The **Dirac equation** for a spin ½ particle
is of the form

For a free particle

with a_{x}^{2}=a_{y}^{2}=a_{z}^{2}=b^{2}=1
and all four quantities a_{x}, a_{y}, a_{z}, and b anti-commuting in pairs.

For example a_{x}a_{y}+a_{y}a_{x}=0.

Since **a** and b
anti-commute, they cannot be numbers. For a spin ½ particle a_{x}, a_{y}, a_{z}, and b are represented by 4´4 matrices.

In compact notation

where each entry is a 2´2 matrix.

Given **a** and b as 4´4 matrices, *y*(* r,t*)
must be a matrix with 4 rows and 1 column.

is equivalent to 4 coupled first order, linear, homogeneous, partial differential
equations for the 4 *y*’s. Plane wave solutions of
the form *y _{j}(r,t)=u_{j}exp(i(k* ×

Substituting these plane wave solutions into yields a set of algebraic equations for the *u _{j}*, where are now numbers, not operators.

.

.

.

.

These equations are homogeneous in the *u _{j}* and have a solution only if
the determinant of the coefficients is zero. This determinant equals
Explicit solutions can be obtained for any

If we choose the negative square root, then we obtain two different solutions, which are written as

Each of these 4 solutions can be normalized in the sense that *yy**=1
by dividing it by

In the non relativistic limit *E _{+}* is close to

The Dirac Hamiltonian for a spin ½ particle in the presence of an electromagnetic field is

and the time independent Dirac equation may be written as

in SI units. For a particle with charge q in a spherically symmetric scalar potential we have

Assume i.e. neglect
the spin of the proton. (*q _{e}*=1.6×10

Assume

Then the orbital angular momentum * L* is not a constant of motion.

In the Heisenberg picture we have

Let us define the operator s’, which is represented by the matrix

The operator s’ is not a constant of motion. We have

But we can see that the operator is a constant of motion. It commutes with *H _{D}*.
This
operator is interpreted as the

Let in compact notation, both
*y*_{1} and *y*_{2}
having two components. We may then write the eigenvalue equation for *H _{D}*
,

as

This yields

and

We can use the second equation to obtain

and substitute this expression for y_{2} into the
first equation. We the obtain

or

In the non relativistic limit

In the above equation * p* is an operator,
Therefore

using the general result

[We can apply this equation to the operators since * s* commutes with

Therefore

For a spherically symmetric potential *V=-q _{e}f*
we have

In the non relativistic limit we also have and Therefore

where

The first and third term give the non relativistic Schroedinger equation, the second
term has the form of the classical relativistic correction, the last term is the
spin-orbit energy, which appears as an automatic consequence of the Dirac equation.
The
fourth term is a similar relativistic correction to the potential energy, which does not
have a classical analog. It is called the Darwin term.

The ratio of < all the correction terms > to < > is on the order of *v ^{2}/c^{2}*.

[Note that *y*_{1} is the
two component spinor as used in the non relativistic description of a spin ½ particle.]

If we do not neglect the spin and the magnetic moment of the proton, we cannot set * A(r,t)=0*.
We now have to replace

We assume that <* q _{e}A* > << <

We have in general Therefore

[ is an operator. When it operates on a wavefunction y we have

]

The new term therefore is

and

The first term contains the spin-orbit interaction energy of the proton and the last
term the spin-spin interaction energy. Both terms are **hyperfine
structure terms**.

The **hyperfine structure Hamiltonian** (neglecting
terms in *(q A)*

It may be rewritten as

since * A* is the vector potential due to the magnetic moment of the proton and (See Cohen-Tannoudji, Complement A

The **fine structure Hamiltonian** is

* *

*H = H _{0} + W_{f} + W_{hf}*.

The ratio is on the order of