Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When the hydrogen atom is solved in QM books spin is usually ignored because its effect is to add tiny piece to the energy. My question is, is there a way to see this in advance, to see that if we included the spin interaction pieces to the Hamiltonian that will contribute so little to the energy spectrum hence we can ignore it for the 1st approximation?

share|cite|improve this question

The electron spin can only enter through magnetic effects, and these are usually smaller than electrostatic effects by about an couple of orders of magnitude. It is therefore a relatively safe assumption when one is trying out the problem for the first time to ignore magnetic, and therefore spin, effects. Of course, there is then the obligation to carry on and compute the fine-structure terms that arise from magnetic and relativistic effects (meaning: you also ignored relativity, without any guarantee that you could. It is also no coincidence that these two terms are about the same order of magnitude), and check that they are indeed just perturbations on the main spectrum.

There is also an additional check you can do before you barge into the electrostatic calculation. Using a semiclassical, Bohr-like model, you can estimate the velocity using the angular momentum, $L\approx\hbar$, and roughly equating the kinetic and Coulomb energies, $\frac{L^2}{2mr^2}\approx \frac{e^2}{r}$, to get $v\approx e^2/\hbar\approx0.007c$ (so $\gamma=1/\sqrt{1-v^2/c^2}\approx1.000025$). This means magnetic and relativistic effects will probably be negligible.

One must also keep in mind, however, that this is valid except when the electrostatic calculation gives 0 as an answer and particularly as an energy difference. The electrostatic model then gives degeneracies that are not actually there - the fine structure calculation breaks the $l$ degeneracy between same-$n$ states. This is usually the case when neglecting small terms - they can always cause finite differences and make exact 0's into approximate ones, which can have important consequences on a theory.

share|cite|improve this answer

By working with dimensionless variables , you can determine what are the dominant terms.

share|cite|improve this answer

Let's start with this: why shouldn't you ignore the spin? The spin magnetic dipole does not couple to the Coulomb potential in the atom, and there's no magnetic field in the hydrogen atom. So why bother with it at all?

And in fact, to a first approximation, this turns out to be pretty good! However, it's not good enough; as you know there is additional structure to the Hydrogen atom, and so we must account for the spin.

So then, we have to grapple with the more difficult question: how the heck does the spin come into play anyways? The answer comes from classical electrodynamics: relativistic covariance reveals a way to transform electric fields into magnetic fields. Of course, this effect is usually only important in the relativistic limit, v/c -> 1. Nevertheless, since the electron moves about the nucleus with some velocity, you can think of a spin-magnetic coupling with this field, which is attenuated by this factor of c. And so this is why the effect is small.

It's hand-wavy, but hopefully somewhat more intuitive.

share|cite|improve this answer Hydrogen atom internal field is about 0.4 tesla. A spin flip is 21 cm, an electron transition is hundreds of nm. Big energy difference. – Uncle Al Feb 8 '14 at 0:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.