diff --git a/PS1/feedback.pdf b/PS1/feedback.pdf deleted file mode 100644 index 818f381..0000000 Binary files a/PS1/feedback.pdf and /dev/null differ diff --git a/PS3/doc.pdf b/PS3/doc.pdf new file mode 100644 index 0000000..fc911b2 Binary files /dev/null and b/PS3/doc.pdf differ diff --git a/PS3/doc.tex b/PS3/doc.tex index ee506e6..6fc32e5 100644 --- a/PS3/doc.tex +++ b/PS3/doc.tex @@ -66,14 +66,14 @@ \newcommand{\angstrom}{\textup{\AA}} -\title{ECE 456 - Problem Set 2} -\date{2021-03-08} +\title{ECE 456 - Problem Set 3 (Part 1)} +\date{2021-03-31} \author{David Lenfesty \\ lenfesty@ualberta.ca \and Phillip Kirwin \\ pkirwin@ualberta.ca} \pagestyle{fancy} -\fancyhead[L]{\textbf{ECE 456} - Problem Set 2} +\fancyhead[L]{\textbf{ECE 456} - Problem Set 3 (Part 1)} \fancyhead[R]{David Lenfesty and Phillip Kirwin} \fancyfoot[C]{Page \thepage} \renewcommand{\headrulewidth}{1pt} @@ -90,670 +90,225 @@ \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \item %1a - Code: - \begin{lstlisting}[language=Matlab] -clear all; -%physical constants in MKS units + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] + \item %1bi + The matrix equation is + \begin{equation*} + [\hat{H}]\left\{\phi\right\} = E \left\{\phi\right\}, + \end{equation*} + where \([\hat{H}]\) is an \(N\)-by-\(N\) matrix with \([\hat{H}]_{nm} = 0\) except for the following elements: + \begin{align*} + &[\hat{H}]_{nn} = 2t_0 + U_n \\ + &[\hat{H}]_{n,n \pm 1} = -t_0 \\ + &[\hat{H}]_{0,N} = [\hat{H}]_{N,0} = -t_0, + \end{align*} + with \(t_0 = \hbar^2/(2ma^2)\) and \(U_n = U(na)\). The \(N\)-vector \(\left\{\phi\right\}\) has elements \(\phi_n\) which each represent + the value of the eigenvector at the point \(na = x_n\). + \item %1bii -hbar = 1.054e-34; -q = 1.602e-19; -m = 9.110e-31; + The expression of the wave function \( \phi (x) \) as a sum of basis functions is as below: -%generate lattice + \begin{equation*} + \phi (x) = \sum _{n = 1} ^{N} \phi _n u_n(x) + \end{equation*} -N = 100; %number of lattice points -n = [1:N]; %lattice points -a = 1e-10; %lattice constant -x = a * n; %x-coordinates -t0 = (hbar^2)/(2*m*a^2)/q; %encapsulating factor -L = a * (N+1); %total length of consideration + The derived matrix equation: -%set up Hamiltonian matrix + \begin{equation*} + [\hat{H}] _u \{ \phi \} = [S]_u \{ \phi \} + \end{equation*} -U = 0*x; %0 potential at all x -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_50 = eigenvectors(:,50); - -% 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); - -% Find first N analytic eigenvalues -E_col_analytic = (1/q) * (hbar^2 * pi^2 * n.*n) / (2*m*L^2); - -% Plot the probability densities for 1st and 50th eigenvectors - -figure(1); clf; h = plot(x,P_1,'kx',x,P_50,'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]); - -% Add analytic eigenvalues to above plot - -hold on; -plot(n,E_col_analytic,'k-'); -legend({'Numerical','Analytical'},'Location','northwest'); - - \end{lstlisting} - % - \begin{minipage}[t]{\linewidth} + Where \( [\hat{H}]_u \) is a matrix with the elements: - \begin{figure}[H] - \centering - \begin{subfigure}{0.5\linewidth} - \centering - \includegraphics[width=\textwidth]{q1a_1and50.png} - \caption{} - \label{fig:q3_ne} - \end{subfigure}% - \begin{subfigure}{0.5\linewidth} - \centering - \includegraphics[width=\textwidth]{q1a_eigenvals.png} - \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} + \begin{equation*} + H_{nm} = \int u _n ^* (x) \hat{H} u_m(x) dx + \end{equation*} + + and \( [S]_u \) is a matrix with the elements: + + \begin{equation*} + S_{nm} = \int u _n ^* (x) u_m(x) dx + \end{equation*} + + \([\hat{H}]_u\) and \([S]_u\) are both of size \(N\)-by-\(N\). The elements + of \(\left\{\phi\right\}\), \(\phi_n\), are the expansion coefficients of \(\phi(x)\). + \end{enumerate} + \item %1b \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] \item %1bi - The analytical solution is: + code: + \begin{lstlisting} +%constants +E1 = -13.6; +R = 0.074; +a0 = 0.0529; - \begin{equation} - \phi(x) = A \sin{\left( \frac{n \pi}{L} x \right)}. - \label{eq:analytical} - \end{equation} +%matrix elements +R0 = R/a0; - In order to normalise this equation it must conform to the following: +a = 2*E1*(1-(1+R0)*exp(-2*R0))/R0; +b = 2*E1*(1+R0)*exp(-R0); +s = exp(-R0)*(1+R0+(R0^2/3)); - \begin{equation} - \int_{0}^{L} \left| \phi(x) \right| ^2 dx = 1. - \label{eq:normalise} - \end{equation} +%matrices +H_u = [E1 + a, E1*s+b; E1*s+b, E1 + a]; +S_u = [1, s; s, 1]; - We use the following identity: +%find eigenvalues and eigenvectors +[vectors,energies] = eig(inv(S_u)*H_u); - \begin{equation} - \int\sin^2{(a x)}\:dx = \frac{1}{2} x -\frac{1}{4a}\sin{(2a x)}. - \label{eq:integral_ident} - \end{equation} - - 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: - - \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)}}. - \end{equation*} - - From this, we find: - - \begin{equation*} - \boxed{A = \sqrt{\frac{2}{L}}.} - \end{equation*} - - \item %1bii - - Starting with the normalization condition for the numerical case: - \begin{align} - 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 \(x = a\ell\), 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), - \label{eq:Riemann_sum} - \end{equation} - where \(\Delta x = \frac{d-c}{n}\) and \(x_i = c + \Delta x \cdot i\). In our case, - \(n = N\), \(i = \ell\), \(c = 0\), \(d = L\), 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_{\ell=1}^N a \cdot \left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 - = a. - \end{equation*} - - Using (\ref{eq:integral_ident}), we have - \begin{equation*} - \int_0^L \left|B\sin{\left(\frac{n\pi}{L}x\right)}\right|^2\:dx - = \frac{1}{2}L - \cancel{\frac{L}{4n\pi}\sin{\left(\frac{2n\pi}{L}L\right)}} - - 0 + 0 - = \frac{a}{B^2}. - \end{equation*} - - This means that \(B\) must be - \begin{equation*} - \boxed{B = \sqrt{\frac{2 a}{L}} = \sqrt{a} \times A.} - \end{equation*} - - \end{enumerate} - - \item %1c - \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] - \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 = 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 - $a = \frac{n \pi a \ell}{L}$ 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)}. - \end{equation*} - - 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).} - \label{eq:numerical_E} - \end{equation} - - \item %1cii - - \begin{minipage}[t]{\linewidth} - - \centering - \adjustbox{valign=t}{ - \includegraphics[width=0.5\textwidth]{q1cii.png} - } - \captionof{figure}{Comparison between analytical result and numerical result. Above \(n=50\), the results diverge substantially.} - \end{minipage} - - We can see here that the "predicted" numerical response matches nearly exactly the actual - calculated numerical solution. - - \item %1ciii - - Applying the approximation $\cos{(\theta)} \approx 1 - \frac{\theta^2}{2}$ 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). - \end{equation*} - - 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 - 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}[b]{\linewidth} - - \begin{figure}[H] - \centering - \begin{subfigure}{0.5\textwidth} - \centering - \includegraphics[width=\textwidth]{q1civ_fig1.png} - \caption{} - \end{subfigure}% - \begin{subfigure}{0.5\textwidth} - \centering - \includegraphics[width=\textwidth]{q1civ_fig2.png} - \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} - - - \end{enumerate} - \item %1d - \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] - \item %1di - - 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: - \begin{lstlisting}[language=Matlab] -% Modify hamiltonian for circular boundary conditions -H(1, N) = -t0; -H(N, 1) = -t0; \end{lstlisting} - + The bonding and antibonding eigenenergies are \(\boxed{\SI{-32.2567}{eV}}\) and \(\boxed{\SI{-15.5978}{eV}}\) respectively. + \item %1bii + Neglecting normalization, we have the following expressions for \(\phi_B(z)\) and \(\phi_A(z)\): + \begin{align*} + \phi_B(z) = u_L(z) + u_R(z) \\ + \phi_A(z) = u_L(z) - u_R(z). + \end{align*} + We obtain the following plot: + \begin{minipage}[t]{\linewidth} - \begin{figure}[H] - \centering - \begin{subfigure}{0.5\textwidth} - \centering - \includegraphics[width=\textwidth]{q1di_fig1.png} - \caption{} - \end{subfigure}% - \begin{subfigure}{0.5\textwidth} - \centering - \includegraphics[width=\textwidth]{q1di_fig2.png} - \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 \boxed{0.06\:\mathrm{eV}}. - These eigenstates are degenerate because they both have the same - eigenvalue/energy. - - \item %1diii - - \begin{minipage}[t]{\linewidth} \centering \adjustbox{valign=t}{ - \includegraphics[width=0.5\textwidth]{q1div.jpg} + \includegraphics[width=0.5\textwidth]{q1bii.png} } - \captionof{figure}{Sketch of degenerate energy levels. In the sketch, closely-spaced - levels are in fact degenerate.} + \captionof{figure}{non-normalized probability densities for bonding and antibonding solutions.} \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 within an - acceptable margin to 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 - 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 - 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=0.6\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} - + \section*{Problem 2} + \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] + \item %2a + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] + \item %2ai + + \begin{minipage}[t]{\linewidth} + + \centering + \adjustbox{valign=t}{ + \includegraphics[width=0.5\textwidth]{q2ai.png} + } + % TODO still don't like this caption + \captionof{figure}{Energy vs. wave vector relationship (note: not discretized to account for N)} + \end{minipage} + + \item %2aii + + Energy values from \( -2 \) eV to \(2\) eV are allowed. + + \item %2aiii + + The vector \( \{ \phi \} \), which is of length \( N \), and has elements \( n \) is: + + \begin{equation*} + \{ \phi \} = C e ^{ i k \cdot n a } + \end{equation*} + + The corresponding wave function is: + + \begin{equation*} + \phi (x) = \sum _ {n = 1} ^ N C e ^ { i k \cdot n a } u_n (x) + \end{equation*} + + There is one wave function and thus one energy level for each value of \( k \). + This means that there is one electronic state per \( k \). + + \end{enumerate} + \item %2b + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] + \item %2bi + \begin{align*} + \left\{\phi\right\} = \begin{bmatrix} + C_Ae^{ika} \\ + C_Be^{ika} \\ + C_Ae^{ik2a} \\ + C_Be^{ik2a} \\ + \vdots \\ + C_Ae^{ikNa} \\ + C_Ae^{ikNa} + \end{bmatrix} + \end{align*} + \item %2bii + \begin{equation*} + \phi(x) = \sum_{n=1}^{N} C_Ae^{ikna} u_{nA}(x) + C_Be^{ikna} u_{nB}(x) + \end{equation*} + \item %2biii + \([h(k)]\) is of size 2-by-2. Thus there will be two values of \(E(k)\) for a fixed \(k\). + This also means there are two \(\phi(x)\) for each \(k\). + \end{enumerate} + \item %2c + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] + \item %2ci + + \begin{align*} + \phi _2 + 2 \phi _3 + \phi _4 &= E \phi_1 \\ + \phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\ + 2 \phi _1 + \phi _2 + \phi _4 &= E \phi_3 \\ + \phi _1 + 2 \phi _2 + \phi _3 &= E \phi_4 \\ + \end{align*} + + \item %2cii + + \begin{align*} + \phi _2 + 2 \phi _3 + \phi _0 &= E \phi_1 \\ + \phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\ + 2 \phi _5 + \phi _2 + \phi _4 &= E \phi_3 \\ + \phi _5 + 2 \phi _6 + \phi _3 &= E \phi_4 \\ + \end{align*} + + \item %2ciii + + A generalized form of the $n$th equation is: + + \begin{equation*} + E \phi_n = \phi_{n-1} + \phi_{n+1} + 2 \phi_{n+2} + \end{equation*} + + \item %2civ + + \begin{align} + E C e ^{ikna} &= Ce ^ {ik (n + 1)a} + Ce^{ik (n - 1)a} + 2 Ce^{ik (n + 2)a} \nonumber \\ + E &= e^{ika} + e^{-ika} + 2 e^{2ika} \nonumber \\ + E(k) &= 2 e^{2ika} + 2\cos(ka) \label{eq:e_k} + \end{align} + + \item %2cv + + Imposing the repeating boundary conditions \( \phi _{n + 4} = \phi _{n} \), We obtain the following relationship: + + \begin{align*} + Ce^{ikna} &= Ce^{ik(n+4)a} \\ + 1 &= e^{i4ka} + \end{align*} + + For this to hold, \(4ka\) must be some multiple of \(2 \pi \), and this mean \( k = \frac{\pi}{2a} \cdot integer \). + + \item %2cvi + + Using the E-k relationship from equation \ref{eq:e_k}, + we know that \( k \) must always be real, so the \( 2\cos(ka) \) portion of the E-k relationship must be + real. As well, if we substitute in the equation for \( k \) we obtained in part \nolinebreak (v), we get the following (partial) expression: + + \begin{equation*} + 2e^{i2ka} = 2 e^{i\pi \cdot integer} + \end{equation*} + + Which we know will always be real (with a value of \( \pm 2 \)). + + \item %2cvii + Since \(e^{2\pi n} = 1\), we have: + \begin{align*} + \phi_n(k+\frac{2\pi}{a}) &= Ce^{i(k+\frac{2\pi}{a} \cdot nA} = Ce^{i(k \cdot nA} \\ + \phi_n(k+\frac{2\pi}{a}) &= \phi_n(k). + \end{align*} + Therefore wavefunctions for which \(k\) is separated by \(\frac{2\pi}{a}\) are equivalent, and we + only need consider the range \(k \in [-\frac{\pi}{a}, \frac{\pi}{a}]\). + \end{enumerate} + \end{enumerate} + \end{document} \ No newline at end of file diff --git a/PS3/q1bi.m b/PS3/q1bi.m new file mode 100644 index 0000000..b44eeba --- /dev/null +++ b/PS3/q1bi.m @@ -0,0 +1,41 @@ +%constants +E1 = -13.6; +R = 0.074; +a0 = 0.0529; + +%matrix elements +R0 = R/a0; + +a = 2*E1*(1-(1+R0)*exp(-2*R0))/R0; +b = 2*E1*(1+R0)*exp(-R0); +s = exp(-R0)*(1+R0+(R0^2/3)); + +%matrices +H_u = [E1 + a, E1*s+b; E1*s+b, E1 + a]; +S_u = [1, s; s, 1]; + +%find eigenvalues and eigenvectors +[vectors,energies] = eig(inv(S_u)*H_u); +z = linspace(-2,2); + +z_L = -0.37; +z_R = 0.37; +a0 = 0.529; + +u_L = exp(-abs(z - z_L)./a0)./sqrt(pi*a0^3); + +u_R = exp(-abs(z - z_R)./a0)./sqrt(pi*a0^3); + +phi_B = u_L + u_R; + +phi_A = u_L - u_R; + +figure(1); + +plot(z, phi_B.^2, 'k-'); +hold on +plot(z, phi_A.^2, 'b-'); +grid on; +xlabel('z [Angstroms]'); +ylabel('probability [arbitrary scaling]'); +legend('Bonding', 'Antibonding'); diff --git a/PS3/q1bii.png b/PS3/q1bii.png new file mode 100644 index 0000000..1f3cc7a Binary files /dev/null and b/PS3/q1bii.png differ diff --git a/PS3/q2.m b/PS3/q2.m new file mode 100644 index 0000000..e4da164 --- /dev/null +++ b/PS3/q2.m @@ -0,0 +1,16 @@ +% Atomic spacing +a = 5e-10; +% +t = -1; +E_0 = 0; + +N = 100; +k = linspace(- pi / a, pi / a, N); + +e_k = E_0 + 2* t * cos(k*a); + +figure(1); +plot(k, e_k, 'k-'); +grid on; +xlabel('k'); +ylabel('E[k] (eV)'); diff --git a/PS3/q2ai.fig b/PS3/q2ai.fig new file mode 100644 index 0000000..5751352 Binary files /dev/null and b/PS3/q2ai.fig differ diff --git a/PS3/q2ai.png b/PS3/q2ai.png new file mode 100644 index 0000000..1106509 Binary files /dev/null and b/PS3/q2ai.png differ