar = 1.054e-34; q = 1.602e-19; m = 9.110e-31; @@ -475,13 +482,14 @@ axis([0 1e-9 0 0.04]); \begin{subfigure}{0.5\textwidth} \centering \includegraphics[width=\textwidth]{q2c_fig1_1s.png} - \caption{1s probability density.} + \caption{} \end{subfigure}% \begin{subfigure}{0.5\textwidth} \centering \includegraphics[width=\textwidth]{q2c_fig2_2s.png} - \caption{2s probability density.} + \caption{} \end{subfigure} + \caption{(a) 1s probability density. (b) 2s probability density.} \end{figure} \end{minipage} \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). \begin{minipage}[t]{\linewidth} \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).} \end{minipage}