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Janita Willumsen
2024-01-01 15:34:19 +01:00
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6 changed files with 75 additions and 37 deletions

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@@ -3,7 +3,8 @@
\begin{document}
\appendix
\section{The Crank-Nicholson method}\label{ap:crank_nicolson}
The Crank-Nicolson (CN) approach consider both the forward difference, an explicit scheme,
The Crank-Nicolson (CN) approach consider both the forward difference, an explicit
scheme,
\begin{equation*}
\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} = F_{\ivec, \jvec}^{n} \ ,
\end{equation*}
@@ -20,9 +21,11 @@ The parameter $\theta$ is introduced for a general approach, however, for CN $\t
\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} &= \frac{1}{2} \bigg[ F_{\ivec, \jvec}^{n+1} + F_{\ivec, \jvec}^{n} \bigg] \\
\end{align*}
We need the first derivative in respect to both time and position, as well as the second derivative in respect to position. Taylor expanding will result in a discretized version, assume this is known...
We need the first derivative in respect to both time and position, as well as the
second derivative in respect to position. Taylor expanding will result in a discretized
version.
Schrödinger contain $i$ at the lhs, factor it as
The Schrödinger equation contains $i$ on the left hand side, we rewrite it as
\begin{align*}
\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} &= \frac{1}{2i} \bigg[ F_{\ivec, \jvec}^{n+1} + F_{\ivec, \jvec}^{n} \bigg] \\
&= -\frac{i}{2} \bigg[ F_{\ivec, \jvec}^{n+1} + F_{\ivec, \jvec}^{n} \bigg] \ ,
@@ -44,8 +47,7 @@ the left hand side, and the terms containing $n$ time step on the right hand sid
&= u_{\ivec, \jvec}^{n} + \frac{i \Delta t}{2 \Delta x^{2}} \big[ u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n} \big] \\
& \quad + \frac{i \Delta t}{2 \Delta y^{2}} \big[ u_{\ivec, \jvec+1}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec, \jvec-1}^{n} \big] - \frac{i \Delta t}{2} v_{\ivec, \jvec} u_{\ivec, \jvec}^{n}
\end{align*}
In addition, since we will use an equal step size $h$ in both $x$ and $y$ direction,
we can use
Since we will use an equal step size $h$ in both $x$ and $y$ direction, we can use
\begin{align*}
\frac{i \Delta t}{2 \Delta h^{2}} = \frac{i \Delta t}{2 \Delta x^{2}} = \frac{i \Delta t}{2 \Delta y^{2}} \ ,
\end{align*}
@@ -53,7 +55,7 @@ and define
\begin{align*}
r \equiv \frac{i \Delta t}{2 \Delta h^{2}}
\end{align*}
Now, the discretized Schrödinger equation can be written as
Now, the discretized Schrödinger equation can be rewritten as
\begin{align*}
& u_{\ivec, \jvec}^{n+1} - r \big[ u_{\ivec+1, \jvec}^{n+1} - 2u_{\ivec, \jvec}^{n+1} + u_{\ivec-1, \jvec}^{n+1} \big] \\
& - r \big[ u_{\ivec, \jvec+1}^{n+1} - 2u_{\ivec, \jvec}^{n+1} + u_{\ivec, \jvec-1}^{n+1} \big] + \frac{i \Delta t}{2} v_{\ivec, \jvec} u_{\ivec, \jvec}^{n+1} \\
@@ -97,6 +99,8 @@ $(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by
\section{Figures}\label{ap:figures}
We created colormap plots of the real and imaginary part of $u_{\ivec, \jvec}$,
in Figure \ref{fig:colormap_real_imag}.
\begin{figure*}
\centering
\begin{subfigure}[b]{0.3\textwidth}
@@ -119,9 +123,7 @@ $(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by
\caption{Re$(u_{\ivec, \jvec})$ at time $t=0.002$.}
\label{fig:colormap_2_real}
\end{subfigure}
\newline
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{images/color_map_0_imag.pdf}
@@ -142,7 +144,17 @@ $(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by
\caption{Im$(u_{\ivec, \jvec})$ at time $t=0.002$.}
\label{fig:colormap_2_imag}
\end{subfigure}
\caption{The time evolution of the probability function $p_{\ivec, \jvec}^{n}$.}
\label{fig:colormap}
\caption{The time evolution of Re($u_{\ivec, \jvec}$) and Im($u_{\ivec, \jvec}$),
for time steps $t=[0, 0.001, 0.002]$.}
\label{fig:colormap_real_imag}
\end{figure*}
% \begin{figure*}
% \centering
% \includegraphics[width=0.9\textwidth]{images/color_map_all.pdf}
% \caption{The time evolution of the probability function $p_{\ivec, \jvec}^{n}$ (top row),
% Re($u_{\ivec, \jvec}^{n}$) (middle row), and Im($u_{\ivec, \jvec}^{n}$) (bottom row).
% Time step $t=0$ in the left column, $t=0.001$ in the middle column, and $t=0.002$
% in the right column.}
% \label{fig:colormap_all}
% \end{figure*}
\end{document}