Schrödinger equation

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2D infinite square well

We are solving a Schrödinger equation for a quantum particle that is confined to a 2-dimensional square box, $0<x<L$ and $0<y<L$. Within this square the particle behaves as a free particle, but the walls are impenetrable, so the wave function is zero at the walls. This is described by the infinite potential

$$V(x)= \begin{cases} 0, &0<x<L,\: 0<y<L, \\ \infty, &{\text{otherwise.}} \end{cases}$$

With this potential Schrodinger equation simplifies to

$$-{\hbar^2 \over 2m} \nabla^2 \Psi(x,y,t) = i\hbar {\partial \over \partial t}\Psi(x,y,t),$$ 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

$$\nabla^2 \Psi = -i{\partial \over \partial t}\Psi \quad \text{in} \quad \Omega,$$ with Dirichlet boundary condition $$\Psi = 0 \quad \text{on} \quad \partial \Omega,$$ 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 $$\Psi(x,y,t=0) = \sin{(\pi x)} \sin{(\pi y)}.$$

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 $psi$ .

 1 Eigen::SparseMatrix<std::complex<double>, Eigen::RowMajor> M(N, N);
 2 // set initial state
 3 for(int i : domain.interior()) {
 4     double x = domain.pos(i, 0);
 5     double y = domain.pos(i, 1);
 6     rhs(i) = sin(PI * x) * sin(PI * y);
 7 }
 8 
 9 // set equation on interior
10 for(int i : domain.interior()) {
11     op.value(i) + (-1.0i) * dt * op.lap(i) = rhs(i);
12 }
13 
14 // set equation on boundary
15 for (int i: domain.boundary()){
16     op.value(i) = 0;
17 }

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.

1 // solve matrix system
2 psi = solver.solve(rhs);
3 
4 // update rhs
5 rhs = psi;

The animation of the real part of solution $\Re \Psi(x,y)$ is presented below.

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