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+%physical constants in MKS units + +hbar = 1.054e-34; +q = 1.602e-19; +m = 9.110e-31; + +%generate lattice + +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 + +%set up Hamiltonian matrix + +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} + + \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} + \item %1b + \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] + \item %1bi + The analytical solution is: + + \begin{equation} + \phi(x) = A \sin{\left( \frac{n \pi}{L} x \right)}. + \label{eq:analytical} + \end{equation} + + In order to normalise this equation it must conform to the following: + + \begin{equation} + \int_{0}^{L} \left| \phi(x) \right| ^2 dx = 1. + \label{eq:normalise} + \end{equation} + + We use the following identity: + + \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} + + \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} + } + \captionof{figure}{Sketch of degenerate energy levels. In the sketch, closely-spaced + levels are in fact degenerate.} + \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} + + +\end{document} \ No newline at end of file