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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!

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    $\begingroup$ 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. $\endgroup$
    – user226006
    Jan 23, 2020 at 23:03
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    $\begingroup$ Are you trying to simulate the nucleus and the electrons or just find the energy levels of the atom? $\endgroup$
    – bemjanim
    Jan 23, 2020 at 23:12
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    $\begingroup$ ah that'll be a complicated hamiltonian then $\endgroup$
    – bemjanim
    Jan 23, 2020 at 23:18
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    $\begingroup$ Nobody does it like this, because the tiny nuclear effects are probably smaller than the numerical errors in solving for the electrons. $\endgroup$
    – G. Smith
    Jan 23, 2020 at 23:44
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    $\begingroup$ Brute force simulation of processes at widely different scales is usually not a productive approach to understanding physical systems. $\endgroup$
    – G. Smith
    Jan 23, 2020 at 23:47

1 Answer 1

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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.

Simulating the nucleus requires quite a lot more elaboration. Now you have 4 particles, making a 12D state space (18 in total). The hamiltonian is about the same but with more interaction terms (see section 3.1), corresponding to nucleon-nucleon potentials that generally look somewhat messy.

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  • $\begingroup$ Thanks a lot for this answer! This was exactly what I was looking for. So I can add the nuclear terms to the Hamiltonian and let it loose, if I understand correctly? $\endgroup$
    – Siddharth Bhat
    Jan 23, 2020 at 23:44
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    $\begingroup$ @SiddharthBhat This equation is already 6-dimensional. Have you ever solved a PDE with 6 or more dimensions on a computer? How many lattice points did you use in each dimension? $\endgroup$
    – G. Smith
    Jan 24, 2020 at 0:16
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    $\begingroup$ @SiddharthBhat How many points per dimension could you fit in your computer for an 18-dimensional computation? Not even 4! $\endgroup$
    – G. Smith
    Jan 24, 2020 at 0:26
  • $\begingroup$ It’s worth mentioning that this Hamiltonian ignores the fact that each electron, because it has spin, also has a (small) magnetic interaction with the proton and with the other electron. $\endgroup$
    – G. Smith
    Jan 24, 2020 at 21:24
  • $\begingroup$ @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. $\endgroup$
    – Ruslan
    Jan 24, 2020 at 22:29

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