To this simple Coulomb problem, we will add several corrections:
Correction (1) comes from relativity.
The electron's velocity in hydrogen is of order
It is not very relativistic but a small correction is in order.
the next order relativistic correction to the kinetic energy
we find the additional term in the Hamiltonian
We can estimate the size of this correction compared to the Hydrogen binding energy by taking the
ratio to the Hydrogen kinetic energy.
(Remember that, in the hydrogen ground state,
The second term, due to Spin-Orbit interactions, is harder to derive correctly. We understand the basis for this term. The magnetic moment from the electron's spin interacts with the B field produced by the current seen in the electron's rest frame from the circulating proton.
We can derive
B from a Lorentz transformation of the E field of a static proton
(We must also add in the Thomas Precession which we will not try to understand here).
the relativity correction in first order perturbation theory .
The first order perturbation energy shift from the spin orbit correction is
for the states of definite .
Actually, the term should give 0 for ! In the above calculation there is an factor which makes the result for undefined. There is an additional Dirac Equation contribution called the ``Darwin term'' which is important for and surprisingly makes the above calculation right, even for !
We will now add these three fine structure corrections together for states of definite . We start with a formula which has slightly different forms for .
We can write
, so that
We have calculated the fine structure effects in Hydrogen. There are, of course, other, smaller corrections to the energies. A correction from field theory, the Lamb Shift, causes states of different to shift apart slightly. Nevertheless, the states of definite total angular momentum are the energy eigenstates until we somehow break spherical symmetry.
Jim Branson 2013-04-22