D.82 Dirac fine structure Hamiltonian

This note derives the fine structure Hamiltonian of the hydrogen atom. This Hamiltonian fixes up the main relativistic errors in the classical solution of chapter 4.3. The derivation is based on the relativistic Dirac equation from chapter 12.12 and uses nontrivial linear algebra.

According to the Dirac equation, the relativistic Hamiltonian and wave
function take the form

where is the mass of the electron when at rest, the speed of light, and the are the 2 2 Pauli spin matrices of chapter 12.10. Similarly the ones and zeros in the shown matrices are 2 2 unit and zero matrices. The wave function is a four-dimensional vector whose components depend on spatial position. It can be subdivided into the two-dimensional vectors and . The two components of correspond to the spin up and spin down components of the normal classical electron wave function; as noted in chapter 5.5.1, this can be thought of as a vector if you want. The two components of the other vector are very small for the solutions of interest. These components would be dominant for states that would have negative rest mass. They are associated with the anti-particle of the electron, the positron.

The Dirac equation is solvable in closed form, but that solution is not something you want to contemplate if you can avoid it. And there is really no need for it, since the Dirac equation is not exact anyway. To the accuracy it has, it can easily be solved using perturbation theory in essentially the same way as in derivation {D.80}. In this case, the small parameter is 1/: if the speed of light is infinite, the nonrelativistic solution is exact. And if you ballpark a typical velocity for the electron in a hydrogen atom, it is only about one percent or so of the speed of light.

So, following derivation {D.80}, take the Hamiltonian apart into
successive powers of 1 as with

and similarly for the wave function vector:

and the energy:

Substitution into the Hamiltonian eigenvalue problem and then collecting equal powers of 1 together produces again a system of successive equations, just like in derivation {D.80}:

The first, order , eigenvalue problem has energy
eigenvalues , in other words, plus or minus the
rest mass energy of the electron. The solution of interest is the
physical one with a positive rest mass, so the desired solution is

Plug that into the order equation to give, for top and bottom subvectors

It follows from the first of those that the first order energy change must be zero because cannot be zero; otherwise there would be nothing left. The second equation gives the leading order values of the secondary components, so in total

where the summation index was renamed to to avoid ambiguity later.

Plug all that in the order equation to give

The first of these two equations is the nonrelativistic Hamiltonian eigenvalue problem of chapter 4.3. To see that, note that in the double sum the terms with pairwise cancel since for the Pauli matrices, 0 when . For the remaining terms in which , the relevant property of the Pauli matrices is that is one (or the 2 2 unit matrix, really,) giving

where is the nonrelativistic hydrogen Hamiltonian of chapter 4.3.

So the first part of the order equation takes the form

The energy will therefore have to be a Bohr energy level and each component of will have to be a nonrelativistic energy eigenfunction with that energy:

The sum multiplying is the first component of vector and the sum multiplying the second. The nonrelativistic analysis in chapter 4.3 was indeed correct as long as the speed of light is so large compared to the relevant velocities that 1 can be ignored.

To find out the error in it, the relativistic expansion must be taken
to higher order. To order , you get for the top vector

Now if is written as a sum of the eigenfunctions of , including , the first term will produce zero times since 0. That means that must be zero. The expansion must be taken one step further to identify the relativistic energy change. The bottom vector gives

To order , you get for the top vector

and that determines the approximate relativistic energy correction.

Now recall from derivation {D.80} that if you do a
nonrelativistic expansion of an eigenvalue problem
, the equations to solve are (D.55) and
(D.56);

The first equation was satisfied by the solution for obtained above. However, the second equation presents a problem. Comparison with the final Dirac result suggests that the fine structure Hamiltonian correction should be identified as

but that is not right, since is not a physical operator, but an energy eigenvalue for the selected eigenfunction. So mapping the Dirac expansion straightforwardly onto a classical one has run into a snag.

It is maybe not that surprising that a two-dimensional wave function cannot
correctly represent a truly four-dimensional one. But clearly,
whatever is selected for the fine structure Hamiltonian must at
least get the energy eigenvalues right. To see how this can be done,
the operator obtained from the Dirac equation will have to be
simplified. Now for any given , the sum over includes a
term , a term , where is
the number following in the cyclic sequence
, and it involves a term
where precedes in the sequence. So the Dirac operator
falls apart into three pieces:

or using the properties of the Pauli matrices that 1, , and for any ,

The approach will now be to show first that the final two terms are
the spin-orbit interaction in the fine structure Hamiltonian. After
that, the much more tricky first term will be discussed. Renotate
the indices in the last two terms as follows:

Since the relative order of the subscripts in the cycle was maintained in the renotation, the sums still contain the exact same three terms, just in a different order. Take out the common factors;

Now according to the generalized canonical commutator of chapter 4.5.4:

where is a constant that produces a zero derivative. So , respectively can be taken to the other side of as long as the appropriate derivatives of are added. If that is done, and cancel since linear momentum operators commute. What is left are just the added derivative terms:

Note that the errant eigenvalue mercifully dropped out. Now the hydrogen potential only depends on the distance from the origin, as 1/, so

and plugging that into the operator, you get

The term between the square brackets can be recognized as the th component of the angular momentum operator; also the Pauli spin matrix is defined as , so

Get rid of using , of using , and using to get the spin-orbit interaction as claimed in the section on fine structure.

That leaves the term

in (D.59). Since , it can be written as

The final term is the claimed Einstein correction in the fine structure Hamiltonian, using to get rid of .

The first term,

is the sole remaining problem. It cannot be transformed into a decent physical operator. The objective is just to get the energy correction right. And to achieve that requires only that the Hamiltonian perturbation coefficients are evaluated correctly at the energy level. Specifically, what is needed is that

for any arbitrary pair of unperturbed hydrogen energy eigenfunctions and with energy . To see what that means, the leading Hermitian operator can be taken to the other side of the inner product, and in half of that result, will also be taken to the other side:

Now if you simply swap the order of the factors in in this expression, you get zero, because both eigenfunctions have energy . However, swapping the order of brings in the generalized canonical commutator that equals . Therefore, writing out the remaining inner product you get

Now, the potential becomes infinite at 0, and that makes mathematical manipulation difficult. Therefore, assume for now that the nuclear charge is not a point charge, but spread out over a very small region around the origin. In that case, the inner product can be rewritten as

and the first term integrates away since vanishes at infinity. In the final term, use the fact that the derivatives of the potential energy give times the electric field of the nucleus, and therefore the second order derivatives give times the divergence of the electric field. Maxwell’s first equation (13.5) says that that is times the nuclear charge density. Now if the region of nuclear charge is allowed to contract back to a point, the charge density must still integrate to the net proton charge , so the charge density becomes where is the three-dimensional delta function. Therefore the Darwin term produces Hamiltonian perturbation coefficients as if its Hamiltonian is

Get rid of using , of using , and using to get the Darwin term as claimed in the section on fine structure. It will give the right energy correction for the nonrelativistic solution. But you may rightly wonder what to make of the implied wave function.