3d schrodinger equation weak form

Question

SCHRODINGER’S EQUATION

$$-ih u_{t}(x,y,z,t) = \frac{h^2}{2m} u_{xx}(x,y,z,t)+ \frac{e^2}{r}u(x,y,z,t)$$ The potential $\frac{e^2}{r}$ is a variable coefficient. So, let’s take the free Schrodinger equation $$-i u_{t}(x,y,z,t) = \frac{1}{2} u_{xx}(x,y,z,t)$$ in three dimensions, where we’ve set $h = m = 1$ and dropped the potential term. It looks like the diffusion equation.

\begin{align*} -i u_{t}(x,y,z,t) &= \frac{1}{2} u_{xx}(x,y,z,t)\\ -i u_{t}(x,y,z,t) - \frac{1}{2} u_{xx}(x,y,z,t)&=0\\ \end{align*} and also setting $k = \frac{i}{2}$ we have the equation: $$ u_{t}(x,y,z,t) - k u_{xx}(x,y,z,t)=0$$ or : $$ u_{t} - k \Delta u =0 $$

with $$u=0, \quad \text{in}\quad \partial \Omega$$ In the weak form : Let $v \in H'_{0}(\Omega)$

\begin{align*} \int_{\Omega}u_{t}v dV - k \int_{\Omega}\Delta u v dV&= 0 \\ \int_{\Omega}u_{t}v dV -k \int_{\Omega}\nabla u \cdot \nabla v dV& - k\int_{\partial \Omega} \nabla u \cdot \vec{n}v dS &= 0 \\ \int_{\Omega}u_{t}v +k\int_{\Omega}\nabla u \cdot \nabla v dV &=0 \end{align*}

The problem is how I treat the $\frac{i}{2}$ ?

Answer 1

The solutions of the Schroedinger equation are complex-valued, so your inner product needs to be $$ (u,v) = \int_\Omega \bar u(x) v(x)\; dx, $$ and the norms then become $$ \|u\| = (u,u)^{1/2}. $$

But it is also worth pointing out that you shouldn't think of the Schroedinger equation as a variation of the heat equation that just happens to have an imaginary diffusion coefficient. In reality, the Schroedinger equation is more akin to a wave equation. I have written the argument for this up in one of the deal.II tutorial programs here: https://dealii.org/developer/doxygen/deal.II/step_58.html (Isn't it funny how there is a deal.II tutorial for nearly every purpose? ;-) )