# How do I simulate an atom?

Let us assume I wish to simulate a Helium atom, since there does not exist a closed-form solution.

However, I presume I would need to simulate the time-dependent Schrodinger wave equation. I would like to know what the time-dependent hamiltonian looks like for this.

I think I would need the terms for electromagnetic, strong, and weak nuclear forces --- I do not think I need relativistic corrections, since I'm only interested in local information.

I wish to simulate an equation on simulation gives me the evolving probability distributions of the electrons.

I understand that this might be computationally infeasible, but I still wish to know what the exact PDE is that I need to solve --- I'm not looking for approximations!

• How much QM background have you got and why do you think you need to include any forces other than electromagnetic? That said, best of luck with it. Jan 23 '20 at 23:03
• Are you trying to simulate the nucleus and the electrons or just find the energy levels of the atom? Jan 23 '20 at 23:12
• ah that'll be a complicated hamiltonian then Jan 23 '20 at 23:18
• Nobody does it like this, because the tiny nuclear effects are probably smaller than the numerical errors in solving for the electrons. Jan 23 '20 at 23:44
• Brute force simulation of processes at widely different scales is usually not a productive approach to understanding physical systems. Jan 23 '20 at 23:47

The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ where the electrons are denoted 1 and 2, and $$r_i$$ is the distance to the nucleus at the origin and $$r_{12}$$ the distance between the electrons. Since the electrons have 3D position vectors $$\mathbf{r_i}$$ this corresponds to 6 degrees of freedom.
One can reformulate the equation to only act in $$r_1,r_2,r_{12}$$ coordinates for the spherically symmetric $$^1S$$ state. With the right approximation methods one can get energy eigenvalues with absurd precision. But I suspect the aim of the question is rather to let loose an electrong wave packet and see how it sloshes around.
• @G.Smith the OP wanted to avoid relativistic corrections, and magnetic term definitely has a $c^{-1}$ factor. Otherwise you'd need to also include all the other relativistic corrections. Jan 24 '20 at 22:29