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412866 words of font info for 64 fonts, out of 3000000 for 9000 - 1141 hyphenation exceptions out of 8191 - 68i,12n,75p,1278b,2408s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on d:/Users\Speedee\Documents\University\Year5\Winter\ECE456\ProblemSets\Git\ECE_456\PS2\doc.pdf (10 pages, 346266 bytes). -PDF statistics: - 112 PDF objects out of 1000 (max. 8388607) - 0 named destinations out of 1000 (max. 500000) - 36 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/PS2/doc.pdf b/PS2/doc.pdf index 03785e4..3997640 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 a2fffc4..5562864 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 98ce2a1..64faed6 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} @@ -16,10 +17,16 @@ \usepackage{lipsum} \usepackage{enumitem} \usepackage{cancel} +\usepackage{siunitx} +\usepackage{adjustbox} \titleformat{\subsection}[runin]{\normalfont \large \bfseries} {\thesubsection}{1em}{} +\captionsetup[figure]{width=0.75\textwidth,labelfont=normalfont,font=it,labelsep=period} + +\DeclareMathOperator*{\sinc}{sinc} +\DeclareMathOperator*{\rect}{rect} \renewcommand{\thesubsection}{\indent(\alph{subsection})} \definecolor{dkgreen}{rgb}{0,0.6,0} @@ -57,6 +64,8 @@ % } +\newcommand{\angstrom}{\textup{\AA}} + \title{ECE 456 - Problem Set 2} \date{2021-03-01} \author{David Lenfesty \\ lenfesty@ualberta.ca @@ -76,7 +85,7 @@ \thispagestyle{empty} \singlespacing \newpage - hello + \section*{Problem 1} \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] @@ -150,28 +159,33 @@ legend({'Numerical','Analytical'},'Location','northwest'); \end{lstlisting} % + \begin{minipage}[t]{\linewidth} + \begin{figure}[H] \centering - \begin{subfigure}{0.4\linewidth} + \begin{subfigure}{0.5\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} + \begin{subfigure}{0.5\linewidth} \centering \includegraphics[width=\textwidth]{q1a_eigenvals.png} - \caption{Comparison of first 101 numerical and analytic eigenvalues.} + \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 \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} @@ -185,45 +199,46 @@ 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} - 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*} - 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*} @@ -235,9 +250,9 @@ 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*} - B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A. + \boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.} \end{equation*} \end{enumerate} @@ -247,76 +262,83 @@ 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 = 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 - $sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$ 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) + \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} - \begin{figure}[H] \centering - \includegraphics[width=\textwidth]{q1cii.png} - \caption{Analytic solution to numeric system, plotted.} - \end{figure} + \adjustbox{valign=t}{ + \includegraphics[width=0.5\textwidth]{q1cii.png} + } + \captionof{figure}{Analytic solution to numeric system, plotted.} + \end{minipage} We can see here that the "predicted" numerical response matches nearly exactly the actual calculated numerical solution. \item %1ciii - Applying the approximation $cos(\Theta) = 1 - \frac{\Theta^2}{2}$ 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 \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 $E$: + We can get our final analytical expression for $E$ by fully substituting the explicit form of $t_0$: \begin{equation*} - 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*} \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}[t]{\linewidth} + \begin{figure}[H] \centering \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} - \end{figure} - % TODO idk if this is enough - 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 - plot in section (a), which has a low-frequency envelope around it. + \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 @@ -325,104 +347,359 @@ legend({'Numerical','Analytical'},'Location','northwest'); 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. + 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{Probability densities for \(n=4\)\\and \(n=5\).} + \caption{} \end{subfigure}% \begin{subfigure}{0.5\textwidth} \centering \includegraphics[width=\textwidth]{q1di_fig2.png} - \caption{Comparison of first 101 numerical and analytic eigenvalues.} + \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 $0.06eV$. - Because these eigenstates both have the same eigenvalue (0.06). + 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 - % TODO add figure for sketch + \begin{minipage}[t]{\linewidth} + \centering + \adjustbox{valign=t}{ + \includegraphics[width=0.5\textwidth]{q1div.jpg} + } + \captionof{figure}{Sketch of degenerate energy levels.} + \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). \end{enumerate} \end{enumerate} - + \newpage \section*{Problem 2} \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \item %2a - \lipsum[1] + 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 - \lipsum[1] - \item %1c - \lipsum[1] - \item %1d - \lipsum[1] - \item %1e - \lipsum[1] - \item %1f - \lipsum[1] + Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be \boxed{0}. + \item %2c + Code: + \begin{lstlisting} +clear all; +%physical constants in MKS units + + +hbar = 1.054e-34; +q = 1.602e-19; +m = 9.110e-31; +epsilon_0 = 8.854e-12; + +%generate lattice + +N = 100; %number of lattice points +n = [1:N]; %lattice points +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} + \begin{minipage}[t]{\linewidth} + + \begin{figure}[H] + \centering + \begin{subfigure}{0.5\textwidth} + \centering + \includegraphics[width=\textwidth]{q2c_fig1_1s.png} + \caption{} + \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} + \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=0.5\textwidth]{q2f_fig.png} + \captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).} + \end{minipage} + + \end{enumerate} - + + \newpage \section*{Problem 3} \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] - \item %2a - \lipsum[1] - \item %2b + \item %3a + + \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) + \end{equation} + + \item %3b + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] - \item %1di - \lipsum[1] - \item %1dii - \lipsum[1] - \item %1diii - \lipsum[1] - \item %1div - \lipsum[1] - \item %1dv - \lipsum[1] - \item %1dvi - \lipsum[1] - \item %1dvii - \lipsum[1] + \item %3bi + + Mapping the provided Fourier identities from \( t \) and \( \omega \) onto + \(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\): + + \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\}. + \end{align*} + + \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*} + \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*} + + + \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 the normalizing portion of the integral. + + % TODO just replace these in the formulas themselves + As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \) + and \( B = \frac{L}{2 \pi \hbar} \). + + \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*} + + Given property (26) of the \( sinc \) function, we can simplify the left and right elements. % TODO better word than elements + Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central element. + + \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] \\ + \end{align} + + \item %3bv + + + \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 wavefunction 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 + \(v = \pm \SI{3.6e4}{m/s}\). + + \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 %1c - \lipsum[1] - \item %1d - \lipsum[1] + + \item %3c + Because the probability density is even about \(p' = 0\), we can surmise that \(\left\langle p'\right\rangle = 0\). + \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) = 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 \( p' = \hbar k' \). + + \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'} + \end{equation*} \end{enumerate} - + \newpage \section*{Problem 4} \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] - \item %2a - \lipsum[1] - \item %2b + \item %4a + + % TODO I really don't like this explanation... + An \( a \) value of \( 0.62 \angstrom \) was chosen, in order to provide an adequately + shaped graph without sacrificing too much computation time. + + % maybe talk about normalization? + + \item %4b \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] - \item %1di - \lipsum[1] - \item %1dii - \lipsum[1] - \item %1diii + \item %4bi \lipsum[1] + \item %4bii + + + + \item %4biii + + Negative Drain Resistance (NDR) is present in this design from a \(V_D\) of + approximately \(0.27V\) to \(0.45V\), with a mostly linear region from + aroung \(0.28V\) to \(0.3V\). + \end{enumerate} \end{enumerate} diff --git a/PS2/problem_3_part_b_v.m b/PS2/problem_3_part_b_v.m new file mode 100644 index 0000000..90d2e05 --- /dev/null +++ b/PS2/problem_3_part_b_v.m @@ -0,0 +1,49 @@ +% A simple MATLAB code to plot the momentum +% wave function for a particle in an escape-proof "box" + +clear all + +hbar = 1.054e-34; +L = 101e-10; + +% Set up k' and p' axes + +k_prime = linspace(-2*pi/L,2*pi/L,100); +dk_prime = k_prime(2) - k_prime(1); +p_prime = hbar * k_prime; +dp_prime = p_prime(2) - p_prime(1); + +% k1 and p1 + +k1 = (pi/L); p1 = hbar * k1; + +% Fourier transform A(k') + +A1 = 0.5*sqrt(L/pi)*( sinc((k_prime+k1)*L/(2*pi)) ); +A2 = 0.5*sqrt(L/pi)*( sinc((k_prime-k1)*L/(2*pi)) ); +A = A1 + A2; +A_sum = sum(abs(A).^2)*dk_prime; + +% Momentum wave function Phi(p') + +Phi1 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime+p1)*L/(2*pi*hbar)) ); +Phi2 = 0.5*sqrt(L/(pi*hbar))*( sinc((p_prime-p1)*L/(2*pi*hbar)) ); +Phi = Phi1 + Phi2; +Phi_sum = sum(abs(Phi).^2)*dp_prime; + +% Normalized p' axis for plotting purposes; the points of classical +% momenta occur when the variable pp_N is plus or minus unity + +pp_N = p_prime / p1; + +figure(1); clf; +h = plot(pp_N,Phi1,'kx',pp_N,Phi2,'ko',pp_N,Phi,'k--'); +grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]'); +ylabel('WAVE FUNCTION [sqrt(s / kg m)]'); +legend('Phi1','Phi2','Phi'); + +figure(2); clf; h = plot(pp_N,abs(Phi).^2,'k-'); +grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('NORMALIZED MOMENTUM [p^{\prime}/p_1]'); +ylabel('PROBABILITY DENSITY [s / kg m]'); diff --git a/PS2/q1div.jpg b/PS2/q1div.jpg index d366334..c2a9ad5 100644 Binary files a/PS2/q1div.jpg and b/PS2/q1div.jpg differ diff --git a/PS2/q1a.asv b/PS2/q2c.m similarity index 63% rename from PS2/q1a.asv rename to PS2/q2c.m index 094029d..d1e40d1 100644 --- a/PS2/q1a.asv +++ b/PS2/q2c.m @@ -4,19 +4,20 @@ clear all; hbar = 1.054e-34; q = 1.602e-19; m = 9.110e-31; +epsilon_0 = 8.854e-12; %generate lattice N = 100; %number of lattice points n = [1:N]; %lattice points -a = 1e-10; %lattice constant -x = a * n; %x-coordinates +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 = 0*x; %0 potential at all x +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 @@ -33,25 +34,25 @@ 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_50 = eigenvectors(:,50); +phi_2 = eigenvectors(:,2); % 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); +P_2 = phi_2 .* conj(phi_2); -% Find first N analytic eigenvalues -E_col_analytic = (1/q) * (hbar^2 * pi^2 * n.*n) / ( -% Plot the probability densities for 1st and 50th eigenvectors +% Plot the probability densities for 1st and 2nd eigenvectors -figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'k-'); +figure(1); clf; h = plot(r,P_1,'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]); +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]); \ No newline at end of file diff --git a/PS2/q2c_fig1_1s.png b/PS2/q2c_fig1_1s.png new file mode 100644 index 0000000..c71b686 Binary files /dev/null and b/PS2/q2c_fig1_1s.png differ diff --git a/PS2/q2c_fig2_2s.png b/PS2/q2c_fig2_2s.png new file mode 100644 index 0000000..6584036 Binary files /dev/null and b/PS2/q2c_fig2_2s.png differ diff --git a/PS2/q2f_fig.png b/PS2/q2f_fig.png new file mode 100644 index 0000000..894d4e3 Binary files /dev/null and b/PS2/q2f_fig.png differ diff --git a/PS2/q3bv_fig1.png b/PS2/q3bv_fig1.png new file mode 100644 index 0000000..e3a93ce Binary files /dev/null and b/PS2/q3bv_fig1.png differ diff --git a/PS2/q3bv_fig2.png b/PS2/q3bv_fig2.png new file mode 100644 index 0000000..e868488 Binary files /dev/null and b/PS2/q3bv_fig2.png differ diff --git a/PS2/q3bvi_fig1.png b/PS2/q3bvi_fig1.png new file mode 100644 index 0000000..a4a7614 Binary files /dev/null and b/PS2/q3bvi_fig1.png differ diff --git a/PS2/q3bvi_fig2.png b/PS2/q3bvi_fig2.png new file mode 100644 index 0000000..551b314 Binary files /dev/null and b/PS2/q3bvi_fig2.png differ diff --git a/PS2/q4a.m b/PS2/q4a.m new file mode 100644 index 0000000..3909d57 --- /dev/null +++ b/PS2/q4a.m @@ -0,0 +1,74 @@ +clear all; + +% Physical constants in MKS units, including free-electron mass m + +hbar = 1.054e-34; +q = 1.602e-19; +m = 9.110e-31; + +% SHO frequency + +w0 = (0.2879*q/hbar); + +% Lattice data: lattice constant "a"; SHO distance parameter beta; +% a discretized spatial grid x; total number of lattice points N; +% Hamiltonian parameter t0, as introduced in class, but now +% divided by q so that it has units of eV; total length of the lattice, +% L = a*(N+1) + +% YOU MUST ENTER AN APPROPRIATE VALUE OF "a" + +a = 12 / sqrt(m*w0/hbar) / 100; +beta = sqrt(m*w0/hbar); +x = [-6/beta:a:6/beta]; +N = length(x); +t0 = (hbar^2)/(2*m*a^2)/q; +L = a*(N+1); + +% Set up the Hamiltonian matrix, in pieces, by constructing square +% matrices containing information for the main +% diagonal, lower diagonal, and upper diagonal, and then adding them +% together; you may need to look up the diag command using help +% to see how this works; U is the potential energy + +U = (1/q)*(0.5*m*w0^2*x.^2); +main_D = diag(2*t0*ones(1,N)+U,0); +lower_D = diag(-t0*ones(1,N-1),-1); +upper_D = diag(-t0*ones(1,N-1),+1); + +H = main_D + lower_D + upper_D; + +% Find the eigenvectors and eigenvalues of H; V will contain the +% eigenvectors in its columns, and D will contain the corresponding +% eigenvalues on its main diagonal + +[V,D] = eig(H); + +% Find the numerical eigenvalues + +E_numerical = diag(D,0); + +% Find the eigenvectors corresponding to sorted eigenvalues 1 and 50; +% we can get these from the matrix V, provided we reference the +% appropriate UNSORTED eigenvalue number + +phi_numerical_1 = V(:,1); +phi_numerical_2 = V(:,2); + +% Find the position-probability densities + +P_numerical_1 = phi_numerical_1 .* conj(phi_numerical_1); +P_numerical_2 = phi_numerical_2 .* conj(phi_numerical_2); + +% Plot out the relevant results + +figure(1); h = plot(x,P_numerical_1,'k-'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]'); +legend('Eigenvector 1'); + +figure(2); h = plot(x,P_numerical_2,'k-'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]'); +legend('Eigenvector 2'); + diff --git a/PS2/q4b_main.m b/PS2/q4b_main.m new file mode 100644 index 0000000..1557ba3 --- /dev/null +++ b/PS2/q4b_main.m @@ -0,0 +1,145 @@ +clear all; + +% Physical constants in MKS units + +hbar = 1.054e-34; +q = 1.602e-19; + +% Energy parameters in eV; included are the single-electron charging +% energy U0; the kBT product; the equilibrium Fermi level mu; and the +% energy levels cal_E1 and cal_E2, where "cal_E" is short-form for "calligraphic E" + +% Please especially note that ALL ENERGY VARIABLES IN THIS CODE +% ARE IN eV (NOT joules); the equations from class must be adjusted +% accordingly, multiplying or dividing appropriate terms by a factor of q + +% YOU MUST ENTER THE APPROPRIATE VALUES OF cal_E1 and cal_E2 + +U0 = 0.025; +kBT = 0.025; +mu = 0; +cal_E1 = 0.2879 / 2; +cal_E2 = 0.2879 * 3 / 2; + +% Capacitance parameters + +alpha_G = 0.5; +alpha_D = 0.5; +alpha_S = 1 - alpha_G - alpha_D; + +% Energy grid in eV, from -1 eV to 1 eV + +NE = 501; +E = linspace(-1,1,NE); +dE = E(2) - E(1); + +% Coupling coefficients, which are now grids over E, with +% gamma_1 equal to 0.005 eV for E > 0, and 0 for E <= 0, and +% with gamma_2 equal to 0.005 eV for all E; the 1e-6 +% term is included to avoid divide-by-zero errors + +gamma_1 = 0.005*(E + abs(E)) ./ (E + E + 1e-6); +gamma_2 = 0.005*ones(1,NE); +gamma = gamma_1 + gamma_2; + +% Reference number of electrons in the channel, assumed to be zero in +% this code + +N0 = 0; + +% Voltage values to consider for the final plots + +NV = 121; +VV = linspace(-0.4,0.8,NV); +dV = VV(2) - VV(1); + +% Loop over voltage values and compute number of electrons and current +% for each voltage value in a self-consistent manner + +for count = 1:NV + + % Set terminal voltages + + VG = 0; + VD = VV(count); + VS = 0; + + % Values of mu1 and mu2; notice that the usual factor of q multiplying + % the voltages is omitted, because in this code, energy is in eV + + mu1 = mu - VS; + mu2 = mu - VD; + + % Compute source and drain Fermi functions + + f1 = 1./(1+exp((E - mu1)./kBT)); + f2 = 1./(1+exp((E - mu2)./kBT)); + + % Value of Laplace potential in eV + + UL = - (alpha_G*VG) - (alpha_D*VD) - (alpha_S*VS); + + % Initial value of Poisson part in eV + + UP = 0; + + % Iterate until self-consistent potential is achieved by monitoring + % the Poisson part (the Laplace part does not change) + + dUP = 1; + while dUP > 1e-6 + + % Lorentzian SHIFTED density of states for levels 1 and 2, + % each normalized so that its integral is unity + + D1 = (0.01/(2*pi))./((E - (UL + UP) - cal_E1).^2+(0.01/2)^2); + D1 = D1./(dE*sum(D1)); + + D2 = (0.01/(2*pi))./((E - (UL + UP) - cal_E2).^2+(0.01/2)^2); + D2 = D2./(dE*sum(D2)); + + % Total density of states + + D = D1 + D2; + + % Compute number of channel electrons + + N(count) = dE*sum( ((gamma_1./gamma).*f1 + (gamma_2./gamma).*f2).*D ); + + % Newly calculated Poisson part of self-consistent potential + + UPnew = U0*( N(count) - N0 ); + + % Change in Poisson part between iterations + + dUP = abs(UP - UPnew); + + % New guess for next iteration, found by adding a fraction of the + % difference between iterations to the old guess + + UP = UP + 0.1*(UPnew - UP); + + end + + % Compute the current in A after the self-consistent potential + % has been achieved; notice the extra factor of q preceding the + % equation, which is needed since the gammas are in eV + + I(count) = q*(q/hbar)... + *dE*sum((f1-f2).*D.*gamma_1.*gamma_2./gamma); + + run q4b_subplots; + +end + +% Plotting commands, including lines to modify the linewidth +% and Fontsize, just to make the plots look nicer; you don't +% need to worry about how these work + +figure(1); h = plot(VV,N,'k'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('DRAIN VOLTAGE [V]'); ylabel('NUMBER OF ELECTRONS'); + +figure(2); h = plot(VV,I,'k'); grid on; +set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]); +xlabel('DRAIN VOLTAGE [V]'); ylabel('CURRENT [A]'); diff --git a/PS2/q4b_subplots.m b/PS2/q4b_subplots.m new file mode 100644 index 0000000..bf9d930 --- /dev/null +++ b/PS2/q4b_subplots.m @@ -0,0 +1,113 @@ +% Code to create the required subplots for Problem Set 2, Problem 5(b) + +% If you've used the same variable names as in the sample code, then you +% should be able to simply insert this into the appropriate spot in your +% own code; otherwise, you'll have to modify this accordingly, which +% should be easy to do---if disaster strikes and it doesn't work, then +% please ask for help + +% The "if" statement is used to choose VD values closest to the required +% values of 0.0, 0.2, 0.3, ..., 0.8 V, and you don't need to +% worry about how this works + +if (abs(VD-0.0) <= dV/2) + figure(3); title('VD = 0.0 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.0 V'); +elseif (abs(VD-0.2) <= dV/2) + figure(4); title('VD = 0.2 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.2 V'); +elseif (abs(VD-0.25) <= dV/2) + figure(5); title('VD = 0.25 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.25 V'); +elseif (abs(VD-0.3) <= dV/2) + figure(6); title('VD = 0.3 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.3 V'); +elseif (abs(VD-0.4) <= dV/2) + figure(7); title('VD = 0.4 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.4 V'); +elseif (abs(VD-0.5) <= dV/2) + figure(8); title('VD = 0.5 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.5 V'); +elseif (abs(VD-0.6) <= dV/2) + figure(9); title('VD = 0.6 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.6 V'); +elseif (abs(VD-0.65) <= dV/2) + figure(10); title('VD = 0.65 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.65 V'); +elseif (abs(VD-0.7) <= dV/2) + figure(11); title('VD = 0.7 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.7 V'); +elseif (abs(VD-0.8) <= dV/2) + figure(12); title('VD = 0.8 V'); + subplot(2,3,1); plot(f1,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,2); plot(D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,3); plot(f2,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f2(E)'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); + subplot(2,3,5); plot(f1-f2,E,'--',D/100,E,'k-'); axis([-0.1 1.1 -1 1]); + xlabel('f1(E)-f2(E), D(E-U)/100'); ylabel('ENERGY [eV]'); title('VD = 0.8 V'); +end diff --git a/PS2/sinc.m b/PS2/sinc.m new file mode 100644 index 0000000..8758064 --- /dev/null +++ b/PS2/sinc.m @@ -0,0 +1,5 @@ +function out = sinc(x) +% Deal with the removable singularity at 0 explicitly. +out = sin(pi*x)./(pi*x); +out(x == 0) = 1; +end \ No newline at end of file