\documentclass[../schrodinger_simulation.tex]{subfiles}

\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,
\begin{equation*}
    \frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} = F_{\ivec, \jvec}^{n} \ ,
\end{equation*}
and the backward difference, an implicit scheme, 
\begin{equation*}
    \frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} = F_{\ivec, \jvec}^{n+1} \ .
\end{equation*}
The result is a linear combination of the explicit and implicit scheme, given by 
\begin{align*}
    \frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} &= \theta F_{\ivec, \jvec}^{n+1} + (1 - \theta) F_{\ivec, \jvec}^{n} \ .
\end{align*}
The parameter $\theta$ is introduced for a general approach, however, for CN $\theta = 1/2$. 
\begin{align*}
    \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...

Schrödinger contain $i$ at the lhs, factor 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] & \text{, where $\frac{1}{i} = -i$}
\end{align*}

Using Equation \eqref{eq:schrodinger_dimensionless}, we get
\begin{align*}
    u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n} & -\frac{i \Delta t}{2} \bigg[ F_{\ivec, \jvec}^{n+1} + F_{\ivec, \jvec}^{n} \bigg] \\
    &= -\frac{i \Delta t}{2} \bigg[ - \frac{u_{\ivec+1, \jvec}^{n+1} - 2u_{\ivec, \jvec}^{n+1} + u_{\ivec-1, \jvec}^{n+1}}{2 \Delta x^{2}} \\
    & \quad - \frac{u_{\ivec, \jvec+1}^{n+1} - 2u_{\ivec, \jvec}^{n+1} + u_{\ivec, \jvec-1}^{n+1}}{2 \Delta y^{2}} + \frac{1}{2} v_{\ivec, \jvec} u_{\ivec, \jvec}^{n+1} \\
    & \quad - \frac{u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n}}{2 \Delta x^{2}} \\
    & \quad - \frac{u_{\ivec, \jvec+1}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec, \jvec-1}^{n}}{2 \Delta y^{2}} + \frac{1}{2} v_{\ivec, \jvec} u_{\ivec, \jvec}^{n} \bigg]
\end{align*}
We rewrite the expression and gather all terms containing the $n+1$ time step on 
the left hand side, and the terms containing $n$ time step on the right hand side. 
\begin{align*}
    & u_{\ivec, \jvec}^{n+1} - \frac{i \Delta t}{2 \Delta x^{2}} \big[ u_{\ivec+1, \jvec}^{n+1} - 2u_{\ivec, \jvec}^{n+1} + u_{\ivec-1, \jvec}^{n+1} \big] \\
    & - \frac{i \Delta t}{2 \Delta y^{2}} \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} \\
    &= 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 
\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*}
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 
\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} \\
    &= u_{\ivec, \jvec}^{n} + r \big[ u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n} \big] \\
    & \quad + r \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*}


\section{Matrix structure}\label{ap:matrix_structure}
For $u$ vector of length $(M-2) = 3$, the matrices $A$ and $B$ have size 
$(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by 
\begin{align*}
    A = 
    \begin{bmatrix}
    a_{0} &  -r  &  0   &  -r   &   0   &  0    &  0   &  0    &  0    \\
    -r  &  a_{1} &  -r  &  0    &   -r  &  0    &  0   &  0    &  0    \\
    0   &  -r  &  a_{2} &  0    &   0   &  -r   &  0   &  0    &  0    \\
    -r  &  0   &  0   &  a_{3}  &   -r  &  0    &  -r  &  0    &  0    \\
    0   &  -r  &  0   &  -r   &   a_{4} &  -r   &  0   &  -r   &  0    \\
    0   &  0   &  -r  &  0    &   -r  &  a_{5}  &  0   &  0    &  -r   \\
    0   &  0   &  0   &  -r   &   0   &  0    &  a_{6} &  -r   &  0    \\
    0   &  0   &  0   &  0    &   -r  &  0    &  -r  &  a_{7}  &  -r   \\
    0   &  0   &  0   &  0    &   0   &  -r   &  0   &  -r   &  a_{8}  \\
    \end{bmatrix}
\end{align*}

\begin{align*}
    B = 
    \begin{bmatrix}
    b_{0} &  r   &  0   &  r    &   0   &  0    &  0   &  0    &  0    \\
    r   &  b_{1} &  r   &  0    &   r   &  0    &  0   &  0    &  0    \\
    0   &  r   &  b_{2} &  0    &   0   &  r    &  0   &  0    &  0    \\
    r   &  0   &  0   &  b_{3}  &   r   &  0    &  r   &  0    &  0    \\
    0   &  r   &  0   &  r    &   b_{4} &  r    &  0   &  r    &  0    \\
    0   &  0   &  r   &  0    &   r   &  b_{5}  &  0   &  0    &  r    \\
    0   &  0   &  0   &  r    &   0   &  0    &  b_{6} &  r    &  0    \\
    0   &  0   &  0   &  0    &   r   &  0    &  r   &  b_{7}  &  r    \\
    0   &  0   &  0   &  0    &   0   &  r    &  0   &  r    &  b_{8}  \\
    \end{bmatrix}
\end{align*}
\end{document}