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 > iħ d/dt p > iħ d/dr
and supplying them with a wave function to operate on.
What if we start with the relativistic expression:
E^{2} = p^{2}c^{2} + m^{2}c^{4}
and apply the same idea.
What we get is known at the KleinGordon wave equation:
δ^{2}ψ/δt^{2} = (del^{2}  m^{2})ψ
This wave equation is suitable for describing spin 0 bosons.
To demonstrate that the KleinGordon 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:
(δ^{2}/δt^{2}  δ^{2}/δr^{2} + m^{2})ψ = 0
Now rewrite the combination of second derivatives in the "covariant" form:
(δ^{μ} δ_{μ} + m^{2})ψ = 0
The Dirac Equation
The KleinGordon 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{p^{2} + m^{2}}.
While this will give an equation that is first order in t, the p^{2} 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):
(α·p + βm)ψ = i(δ/δt) ψ
In this expression, α behaves like a vector, and β is a scalar so that the whole thing is rotationally invariant.
The right hand side is the energy operator (with ħ=1), and the left hand side is called the Dirac Hamiltonian:
H_{Dirac} = α·p + βm
The energy and momentum must still be related by E^{2} = p^{2} + m^{2}, which must be the result when the Dirac Hamiltonian is applied twice:
H_{Dirac}H_{Dirac} = E^{2} = p^{2} + m^{2} = (α·p + βm)(α·p + βm)
Expanding this expression, with an obvious notation for the components of α yields
(α_{1}p_{1} + α_{2}p_{2} + α_{3}p_{3} + βm)(α_{1}p_{1} + α_{2}p_{2} + α_{3}p_{3} + βm) = p_{1}^{2} + p_{2}^{2} + p_{3}^{2} + m^{2}
Looking at products like p_{i}^{2} and m^{2} on the left, we readily see that the α_{1}^{2} = 1 and β^{2} = 1.
Next look at cross terms like p_{1}p_{2} which are zero on the right hand side:
(α_{1}α_{2} + α_{2}α_{1})p_{1}p_{2} = 0.
Recall that p represents an operator, so p_{1}p_{2} is also just an operator.
The only way the expression can be zero is if we require:
α_{1}α_{2} + α_{2}α_{1} = 0
Look closely and you will see that similar expressions occur for the other cross terms.
This equation cannot be satisfied if β and the α_{i} 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 β and the α_{i} 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:
β = γ_{0}, α_{1} = γ_{1}, α_{2} = γ_{2}, α_{3} = γ_{3}
so that the collection of gamma matrices can be treated as a (constant) 4vector.
The gamma matrices are 4×4 matrices with zero trace.
Written out explicitly, they are:
γ_{0} =  1  0  0  0 
0  1  0  0 
0  0  1  0 
0  0  0  1 


γ_{1} =  0  0  0  1 
0  0  1  0 
0  1  0  0 
1  0  0  0 


γ_{2} =  0  0  0  i 
0  0  i  0 
0  i  0  0 
i  0  0  0 


γ_{3} =  0  0  1  0 
0  0  0  1 
1  0  0  0 
0  1  0  0 

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:
(i γ_{μ} δ/δx_{μ}  m)ψ = 0
With the addition of matrices to the wave equation, the wave function ψ is now a fourcomponent column vector!
This column vector is often called a spinor.
To span the space of the fourcomponent 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
H_{Dirac} = γ_{0} m =  m  0  0  0 
0  m  0  0 
0  0  m  0 
0  0  0  m 
Now apply this simple Hamiltonian to a basis state, to find:
H  1  =  m  0  0  0  ·  1  =  m  = m  1 
0  0  m  0  0  0  0  0 
0  0  0  m  0  0  0  0 
0  0  0  0  m  0  0  0 
Therefore, the first basis spinor is an eigenstate of the zeromomentum 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 zeromomentum Hamiltonian.
The second also has energy m, while the last two have energy m.
This raises two questions:
 Why are there four eigenstates for a particle at rest?
 Why do two of the states have negative energy?
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 spin½ 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 particleantiparticle 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 2005