diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..5fd6a19 --- /dev/null +++ b/.gitignore @@ -0,0 +1,6 @@ +PS2/doc.aux +PS2/doc.fdb_latexmk +PS2/doc.fls +PS2/doc.log +PS2/doc.synctex.gz +PS2/doc.synctex.gz diff --git a/PS2/doc.pdf b/PS2/doc.pdf index 6319256..0bb9a36 100644 Binary files a/PS2/doc.pdf and b/PS2/doc.pdf differ diff --git a/PS2/doc.tex b/PS2/doc.tex index 2434949..ee506e6 100644 --- a/PS2/doc.tex +++ b/PS2/doc.tex @@ -67,7 +67,7 @@ \newcommand{\angstrom}{\textup{\AA}} \title{ECE 456 - Problem Set 2} -\date{2021-03-01} +\date{2021-03-08} \author{David Lenfesty \\ lenfesty@ualberta.ca \and Phillip Kirwin \\ pkirwin@ualberta.ca} @@ -292,7 +292,7 @@ legend({'Numerical','Analytical'},'Location','northwest'); \adjustbox{valign=t}{ \includegraphics[width=0.5\textwidth]{q1cii.png} } - \captionof{figure}{Analytic solution to numeric system, plotted.} + \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 @@ -309,9 +309,9 @@ legend({'Numerical','Analytical'},'Location','northwest'); 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*} + \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 @@ -319,7 +319,7 @@ legend({'Numerical','Analytical'},'Location','northwest'); now a constant-amplitude wave, which corresponds to the expected analytic result, in contrast to the plot in section (a), which has a low-frequency envelope around it. - \begin{minipage}[t]{\linewidth} + \begin{minipage}[b]{\linewidth} \begin{figure}[H] \centering @@ -385,15 +385,16 @@ H(N, 1) = -t0; \adjustbox{valign=t}{ \includegraphics[width=0.5\textwidth]{q1div.jpg} } - \captionof{figure}{Sketch of degenerate energy levels.} + \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 very closely, to within - acceptable margin of the numerical results from part (ii). + indices \(n=0\), \(1\), and \(2\), respectively. These match within an + acceptable margin to the numerical results from part (ii). \end{enumerate} @@ -528,7 +529,7 @@ axis([0 1e-9 0 0.04]); which corresponds to the analytical result for the energy being slightly greater in magnitude (\(-13.6\) eV versus \(-13.4978\) eV). \begin{minipage}[t]{\linewidth} \centering - \includegraphics[width=0.5\textwidth]{q2f_fig.png} + \includegraphics[width=0.6\textwidth]{q2f_fig.png} \captionof{figure}{Numerical result (black line) and analytical solution scaled by \(a\) (orange circles).} \end{minipage} @@ -542,10 +543,15 @@ axis([0 1e-9 0 0.04]); \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] \item %3a - + Recalling the identity \begin{equation} - \theta (x') = \sqrt{\frac{2}{L}} sin \left( \frac{\pi (x' + L/2)}{L} \right) = \sqrt{\frac{2}{L}} cos \left( \frac{\pi x'}{L} \right) + \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 @@ -553,13 +559,18 @@ axis([0 1e-9 0 0.04]); \item %3bi Mapping the provided Fourier identities from \( t \) and \( \omega \) onto - \(x'\) and \(k'\), we can evaluate the Fourier transform \(A(k')\): + \(x'\) 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(\frac{x'}{L}) \right] &= \frac{L}{\sqrt{2 \pi}} sinc \left( \frac{k' L}{2 \pi} \right) \\ - \mathcal{F}\left[f(x')cos(\frac{\pi}{L}x')\right] &= \frac{1}{2} \left[ F(k' + k_1) + F(k' - k_1) \right] \\ - A(k') &= \frac{1}{2} \sqrt{\frac{L}{\pi}} \left\{ \sinc \left( \frac{L}{2\pi} (k' + k_1) \right) + \sinc \left( \frac{L}{2\pi} (k' - k_1) \right) \right\}. + \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 @@ -568,8 +579,9 @@ axis([0 1e-9 0 0.04]); \end{equation*} we can obtain \begin{equation*} - \Phi(p') = \frac{1}{2} \sqrt{\frac{L}{\pi\hbar}} \left\{ \sinc \left( \frac{L}{2\pi\hbar} (p' + p_1) \right) + \sinc \left( \frac{L}{2\pi\hbar} (p' - p_1) \right) \right\}. + \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 @@ -580,73 +592,78 @@ axis([0 1e-9 0 0.04]); \item %3biv - \( sinc \) is a purely real function, so we can ignore the normalizing portion of the integral. - As well, to simplify the interim equations we will assign the constants \( A = \frac{1}{2} \sqrt{\frac{L}{\pi \hbar}} \) - and \( B = \frac{L}{2 \pi \hbar} \). + \( sinc \) 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{equation*} - \int _{-\infty} ^{\infty} \Phi (p')^2 dp = \int _{-\infty} ^{\infty} A^2 \left[ sinc\left( B(p'+p_1) \right)^2 + 2 sinc \left( B(p'+p_1) \right) sinc \left( B(p'-p_1) \right) + sinc \left( B(p'-p_1) \right) \right] dp - \end{equation*} + \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, we can simplify the left and right terms. - Using a change of variable \( p'' = p_1 - p' \), and properties (27) and (28), we can further evaluate the central term. + 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 = \int _{-\infty} ^{\infty} A^2 \left[2 sinc \left( B(2p_1 - p'') \right) sinc \left( B(-p'') \right) \right] dp + A^2 \frac{2}{B} \\ - &= \int _{-\infty} ^{\infty} A^2 \left[ 2 sinc \left(B(2p_1 - p'') \right) sinc(B p'') \right] dp + A^2 \frac{2}{B} \\ - &= \frac{A^2}{B} \left[ sinc(2p_1) + 2 \right] \\ - &= \frac{1}{2} \left[ sinc(2p_1) + 2 \right] - \end{align} + \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{minipage}[t]{\linewidth} - \begin{figure}[H] - \centering - \begin{subfigure}{0.5\textwidth} + \begin{figure}[H] \centering - \includegraphics[width=\textwidth]{q3bv_fig1.png} - \caption{} - \end{subfigure}% - \begin{subfigure}{0.5\textwidth} - \centering - \includegraphics[width=\textwidth]{q3bv_fig2.png} - \caption{} - \end{subfigure} - \caption{(a) momentum wavefunction versus normalized momentum. (b) Probability density versus normalized momentum.} - \end{figure} - \end{minipage} + \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 - \(v = \pm \SI{3.6e4}{m/s}\). + \(\boxed{v = \pm \SI{3.6e4}{m/s}}\). - \begin{minipage}[t]{\linewidth} + \begin{minipage}[t]{\linewidth} - \begin{figure}[H] - \centering - \begin{subfigure}{0.5\textwidth} + \begin{figure}[H] \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. + \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 \(\left\langle p'\right\rangle = 0\). + 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 @@ -659,13 +676,13 @@ axis([0 1e-9 0 0.04]); \begin{align*} \left\langle p'\right\rangle\ &= \left.\frac{-i\hbar}{L}\sin^2{\left(\frac{\pi}{L}x'\right)}\right|_{-L/2}^{L/2}\\ - \left\langle p'\right\rangle\ &= \frac{-i\hbar}{L}\left(1 - 1\right) = 0, + \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 wavefunction \( \theta (x') = e^{ik'x'} \) is sharp, - and the corresponding value is \( p' = \hbar k' \). + 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'} @@ -692,12 +709,12 @@ axis([0 1e-9 0 0.04]); \includegraphics[width=\textwidth]{q4a_e2.png} \caption{} \end{subfigure} - \caption{Probability Density Plots for first two energy levels.} + \caption{Probability densities versus position for first two energy levels.} \end{figure} \end{minipage} - An \( a \) value of \( 0.53 \angstrom \) was chosen in order to provide an adequately - shaped graph without sacrificing too much computation time and ensure that the first two + 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 \(\SI{0.43140}{\electronvolt} \), @@ -710,29 +727,29 @@ axis([0 1e-9 0 0.04]); \begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)] \item %4bi - The energies were used were \( 0.14395eV \) and \( 0.43185eV \) for the first and second energy levels, + 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=\textwidth]{q4bi_I-V.png} - \captionof{figure}{Current vs. Voltage.} + \includegraphics[width=0.5\textwidth]{q4bi_I-V.png} + \captionof{figure}{Current-voltage characteristic of a 2-level molecule.} \end{minipage} \item %4bii - Between \( 0V \) and \( 0.25V \), only the first energy level is carrying any current. - This current drops to 0 above \( 0.25V\) because the coupling between the contacts and that + 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 \(0.4V\) and \(0.65V\), only the second energy level is carrying current. - This energy level stops conducting current because it's shifted energy drops below + 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 Drain Resistance (NDR) is present in this design from a \(V_D\) of - approximately \(0.27V\) to \(0.45V\), as well as from \(0.65V\) to \(0.8V\) + 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}