Solving Schrodinger Equation with finite element and Crank-Nicolson?

Question

I have asked this in Mathematic section, but received no reply. Please let me ask here to see if threr is any difference.

The Schrodinger equation without potential has the following form: $$\partial_t\psi =i\nabla^2\psi.$$

The finite element software that I'm using does not handle complex function directly, so I solve it by splitting the equation into real and imaginary parts: $$\partial_t\psi_R = -\nabla^2\psi_I, \quad\quad \partial_t\psi_I = \nabla^2\psi_R .$$ Now if I apply finite element method, the equations become $$M\dot{\xi_R} = -A\xi_I, \quad\quad M\dot{\xi_I} = A\xi_R.$$

Here $M$ is the mass matrix, and $A$ is the diffusion matrix, and we solve for $\xi_R$ and $\xi_I$. Assuming periodic boundary here. If I apply the Crank-Nicolson Method, it becomes $${\xi_{R,n+1}} = \left(\dfrac{2M-\Delta tA}{2M+\Delta tA}\right)\xi_{I,n}, \quad\quad {\xi_{I,n+1}} =\left(\dfrac{2M+\Delta tA}{2M-\Delta tA}\right)\xi_{R,n}.$$

I am following approximately this paper to derive the equations.

Problem: After one time step, the function evolves correctly, but as it evolves to further timestep, the solusion simply goes back and forth between 0 and 1 timestep, because $$\left(\dfrac{2M+\Delta tA}{2M-\Delta tA}\right)\left(\dfrac{2M-\Delta tA}{2M+\Delta tA}\right) = I.$$ so $$\xi_{R,n+2} = I\xi_{R,n}$$

Does any expert here know how to resolve this problem, so I can go to further timestep? Thanks.

Answer 1

You made an error in the indices for the real and imaginary part. $$ M\frac{\xi_{R,n+1}-\xi_{R,n}}{Δt}=-A\frac{\xi_{I,n+1}+\xi_{I,n}}2 \\ M\frac{\xi_{I,n+1}-\xi_{I,n}}{Δt}=A\frac{\xi_{R,n+1}+\xi_{R,n}}2 $$ has all 4 vectors in each equation, it does not factorize the way you wrote. $$ \begin{bmatrix} 2M&Δt\,A\\ -Δt\,A&2M \end{bmatrix} \begin{bmatrix} \xi_{R,n+1}\\\xi_{I,n+1} \end{bmatrix} = \begin{bmatrix} 2M&-Δt\,A\\ Δt\,A&2M \end{bmatrix} \begin{bmatrix} \xi_{R,n+1}\\\xi_{I,n+1} \end{bmatrix} $$