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I'm studying up on methods for numerically solving the Schrodinger equation. The Schrodinger equation with a zero potential is formally identical to the heat equation in the sense that we just make one of the coefficients in the equation imaginary.

A practitioner of finite-difference methods for the heat equation has to be aware of certain things. From a casual reading of some Wikipedia articles, the main things I've learned are that:

  1. the forward method can be unstable;

  2. Crank-Nicolson has better asymptotics; and

  3. I need to satisfy the Courant condition in the Crank-Nicolson method.

Do any of these facts change if I switch to the Schrodinger equation with a nonzero potential?

In the Schrodinger equation, a kink in the wavefunction contains arbitrarily short wavelengths and therefore has components that propagate with arbitrarily high velocities. Does this mean that the Courant condition can never be satisfied if there is a kink? In a discretized representation, however, there is no clear distinction between a kink and a differentiable point, so is there some adaptive criterion that can be used to set an appropriate time step?

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  • $\begingroup$ I think we'll need to review some of those "facts" first. What do you mean by "the forward method"? What do you mean by "Crank-Nicolson has better asymptotics"? Where did you find on Wikipedia that Crank-Nicolson has the Courant condition? We should edit that article. Why do you think a wavefunction can have a kink? That would mean infinite kinetic energy, unless the wavefunction vanishes at that point. If there are no delta-functions in the potential, the wavefunction should be smooth. $\endgroup$ – Maxim Umansky Jul 3 at 4:32
  • $\begingroup$ What do you mean by "the forward method"? en.wikipedia.org/wiki/Finite_difference_method#Explicit_method What do you mean by "Crank-Nicolson has better asymptotics"? Its errors are proportional to the square of the time step, whereas the forward method's errors are proportional to the time step. Where did you find on Wikipedia that Crank-Nicolson has the Courant condition? Same article: "The von Neumann and Courant-Friedrichs-Lewy criteria are often evaluated to determine the numerical model stability." If I'm interpreting this wrong, it would be good to know that. $\endgroup$ – Ben Crowell Jul 3 at 17:24
  • $\begingroup$ I think that you could add some clarification to your question. Comments are not intended to discuss but to clarify questions. For example, you could say "explicit Euler" or "forward Euler" instead of "forward method". $\endgroup$ – nicoguaro Jul 3 at 18:38
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The potential does not pose any issues in practice. It's the Laplace term. The nonlinear term is also awkward if you have one.

I've written a lengthy discussion of how one solves the (nonlinear) Schroedinger equation here. You will see that the statement "The Schrodinger equation with a zero potential is formally identical to the heat equation in the sense that we just make one of the coefficients in the equation imaginary." might be "formally" true, but that doesn't mean that any of the properties of the heat equation carry over. In particular, while the heat equation is diffusive, the Schroedinger equation is not and instead has a wave equation character.

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