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@ -7,7 +7,8 @@
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\newlabel{eq:analytical}{{1}{3}}
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entering extended mode
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@ -332,19 +332,36 @@ File: q1a_eigenvals.png Graphic file (type png)
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@ -155,7 +155,7 @@ legend({'Numerical','Analytical'},'Location','northwest');
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\begin{subfigure}{0.4\linewidth}
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\centering
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\includegraphics[width=\textwidth]{q1a_1and50.png}
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\caption{probability densities for \(n=1\)\\and \(n=50\).}
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\caption{Probability densities for \(n=1\)\\and \(n=50\).}
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\label{fig:q3_ne}
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\end{subfigure}%
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\begin{subfigure}{0.4\linewidth}
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@ -206,29 +206,38 @@ legend({'Numerical','Analytical'},'Location','northwest');
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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_{l=1}^N|\phi_l|^2 = a \\
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a\sum_{l=1}^N|B\sin{(\frac{n\pi}{L}x_l)}|^2 = a,
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\end{align*}
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\begin{align}
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a\sum_{l=1}^N\left|\phi_l\right|^2 = a \\
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a\sum_{l=1}^N\left|B\sin{\left(\frac{n\pi}{L}x_l\right)}\right|^2 = a,
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\label{eq:numerical_norm}
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\end{align}
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recalling that \(x = al\), 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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\(n = N\), \(i = l\), \(c = 0\), \(d = L\), and \(\Delta x = a\), \(x_i = x_l\),
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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_a^bf(x)\:dx = \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i).
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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_{l=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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The output of the numerical sum will change, as the value of $\alpha$ changes, by a factor of $\alpha$.
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To correct for this, we need to factor the result of the summation by $\alpha$, which corresponds to
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% Idk how to explain this at all, I need to chew on it internally and come back later.
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Using (\ref{eq:integral_ident}), we have
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\begin{equation*}
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foo
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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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B = \sqrt{\frac{2 \alpha}{L}}
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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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\centering
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\includegraphics[width=\textwidth]{q1cii.png}
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\caption{Analytic solution to numeric system, plotted.}
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\label{fig:q1b_electrons}
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\end{figure}
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We can see here that the "predicted" numerical response matches nearly exactly the actual
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\end{equation*}
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\item %1civ
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\lipsum[1]
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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{Probability densities for \(n=1\)\\and \(n=50\).}
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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{Comparison of first 101 numerical and analytic eigenvalues.}
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\end{subfigure}
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\end{figure}
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% TODO idk if this is enough
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With the decreased lattice spacing, and increased number of points, we can see the numerical solution
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of eigenvalues matches the analytical solution much closer. As well, we can see that the probability
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density function appears to be "squeezed" in the centre, for the $n = 50$ case.
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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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\lipsum[1]
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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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% TODO do we need a listing here
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\item %1dii
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\lipsum[1]
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\item %1diii
|
||||
\lipsum[1]
|
||||
|
||||
|
||||
\item %1div
|
||||
\lipsum[1]
|
||||
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
60
PS2/q1d.m
Normal file
60
PS2/q1d.m
Normal file
@ -0,0 +1,60 @@
|
||||
clear all;
|
||||
%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
|
||||
|
||||
% Modify hamiltonian for circular boundary conditions
|
||||
H(1, N) = -t0;
|
||||
H(N, 1) = -t0;
|
||||
|
||||
[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_4 = eigenvectors(:,4);
|
||||
phi_5 = eigenvectors(:,5);
|
||||
|
||||
% find the probability densities of position for 1st and 50th eigenvectors
|
||||
|
||||
P_4 = phi_4 .* conj(phi_4);
|
||||
P_5 = phi_5 .* conj(phi_5);
|
||||
|
||||
% 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_4,'kx',x,P_5,'k-');
|
||||
grid on; set(h,'linewidth',[2.0]); set(gca,'Fontsize',[18]);
|
||||
xlabel('POSITION [m]'); ylabel('PROBABILITY DENSITY [1/m]');
|
||||
legend('n=4','n=5');
|
||||
|
||||
% 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]');
|
||||
Loading…
Reference in New Issue
Block a user