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.


1 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. $$ \begin{bmatrix} 2M&Δt\,A\\ -Δt\,A&2M \end{bmatrix} \begin{bmatrix} \xi_{R,n+1}\\\xi_{I,n+1} \end{bmatrix} = \begin{bmatrix} 2M&-Δt\,A\\ Δt\,A&2M \end{bmatrix} \begin{bmatrix} \xi_{R,n+1}\\\xi_{I,n+1} \end{bmatrix} $$

  • $\begingroup$ Thank you so much! I can't believe I made such mistakes... One more question. Is there a simple way to solve this problem? Sinec M and A are matrix themselves, it looks difficult to solve matrix of a matrix.... $\endgroup$ Dec 8, 2021 at 13:19
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    $\begingroup$ No, apart from returning to the complex domain, I see no simplifications. But the matrices from a finite-element method are usually sparse, this should reduce the effort, in space more than in time, to construct the large matrix and solve the system. $\endgroup$ Dec 8, 2021 at 14:07
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    $\begingroup$ You think of it as a matrix of matrices, but the way you should really think about it is as one large matrix that you can partition into the block structure you see here. It isn't a matrix of matrices, but just one big matrix. $\endgroup$ Dec 9, 2021 at 3:35

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