759 lines
33 KiB
TeX
759 lines
33 KiB
TeX
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\newcommand{\angstrom}{\textup{\AA}}
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\title{ECE 456 - Problem Set 2}
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\date{2021-03-08}
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\author{David Lenfesty \\ lenfesty@ualberta.ca
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\and Phillip Kirwin \\ pkirwin@ualberta.ca}
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\pagestyle{fancy}
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\fancyhead[L]{\textbf{ECE 456} - Problem Set 2}
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\fancyhead[R]{David Lenfesty and Phillip Kirwin}
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\fancyfoot[C]{Page \thepage}
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\renewcommand{\headrulewidth}{1pt}
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\renewcommand{\footrulewidth}{1pt}
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\begin{document}
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\doublespacing
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\maketitle
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\thispagestyle{empty}
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\singlespacing
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\newpage
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\section*{Problem 1}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %1a
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Code:
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\begin{lstlisting}[language=Matlab]
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clear all;
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%physical constants in MKS units
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hbar = 1.054e-34;
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q = 1.602e-19;
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m = 9.110e-31;
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%generate lattice
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N = 100; %number of lattice points
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n = [1:N]; %lattice points
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a = 1e-10; %lattice constant
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x = a * n; %x-coordinates
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t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor
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L = a * (N+1); %total length of consideration
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%set up Hamiltonian matrix
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U = 0*x; %0 potential at all x
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main_diag = diag(2*t0*ones(1,N)+U,0); %create main diagonal matrix
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lower_diag = diag(-t0*ones(1,N-1),-1); %create lower diagonal matrix
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upper_diag = diag(-t0*ones(1,N-1),+1); %create upper diagonal matrix
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H = main_diag + lower_diag + upper_diag; %sum to get Hamiltonian matrix
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[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each
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%column is an eigenvector
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%"E_diag" is a diagonal matrix where the
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%corresponding eigenvalues are on the
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%diagonal.
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E_col = diag(E_diag); %folds E_diag into a column vector of eigenvalues
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% return eigenvectors for the 1st and 50th eigenvalues
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phi_1 = eigenvectors(:,1);
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phi_50 = eigenvectors(:,50);
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% find the probability densities of position for 1st and 50th eigenvectors
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P_1 = phi_1 .* conj(phi_1);
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P_50 = phi_50 .* conj(phi_50);
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% Find first N analytic eigenvalues
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E_col_analytic = (1/q) * (hbar^2 * pi^2 * n.*n) / (2*m*L^2);
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% Plot the probability densities for 1st and 50th eigenvectors
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figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'k-');
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grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
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xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
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legend('n=1','n=50');
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% Plot numerical eigenvalues
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figure(2); clf; h = plot(n,E_col,'kx'); grid on;
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set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
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xlabel('EIGENVALUE NUMBER'); ylabel('ENERGY [eV]');
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axis([0 100 0 40]);
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% Add analytic eigenvalues to above plot
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hold on;
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plot(n,E_col_analytic,'k-');
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legend({'Numerical','Analytical'},'Location','northwest');
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\end{lstlisting}
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%
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\begin{minipage}[t]{\linewidth}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q1a_1and50.png}
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\caption{}
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\label{fig:q3_ne}
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\end{subfigure}%
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\begin{subfigure}{0.5\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q1a_eigenvals.png}
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\caption{}
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\label{fig:q3_current}
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\end{subfigure}
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\caption{(a) Probability densities for \(n=1\) and \(n=50\).
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(b) Comparison of first 101 numerical and analytic eigenvalues.}
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\end{figure}
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\end{minipage}
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\item %1b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1bi
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The analytical solution is:
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\begin{equation}
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\phi(x) = A \sin{\left( \frac{n \pi}{L} x \right)}.
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\label{eq:analytical}
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\end{equation}
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In order to normalise this equation it must conform to the following:
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\begin{equation}
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\int_{0}^{L} \left| \phi(x) \right| ^2 dx = 1.
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\label{eq:normalise}
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\end{equation}
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We use the following identity:
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\begin{equation}
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\int\sin^2{(a x)}\:dx = \frac{1}{2} x -\frac{1}{4a}\sin{(2a x)}.
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\label{eq:integral_ident}
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\end{equation}
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Given that the sine of a real value is always real, we can disregard the norm operation, and directly relate (\ref{eq:analytical})
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to the above identity.
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Evaluating the integral gives us the following relationship:
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\begin{equation*}
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\frac{1}{A^2} = \frac{1}{2} L - \cancel{\frac{L}{4n \pi} \sin{\left( \frac{2n \pi}{L} L \right)} }
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- \cancel{\frac{1}{2} \cdot 0} + \cancel{\frac{L}{4n \pi} \sin{(0)}}.
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\end{equation*}
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From this, we find:
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\begin{equation*}
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\boxed{A = \sqrt{\frac{2}{L}}.}
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\end{equation*}
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\item %1bii
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Starting with the normalization condition for the numerical case:
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\begin{align}
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a\sum_{\ell=1}^N\left|\phi_l\right|^2 &= a \\
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a\sum_{\ell=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_\ell\right)}\right|^2 &= a, \nonumber
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\label{eq:numerical_norm}
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\end{align}
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recalling that \(x = a\ell\), and allowing \(a \to 0\), while holding \(L\)
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constant, implies that \(N \to \infty\), since \(a=\frac{L}{N}\).
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An integral is defined as the limit of a Riemann sum as follows:
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\begin{equation}
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\int_c^df(x)\:dx \equiv \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i),
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\label{eq:Riemann_sum}
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\end{equation}
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where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case,
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\(n = N\), \(i = \ell\), \(c = 0\), \(d = L\), and \(\Delta x = a\), \(x_i = x_\ell\),
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\(f(x) = \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\). Therefore we can write
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\begin{equation*}
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\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
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= \lim_{N \to \infty}\sum_{\ell=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2
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= a.
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\end{equation*}
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Using (\ref{eq:integral_ident}), we have
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\begin{equation*}
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\int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx
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= \frac{1}{2}L - \cancel{\frac{L}{4n\pi}\sin{\left(\frac{2n\pi}{L}L\right)}}
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- 0 + 0
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= \frac{a}{B^2}.
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\end{equation*}
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This means that \(B\) must be
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\begin{equation*}
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\boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.}
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\end{equation*}
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\end{enumerate}
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\item %1c
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1ci
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% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
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From the base form of $\phi _\ell = B\sin{\left( \frac {n \pi}{L} a \ell \right)} $, we can see that
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$\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
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$\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$ and
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$\sin{(a + B)} = \sin{(a)}\cos{(B)} + \cos{(a)}\sin{(B)}$, respectively, where
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$a = \frac{n \pi a \ell}{L}$ and $B = \frac{n \pi a}{L}$.
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Plugging these identities into equation (7) from the assignment and simplifying, we get to this equation:
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\begin{equation*}
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-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)}.
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\end{equation*}
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At this point, we notice that $\phi _ \ell = B \sin{\left( \frac {n \pi}{L} a \ell \right)} $,
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so we can factor it out.
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With some minor rearranging, this leaves us with the final expression for $E$:
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\begin{equation}
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\boxed{E = 2t_0 \left( 1 - \cos{\left( \frac{n \pi a}{L} \right)} \right).}
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\label{eq:numerical_E}
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\end{equation}
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\item %1cii
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\begin{minipage}[t]{\linewidth}
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\centering
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\adjustbox{valign=t}{
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\includegraphics[width=0.5\textwidth]{q1cii.png}
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}
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\captionof{figure}{Comparison between analytical result and numerical result. Above \(n=50\), the results diverge substantially.}
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\end{minipage}
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We can see here that the "predicted" numerical response matches nearly exactly the actual
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calculated numerical solution.
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\item %1ciii
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Applying the approximation $\cos{(\theta)} \approx 1 - \frac{\theta^2}{2}$ for small $\theta$
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on equation (\ref{eq:numerical_E}), we get the following expression:
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\begin{equation*}
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E = 2 t_0 \left( \frac{n^2 \pi^2 a^2}{2 L^2} \right).
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\end{equation*}
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We can get our final analytical expression for $E$ by fully substituting the explicit form of $t_0$:
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\begin{equation}
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\boxed{E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }.}
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\end{equation}
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\item %1civ
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With the decreased lattice spacing and increased number of points we can see the numerical solution
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more closely matches the analytical solution. As well, the \(n=50\) case is
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now a constant-amplitude wave, which corresponds to the expected analytic result, in contrast to the
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plot in section (a), which has a low-frequency envelope around it.
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\begin{minipage}[b]{\linewidth}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1civ_fig1.png}
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\caption{}
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1civ_fig2.png}
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\caption{}
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\end{subfigure}
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\caption{(a) Probability densities for \(n=1\) and \(n=50\). (b) Comparison of first 101 numerical and analytic eigenvalues.}
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\end{figure}
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\end{minipage}
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\end{enumerate}
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\item %1d
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1di
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In order to modify the computations to those for a particle in a "ring"
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we simply had to add $-t_0$ elements as the "corner" elements of the
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hamiltonian operator array:
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\begin{lstlisting}[language=Matlab]
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% Modify hamiltonian for circular boundary conditions
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H(1, N) = -t0;
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H(N, 1) = -t0;
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\end{lstlisting}
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\begin{minipage}[t]{\linewidth}
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1di_fig1.png}
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\caption{}
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\end{subfigure}%
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=\textwidth]{q1di_fig2.png}
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\caption{}
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\end{subfigure}
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\caption{(a) Probability densities for \(n=4\) and \(n=5\). (b) Comparison of first 101 numerical and analytic eigenvalues.}
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\end{figure}
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\end{minipage}
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\item %1dii
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The energy levels for eigenvalues number 4 and 5 are both \boxed{0.06\:\mathrm{eV}}.
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These eigenstates are degenerate because they both have the same
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eigenvalue/energy.
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\item %1diii
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\begin{minipage}[t]{\linewidth}
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\centering
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\adjustbox{valign=t}{
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\includegraphics[width=0.5\textwidth]{q1div.jpg}
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}
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\captionof{figure}{Sketch of degenerate energy levels. In the sketch, closely-spaced
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levels are in fact degenerate.}
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\end{minipage}
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\item %1div
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Plugging the valid levels for $n$ into equation (10) from the assignment (and dividing
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by the requisite $q$), we get energy levels of \(0\) eV, \(0.0147\) eV, and \(0.0589\) eV for the
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indices \(n=0\), \(1\), and \(2\), respectively. These match within an
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acceptable margin to the numerical results from part (ii).
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\end{enumerate}
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\end{enumerate}
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\newpage
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\section*{Problem 2}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %2a
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Since \(t_0 = \frac{\hbar^2}{2ma^2}\), and \(U_l = -\frac{q^2}{4\pi\varepsilon_0r_l}
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+ \frac{l_o(l_o + 1)\hbar^2}{2mr_l^2}\), the middle diagonal elements will have values
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\begin{equation*}
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\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},}
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\end{equation*}
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and the upper and lower diagonals elements will have values
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\begin{equation*}
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\boxed{\hat{H}_{l(l\pm1)} = -\frac{\hbar^2}{2ma^2}.}
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\end{equation*}
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\item %2b
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Homogenous boundary conditions imply that the corner entries of \(\hat{H}\) will be \boxed{0}.
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\item %2c
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Code:
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\begin{lstlisting}
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clear all;
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%physical constants in MKS units
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hbar = 1.054e-34;
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q = 1.602e-19;
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m = 9.110e-31;
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epsilon_0 = 8.854e-12;
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%generate lattice
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N = 100; %number of lattice points
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n = [1:N]; %lattice points
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a = 0.1e-10; %lattice constant
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r = a * n; %x-coordinates
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t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor
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L = a * (N+1); %total length of consideration
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%set up Hamiltonian matrix
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U = -q^2./(4*pi*epsilon_0.*r) * (1/q); %potential at r in [eV]
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main_diag = diag(2*t0*ones(1,N)+U,0); %create main diagonal matrix
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lower_diag = diag(-t0*ones(1,N-1),-1); %create lower diagonal matrix
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upper_diag = diag(-t0*ones(1,N-1),+1); %create upper diagonal matrix
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H = main_diag + lower_diag + upper_diag; %sum to get Hamiltonian matrix
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[eigenvectors,E_diag] = eig(H); %"eigenvectors" is a matrix wherein each column is an eigenvector
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%"E_diag" is a diagonal matrix where the
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%corresponding eigenvalues are on the
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%diagonal.
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E_col = diag(E_diag); %folds E_diag into a column vector of eigenvalues
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% return eigenvectors for the 1st and 50th eigenvalues
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phi_1 = eigenvectors(:,1);
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phi_2 = eigenvectors(:,2);
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% find the probability densities of position for 1st and 50th eigenvectors
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|
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=0.6\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 %3a
|
|
Recalling the identity
|
|
\begin{equation}
|
|
\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
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %3bi
|
|
|
|
Mapping the provided Fourier identities from \( t \) and \( \omega \) onto
|
|
\(x'\) 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{\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
|
|
\begin{equation*}
|
|
\Phi(p') \equiv \frac{1}{\sqrt{\hbar}} A \left(\frac{p'}{\hbar}\right),
|
|
\end{equation*}
|
|
we can obtain
|
|
\begin{equation*}
|
|
\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
|
|
\(\left|\Phi(p')\right|^2\) 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 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{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 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 &= 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
|
|
|
|
|
|
\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 wave function 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
|
|
\(\boxed{v = \pm \SI{3.6e4}{m/s}}\).
|
|
|
|
\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 %3c
|
|
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
|
|
\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) = \boxed{0},
|
|
\end{align*}
|
|
which verifies our above inference.
|
|
\item %3d
|
|
|
|
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'}
|
|
\end{equation*}
|
|
|
|
\end{enumerate}
|
|
\newpage
|
|
\section*{Problem 4}
|
|
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
|
|
|
|
\item %4a
|
|
\begin{minipage}[t]{\linewidth}
|
|
|
|
\begin{figure}[H]
|
|
\centering
|
|
\begin{subfigure}[b]{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q4a_e1.png}
|
|
\caption{}
|
|
\end{subfigure}%
|
|
\begin{subfigure}[b]{0.5\textwidth}
|
|
\centering
|
|
\includegraphics[width=\textwidth]{q4a_e2.png}
|
|
\caption{}
|
|
\end{subfigure}
|
|
\caption{Probability densities versus position for first two energy levels.}
|
|
\end{figure}
|
|
\end{minipage}
|
|
|
|
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 \(\SI{0.43140}{\electronvolt} \),
|
|
which are in agreement.
|
|
|
|
|
|
|
|
|
|
\item %4b
|
|
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
|
|
\item %4bi
|
|
|
|
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=0.5\textwidth]{q4bi_I-V.png}
|
|
\captionof{figure}{Current-voltage characteristic of a 2-level molecule.}
|
|
\end{minipage}
|
|
|
|
\item %4bii
|
|
|
|
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 \(\SI{0.4}{V}\) 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 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}
|
|
|
|
\end{enumerate}
|
|
|
|
|
|
\end{document} |