98 lines
5.9 KiB
TeX
98 lines
5.9 KiB
TeX
\documentclass[../schrodinger_simulation.tex]{subfiles}
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\begin{document}
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\appendix
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\section{The Crank-Nicholson method}\label{ap:crank_nicolson}
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The Crank-Nicolson (CN) approach consider both the forward difference, an explicit scheme,
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\begin{equation*}
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\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} = F_{\ivec, \jvec}^{n} \ ,
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\end{equation*}
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and the backward difference, an implicit scheme,
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\begin{equation*}
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\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} = F_{\ivec, \jvec}^{n+1} \ .
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\end{equation*}
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The result is a linear combination of the explicit and implicit scheme, given by
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\begin{align*}
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\frac{u_{\ivec, \jvec}^{n+1} - u_{\ivec, \jvec}^{n}}{\Delta t} &= \theta F_{\ivec, \jvec}^{n+1} + (1 - \theta) F_{\ivec, \jvec}^{n} \ .
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\end{align*}
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The parameter $\theta$ is introduced for a general approach, however, for CN $\theta = 1/2$.
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\begin{align*}
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\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] \\
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\end{align*}
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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...
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Schrödinger contain $i$ at the lhs, factor it as
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\begin{align*}
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\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] \\
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&= -\frac{i}{2} \bigg[ F_{\ivec, \jvec}^{n+1} + F_{\ivec, \jvec}^{n} \bigg] & \text{, where $\frac{1}{i} = -i$}
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\end{align*}
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Using Equation \eqref{eq:schrodinger_dimensionless}, we get
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\begin{align*}
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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] \\
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&= -\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}} \\
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& \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} \\
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& \quad - \frac{u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n}}{2 \Delta x^{2}} \\
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& \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]
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\end{align*}
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We rewrite the expression and gather all terms containing the $n+1$ time step on
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the left hand side, and the terms containing $n$ time step on the right hand side.
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\begin{align*}
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& 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] \\
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& - \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} \\
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&= 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] \\
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& \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}
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\end{align*}
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In addition, since we will use an equal step size $h$ in both $x$ and $y$ direction,
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we can use
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\begin{align*}
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\frac{i \Delta t}{2 \Delta h^{2}} = \frac{i \Delta t}{2 \Delta x^{2}} = \frac{i \Delta t}{2 \Delta y^{2}} \ ,
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\end{align*}
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and define
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\begin{align*}
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r \equiv \frac{i \Delta t}{2 \Delta h^{2}}
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\end{align*}
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Now, the discretized Schrödinger equation can be written as
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\begin{align*}
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& 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] \\
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& - 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} \\
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&= u_{\ivec, \jvec}^{n} + r \big[ u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n} \big] \\
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& \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} \ .
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\end{align*}
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\section{Matrix structure}\label{ap:matrix_structure}
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For $u$ vector of length $(M-2) = 3$, the matrices $A$ and $B$ have size
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$(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by
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\begin{align*}
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A =
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\begin{bmatrix}
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a_{0} & -r & 0 & -r & 0 & 0 & 0 & 0 & 0 \\
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-r & a_{1} & -r & 0 & -r & 0 & 0 & 0 & 0 \\
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0 & -r & a_{2} & 0 & 0 & -r & 0 & 0 & 0 \\
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-r & 0 & 0 & a_{3} & -r & 0 & -r & 0 & 0 \\
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0 & -r & 0 & -r & a_{4} & -r & 0 & -r & 0 \\
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0 & 0 & -r & 0 & -r & a_{5} & 0 & 0 & -r \\
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0 & 0 & 0 & -r & 0 & 0 & a_{6} & -r & 0 \\
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0 & 0 & 0 & 0 & -r & 0 & -r & a_{7} & -r \\
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0 & 0 & 0 & 0 & 0 & -r & 0 & -r & a_{8} \\
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\end{bmatrix}
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\end{align*}
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\begin{align*}
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B =
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\begin{bmatrix}
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b_{0} & r & 0 & r & 0 & 0 & 0 & 0 & 0 \\
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r & b_{1} & r & 0 & r & 0 & 0 & 0 & 0 \\
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0 & r & b_{2} & 0 & 0 & r & 0 & 0 & 0 \\
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r & 0 & 0 & b_{3} & r & 0 & r & 0 & 0 \\
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0 & r & 0 & r & b_{4} & r & 0 & r & 0 \\
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0 & 0 & r & 0 & r & b_{5} & 0 & 0 & r \\
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0 & 0 & 0 & r & 0 & 0 & b_{6} & r & 0 \\
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0 & 0 & 0 & 0 & r & 0 & r & b_{7} & r \\
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0 & 0 & 0 & 0 & 0 & r & 0 & r & b_{8} \\
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\end{bmatrix}
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\end{align*}
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\end{document}
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