Schrödinger equation - Revision history

APirnat at 14:19, 19 July 2019

2019-07-19T14:19:31Z

APirnat at 14:35, 15 July 2019

2019-07-15T14:35:30Z

APirnat at 14:21, 15 July 2019

2019-07-15T14:21:31Z

APirnat at 14:19, 15 July 2019

2019-07-15T14:19:00Z

APirnat at 13:17, 15 July 2019

2019-07-15T13:17:39Z

APirnat at 20:33, 17 June 2019

2019-06-17T20:33:35Z

APirnat at 20:32, 17 June 2019

2019-06-17T20:32:14Z

APirnat at 20:22, 17 June 2019

2019-06-17T20:22:53Z

APirnat at 09:44, 17 June 2019

2019-06-17T09:44:43Z

APirnat at 09:39, 17 June 2019

2019-06-17T09:39:21Z

@@ Line 75: / Line 75: @@
 [[File:infinite_well_2D_error.png|400px]]
 [[File:infinite_well_2D_error_fit.png|400px]]
 == 3D infinite cubic well ==
@@ Line 93: / Line 95: @@
 [[File:infinite_well_3D_rsol.png|400px]]
 [[File:infinite_well_3D_csol.png|400px]]
 Go back to [[Medusa#Examples|Examples]].

@@ Line 1: / Line 1: @@
 Go back to [[Medusa#Examples|Examples]].
 == 2D infinite square well ==
@@ Line 83: / Line 85: @@
 \]
-remain unchanged, only the domain changes to unit cube, $\Omega=[0, 1]\times[0, 1]\times [0, 1]$.
+remain unchanged, merely the domain changes to unit cube, $\Omega=[0, 1]\times[0, 1]\times [0, 1]$. The only step left is to select an initial state of the wave function, we choose
 \[
 \Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y) \sin(\pi z)}.
 \]
-The plot of the real and imaginary part of the solution $\Psi(x, y, z)$ at time $T=0.1$ is presented bellow.
+The plot of the real and imaginary part of the solution $\Psi(x, y, z)$ at time $T=0.1$ is shown bellow.
 [[File:infinite_well_3D_rsol.png|400px]]

@@ Line 75: / Line 75: @@
 == 3D infinite cubic well ==
-Now we are dealing with a problem very similar to the previus one, the only diffrenece is, that in this case we have a quantom particle in a 3-dimensional box. Equation and the boundary condition
+Now we are dealing with a problem very similar to the previous one, the only difference is, that in this case we have a quantum particle in a 3-dimensional box. The equation and the boundary condition
 \[
@@ Line 86: / Line 86: @@
 For initial state $\Psi(x, y, z,t= 0)$ we choose
 \[
-\Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y) \sin(\piz)}.
+\Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y) \sin(\pi z)}.
 \]
-The plot of the real and imaginary part of the solution $\Psi(x, y, z)$ at time $T=0.1$ is plotted bellow.
+The plot of the real and imaginary part of the solution $\Psi(x, y, z)$ at time $T=0.1$ is presented bellow.
 [[File:infinite_well_3D_rsol.png|400px]]

@@ Line 73: / Line 73: @@
 [[File:infinite_well_2D_error.png|400px]]
 [[File:infinite_well_2D_error_fit.png|400px]]
 Go back to [[Medusa#Examples|Examples]].

@@ Line 13: / Line 13: @@
 \]
-With this potential Schrodinger equation simplifies to
+With this potential Schrödinger equation simplifies to
 \[
 -{\hbar^2 \over 2m} \nabla^2 \Psi(x,y,t) = i\hbar {\partial \over \partial t}\Psi(x,y,t),
@@ Line 27: / Line 27: @@
 \Psi = 0 \quad \text{on} \quad \partial \Omega,
 \]
-where $\Omega=[0, 1]\times[0, 1]$ denotes the square box domain and $\partial \Omega$ is the boundary of the domain. In order to get the unique solution for the time propagation of the wave function the initial state $\Psi(x, y,t= 0)$ has to be selected. We choose
+where $\Omega=[0, 1]\times[0, 1]$ denotes the square box domain and $\partial \Omega$ is the boundary of the domain. In order to get the unique solution for the time propagation of the wave function, the initial state $\Psi(x, y,t= 0)$ has to be selected. We choose
 \[
 \Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y)}.
 \]
-Once we prepare the domain and operators, we can turn the Schrodinger equation into code. As we are solving the time propagation implicitly, we will have to solve a matrix equation for every step. For this reason a space matrix $M$ is constructed. Next the equations are implemented, $psi$ denotes the matrix solution for each step and $rhs$ the right-hand side of the matrix equation which is in our case equivalent to the previous state $psi$ .
+Once we prepare the domain and operators, we can turn the Schrödinger equation into code. As we are solving the time propagation implicitly, we will have to solve a matrix equation for every step. For this reason a space matrix $M$ is constructed. Next the equations are implemented, $psi$ denotes the matrix solution for each step and $rhs$ the right-hand side of the matrix equation, which is in our case equivalent to the previous state $psi$.
-<syntaxhighlight lang="c++" line>
+<syntaxhighlight lang="cpp">
-Eigen::SparseMatrix<std::complex<double>, Eigen::RowMajor> M(N, N);
+SparseMatrix<complex<double>, RowMajor> M(N, N);
-// set initial state
+// Set initial state.
 for(int i : domain.interior()) {
      double x = domain.pos(i, 0);
@@ Line 44: / Line 44: @@
+}
-// set equation on interior
+// Set equation on interior.
 for(int i : domain.interior()) {
-     op.value(i) + (-1.0i) * dt * op.lap(i) = rhs(i);
+     op.value(i) + (-1.0_i) * dt * op.lap(i) = rhs(i);
+}
-// set equation on boundary
+// Set equation on boundary.
 for (int i: domain.boundary()){
      op.value(i) = 0;
@@ Line 55: / Line 55: @@
 </syntaxhighlight>
-At each time step the matrix equation is solved using an Eigen linear equation solver, in this case, the BICGStab algorithm, and $rhs$ is updated.
+At each time step the matrix equation is solved using an Eigen linear equation solver, in this case the BICGStab algorithm, and $rhs$ is updated.
-<syntaxhighlight lang="c++" line>
+<syntaxhighlight lang="cpp">
-// solve matrix system
+// Solve matrix system.
 psi = solver.solve(rhs);
-// update rhs
+// Update rhs.
 rhs = psi;
 </syntaxhighlight>
-Time propagation of the real part of the solution $\Psi(x,y; t)$ from times 0 to $T=0.63$ is presented below. Time step used was $10^{-5}$.
+Time propagation of the real part of the solution $\Psi(x,y,t)$ from times 0 to $T=0.63$ is presented below. Time step used was $10^{-5}$.
 [[File:video (1).mp4|400px]]
-As analytical solutions for infinite potential well are known this enables comparison between analytical and numerical solution computed with Medusa. On the graph bellow the error of the numerical solution is presented as maximal absolute error, meaning the largest absolute difference between numeric and its analytical solution in node at specific time. Expected linear convergence of error on a log - log scale can be observed.
+As analytical solutions for infinite potential well are known this enables comparison between analytical and numerical solution computed with Medusa. On the graph bellow the error of the numerical solution is presented as maximal absolute error $e_{\infty}$, meaning the largest absolute difference between numeric and its analytical solution in node at specific time. Expected linear convergence of error on a log - log scale can be observed.
-[[File:infinite_well_2D_convergence.png|400px]]
+[[File:infinite_well_2D_error.png|400px]]
-[[File:infinite_well_2D_convergence_fit.png|400px]]
+[[File:infinite_well_2D_error_fit.png|400px]]
 Go back to [[Medusa#Examples|Examples]].

@@ Line 15: / Line 15: @@
 With this potential Schrodinger equation simplifies to
 \[
--{\hbar^2 \over 2m} \nabla^2 \Psi(x,y;t) = i\hbar {\partial \over \partial t}\Psi(x,y;t),
+-{\hbar^2 \over 2m} \nabla^2 \Psi(x,y,t) = i\hbar {\partial \over \partial t}\Psi(x,y,t),
 \]
-where $m$ is the mass of the particle, $\hbar$ is reduced Planck constant and $\Psi(x,y;t)$ is the wave function. We set $\hbar / 2m = 1$ and $L=1$.
+where $m$ is the mass of the particle, $\hbar$ is reduced Planck constant and $\Psi(x,y,t)$ is the wave function. We set $\hbar / 2m = 1$ and $L=1$.
-The solution $\Psi(x,y;t)$ must therefore satisfy
+The solution $\Psi(x,y,t)$ must therefore satisfy
 \[
 \nabla^2 \Psi = -i{\partial \over \partial t}\Psi \quad \text{in} \quad \Omega,
@@ Line 27: / Line 27: @@
 \Psi = 0 \quad \text{on} \quad \partial \Omega,
 \]
-where $\Omega=[0, 1]\times[0, 1]$ denotes the square box domain and $\partial \Omega$ is the boundary of the domain. In order to get the unique solution for the time propagation of the wave function the initial state $\Psi(x, y,\, 0)$ has to be selected. We choose
+where $\Omega=[0, 1]\times[0, 1]$ denotes the square box domain and $\partial \Omega$ is the boundary of the domain. In order to get the unique solution for the time propagation of the wave function the initial state $\Psi(x, y,t= 0)$ has to be selected. We choose
 \[
-\Psi(x,y;t=0) = \sin{(\pi x)} \sin{(\pi y)}.
+\Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y)}.
 \]
@@ Line 64: / Line 64: @@
 </syntaxhighlight>
-The animation of the real part of solution $\Re \Psi(x,y; t)$ is presented below.
+The animation of the real part of solution $\Re \Psi(x,y)$ is presented below.
 [[File:video (1).mp4|400px]]
 Go back to [[Medusa#Examples|Examples]].