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pkirwin 2021-03-01 16:18:33 -07:00
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\newlabel{eq:analytical}{{1}{3}}
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\newlabel{eq:integral_ident}{{3}{3}}
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@ -205,9 +205,20 @@ legend({'Numerical','Analytical'},'Location','northwest');
\item %1bii
{\em I really don't like this explanation}.
The output of the numerical sum will change, as the value of $\alpha$ changes, by a factor if $\alpha$.
Starting with the normalization condition for the numerical case:
\begin{align*}
a\sum_{l=1}^N|\phi_l|^2 = a \\
a\sum_{l=1}^N|B\sin{(\frac{n\pi}{L}x_l)}|^2 = a,
\end{align*}
recalling that \(x = al\), 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_a^bf(x)\:dx = \lim_{n \to \infty}\sum_{i=1}^n\Delta x\cdot f(x_i).
\end{equation*}
The output of the numerical sum will change, as the value of $\alpha$ changes, by a factor of $\alpha$.
To correct for this, we need to factor the result of the summation by $\alpha$, which corresponds to
% Idk how to explain this at all, I need to chew on it internally and come back later.
@ -226,18 +237,56 @@ legend({'Numerical','Analytical'},'Location','northwest');
\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
\item %1ci
From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} a \ell \right)$, we can see that
% TODO check that B is expressed correctly in all these equations (alpha, not ell, wait for pdf gen)
From the base form of $\phi _\ell = Bsin \left( \frac {n \pi}{L} \alpha \ell \right)$, we can see that
$\phi _ {\ell + 1}$ and $\phi _ {\ell - 1}$ correspond to the trigonometric identities
$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$ and
$sin(\alpha + B) = sin(\alpha)cos(B) + cos(\alpha)sin(B)$, respectively, where
$\alpha = \frac{n \pi \alpha \ell}{L}$ and $B = \frac{n \pi \ell}{L}$.
$\alpha = \frac{n \pi \alpha \ell}{L}$ and $B = \frac{n \pi \alpha}{L}$.
Fro
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 \alpha \ell}{L} \right) + 2 t_0 \phi _{\ell} - t_0 sin \left( \frac{n \pi \alpha \ell}{L} \right)
\end{equation*}
At this point, we notice that $\phi _ \ell = B sin \left( \frac {n \pi}{L} \alpha \ell \right)$,
so we can factor it out.
With some minor rearranging, this leaves us with the final expression for $E$:
\begin{equation}
E = 2t_0 \left( 1 - cos \left( \frac{n \pi \alpha}{L} \right) \right)
\label{eq:numerical_E}
\end{equation}
\item %1cii
\lipsum[1]
\begin{figure}[H]
\centering
\includegraphics[width=\textwidth]{q1cii.png}
\caption{Analytic solution to numeric system, plotted.}
\label{fig:q1b_electrons}
\end{figure}
We can see here that the "predicted" numerical response matches nearly exactly the actual
calculated numerical solution.
\item %1ciii
\lipsum[1]
Applying the approximation $cos(\Theta) = 1 - \frac{\Theta^2}{2}$ 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 \alpha^2}{2 L^2} \right)
\end{equation*}
Fully substituting the known value of $t_0$, and we can get our final expression for $E$:
\begin{equation*}
E = \frac{ \hbar^2 n^2 \pi^2 }{ 2 m L^2 }
\end{equation*}
\item %1civ
\lipsum[1]
\end{enumerate}