When solving the Schrodinger equation for the hydrogen atom, the Coulomb potential $V = \frac{e^2}{4 \pi \epsilon_0 r}$ is used.

  1. The Coulomb potential comes from classical electrodynamics, so why do we use it when solving the non-relativistic hydrogen atom?

  2. Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics?

  • $\begingroup$ As a side note, Maxwell's equations come from classical electrodynamics but are used in quantum electrody amics. $\endgroup$
    – jinawee
    Dec 9 '14 at 16:57

You say:

The Coulomb potential comes from classical electrodynamics

but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square law. However if you look at the scattering between e.g. two electrons and calculate the low energy limit you find the result is the Coulomb potential. You'll find the calculation in most QFT textbooks, though it's likely to be completely opaque to non-nerds.

So we expect the Coulomb potential to be an excellent approximation as long as the energies involved are low. As a rough guide we expect relativistic effects to become important when the energies are comparable to the rest mass of the charged particles, so for electrons we expect deviations from the Coulomb law at energies of around 1MeV. If you look at a hydrogen atom the lowest energy orbital is only 13.6eV or about a factor of 100,000 times less than the relativistic energy, and that's why we can use the Coulomb potential without worrying.

The very heaviest atoms, e.g. the actinides, have $1s$ electron energies greater than 0.1MeV, and for these atoms relativistic corrections are indeed significant. However they are still small enough that we start with a simple Coulombic description and then treat the relativistic effects as perturbations.

  • 1
    $\begingroup$ It's unsurprising that the Coulomb potential falls out of quantum electrodynamics in the low energy limit. But quantum electrodynamics came after non-relativistic quantum mechanics a la Schrodinger (wave mechanics) and Heisenberg (matrix mechanics). So why was the Coulomb potential used by Schrodinger when the justification for using the Coulomb potential in the low energy limit came later? Was Schrodinger just doing the best he could with what was known at the time? $\endgroup$
    – Cody
    Dec 9 '14 at 21:37
  • 1
    $\begingroup$ At the time the nature of the Coulomb force wasn't understood. Maxwell had given a description of it, but not a fundamental mechanism. There was no reason to suppose it wasn't universal. $\endgroup$ Dec 10 '14 at 7:01

Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics?

Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework.

Now it is time to specialize to the hydrogen atom for which

$$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$

(more than halfway down the article.)

The Coulomb potential is an experimental fact. That it is successful in describing atomic physics interactions shows that it is still an experimental fact.


No, it is an approach called semiclassical quantum Mechanics. They mix the Schrödinger equation in which quantum particles interact with the classical potentials, to see what the calculations does. And it happens that it gives good results. So even if flawed, it is still predictive enough to be considered good. For instance, it predicts the fine structure of hydrogen!


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