Much of this lecture comes from The Fundamental Particles and Their Interactions by W. Rolnick.

Recall from last lecture:

Free Particle Wave Equations

then replacing E and p by their quantum mechanical operators,

E --> ihbar d/dt      p --> -ihbar d/dr
and supplying them with a wave function to operate on. What if we start with the relativistic expression:
E2 = p2c2 + m2c4
and apply the same idea. What we get is known at the Klein-Gordon wave equation:
d2 y/dt2 = (del2 - m2)y
This wave equation is suitable for describing spin 0 bosons.

To demonstrate that the Klein-Gordon equation is relativistically invariant (the fancy term is "manifestly covariant"), we rewrite it such that the time and space derivatives are on equal footing. First rearrange terms to get:

(d2/dt2 - d2/dr2 + m2)y = 0
Now rewrite the combination of second derivatives in the "covariant" form:
(dm dm + m2)y = 0

The Dirac Equation

The Klein-Gordon equation was known in 1926, but discarded because it presents problems. These problems arise because it is a second order differential equation. It can be shown that, in a general quantum mechanical context, the time evolution of a wave function should be governed by a differential equation that is first order in time.

Dirac went looking for a first order relativistic wave equation. One idea is to try starting from the expression E = sqrt{p2 + m2}. While this will give an equation that is first order in t, the p2 term still results in a second derivative in the spatial dimensions. Dirac tried an expression of the form (he probably tried many things, but this is the path that led to the solution):

(a·p + bm)y = id/dty
In this expression, a behaves like a vector, and b is a scalar so that the whole thing is rotationally invariant. The right hand side is the energy operator (with hbar=1), and the left hand side is called the Dirac Hamiltonian:
HDirac = a·p + bm
The energy and momentum must still be related by E2 = p2 + m2, which must be the result when the Dirac Hamiltonian is applied twice:
HDiracHDirac = E2 = p2 + m2 = (a·p + bm)(a·p + bm)
Expanding this expression, with an obvious notation for the components of a yields
(a1p1 + a2p2 + a3p3 + bm)(a1p1 + a2p2 + a3p3 + bm) = p12 + p22 + p32 + m2
Looking at products like pi2 and m2 on the left, we readily see that the a12 = 1 and b2 = 1.

Next look at cross terms like p1p2 which are zero on the right hand side:

(a1a2 + a2a1)p1p2 = 0.
Recall that p represents an operator, so p1p2 is also just an operator. The only way the expression can be zero is if we require:
a1a2 + a2a1 = 0
Look closely and you will see that similar expressions occur for the other cross terms. This equation cannot be satisfied if b and the ai are merely numbers, they must be something else. The relation above is an anticommutator relation, similar to the standard commutator relations of quantum mechanics. We normally see these in quantum mechanics for operators such as x, or p. But b and the ai cannot be operators -- at least the argument we are following doesn't make sense if they are. They must be constants, and the kind that anticommute. Matrices satisfy these requirements.

The usual choice for the matrices are the so called gamma matrices, with:

b = g0, a1 = g1, a2 = g2, a3 = g3
so that the collection of gamma matrices can be treated as a (constant) 4-vector. The gamma matrices are 4×4 matrices with zero trace. Written out explicitly, they are:
g0 = 1000
0100
00-10
000-1
   
g1 = 0001
0010
0100
1000
 
g2 = 000-i
00i0
0-i00
i000
   
g3 = 0010
000-1
1000
0-100
You can check that these agree with the shorthand notation given in Eq. 1.20c in Perkins.

Now we can express the Dirac equation in the manifestly covariant form:

(igm d/dxm - m)y = 0
With the addition of matrices to the wave equation, the wave function y is now a four-component column vector! This column vector is often called a spinor.

To span the space of the four-component spinor, we need four basis vectors. Let's make the simple choice of the four spinors with three elements equal to 0 and one equal to 1. Further, consider the situation where a particle is at rest so that

HDirac = g0 m = m000
0m00
00-m0
000-m
Now apply this simple Hamiltonian to a basis state, to find:
1  =  m000 · 1  =  m  = m  1
00m00000
000-m0000
0000-m000
Therefore, the first basis spinor is an eigenstate of the zero-momentum Hamiltonian with energy m. This is generally what we would hope for. But the Dirac equation has three more basis states, which it is easy to see are also eigenstates of the zero-momentum Hamiltonian. The second also has energy m, while the last two have energy -m.

This raises two questions:

First, let's consider the two states of positive energy. It can be shown that they correspond to particles of opposite spin. Without inserting anything by hand, the Dirac equation "predicts" spin, specifically, spin ½.

The two negative energy states also correspond to particles of opposite spin. But what is a negative energy particle? We don't know of any negative energy particles, so can we just toss out those two components? Without those two components, the Dirac equation no longer works properly, so they must be kept. Dirac borrowed an idea from solid state physics to explain the (non)existence of negative energy particles. He suggested that there is a negative energy sea that is normally completely filled. Since spin-½ particles are fermions, a positive energy fermion cannot fall into the negative energy sea, the same way that an electron orbiting an atom can't fall into an occupied orbital. However if enough energy were supplied to an electron in the negative sea, it could be raised to a state of positive energy, creating a positive energy electron and leaving behind a vacancy (hole) in the negative energy sea. The vacancy would behave like a particle with identical mass but opposite charge to the electron, i.e. an antiparticle of the electron.

The Dirac equation "predicts" the creation and annihilation of particle-antiparticle pairs. This can be seen when, for example, an electron impinges on a potential barrier. For some incoming energies, the probability for an electron to be reflected from the barrier is greater than unity. This fact connects the Dirac wave equation to further refinements that we will discuss next.

For a detailed derivation of the Dirac equation refer to a text on advanced quantum mechanics or field theory, a number of which are listed in the references page.


Copyright © Robert Harr 2003