The Dirac Equation

In classical physics we devise equations of motion that we can solve to obtain a description of a particle’s motion through space with the elapse of time. Those equations and their solutions conform well to our intuitions about force and its effect on the motions of bodies. In quantum mechanics we devise equations of evolution that we can solve to obtain a description of how a partial-probability field changes with the elapse of time. One of those fields, the matter field, connects quantum mechanics with classical mechanics by way of Born’s theorem.

Those equations come to us by way of the Euler-Lagrange equations, from the principle of least action (or of least action density in the quantum case). That principle mandates finding the minimum value taken by the time integral of a Lagrangian function, which is a Lorentz invariant, the dot product of two four-vectors. In the classical case we get the Lagrangian by multiplying the momentum-energy vector by the space-time vector and we get the Lagrangian as the kinetic energy minus the potential energy of the system. In the quantum theory we cobble together Lagrangian densities by a similar procedure and then use the Euler-Lagrange equations to devise the desired equations of evolution.

In the present case we multiply the momentum-energy four-vector acting on the partial probability by an unknown four-vector-like object acting on the adjoint of the partial probability, as if we were about to calculate an expectation value through Born’s theorem. That gives us the kinetic energy part of the Lagrangian density. Then we subtract out the rest mass-energy, which we treat as a potential energy because it represents energy that does not involve actual motion.

(Eq’n 1)

In carrying out the dot multiplication I used the Minkowski space-time metric tensor in the form

(Eq’n 2)

That tensor tells us that time is the zeroth coordinate, instead of the fourth, and it’s necessary that we have it that way in order to make the kinetic energy come out with a positive value. Applying the Euler-Lagrange equations with respect to the over-barred psi to Equation 1 gives us

(Eq’n 3)

We can rewrite the operator part of that equation as

(Eq’n 4)

But we know from relativistic dynamics that

(Eq’n 5)

so we also know that

(Eq’n 6)

must stand in direct proportion to it.

No elements of the infinite set of the complex numbers can stand in that equation for alpha, beta, and/or eta. The result of such a substitution would not conform to Equation 5. In particular, the cross terms wouldn’t vanish, as they must. In order to have

(Eq’n 7)

we must have entities that obey an anticommutation rule when multiplied
together: we must have matrices. If we equate Equations 5 and 6 to each other,
then we must have β^{2}=η^{2}=I,
the identity matrix, because beta does not equal eta. For simplicity we make the
substitutions γ^{0}=η
and (γ^{1},γ^{2},γ^{3})=α,
so that we have the anticommutator relation

(Eq’n 8)

But the metric tensor is a 4x4 matrix, so the gammas must also be 4x4 matrices. The set of the gamma matrices is not unique: a number of quartets will satisfy Equation 8. For convenience I’ll use the Bjorken and Drell convention used by most physicists; that is, I’ll use the matrices

(Eq’ns 9)

in which the sigmas represent the 2x2 Pauli spin matrices.

We can substitute these matrices into Equation 3 and also make the usual quantum-theory substitutions

(Eq’ns 10)

in the form

(Eq’n 11)

and we get

(Eq’n 12)

That’s the Dirac equation and it’s clear that the state function that solves it must be a four-component column matrix known as a Dirac spinor.

Because the Dirac equation contains only first-order derivatives, we can solve it readily. Assume that the spatial derivatives of the state functions equal zero; that is, assume that the particle under consideration has no linear momentum. The Dirac equation then becomes

(Eq’n 13)

In that equation the state function must have the form of a 1x4 column vector
with components (ψ_{1},ψ_{2},ψ_{3},ψ_{4}).
We carry out the indicated matrix multiplication and get four equations:

(Eq’ns 14)

If we divide each of those equations by its state function, multiply it by dt, and carry out the implied integration, then we can work out the solutions:

(Eq’ns 15)

In those equations the subscripted upper-case cees represent amplitude factors that do not depend upon the elapse of time, but which may depend upon other variables. The exponentials are simply the energy-time wave functions that we see in solutions of Schrödinger’s equation, but the exponentials in the third and fourth of Equations 15 represent states with negative energy, something that cannot legitimately exist.

In an electric field the particles described by the first two solutions accelerate in one direction and the particles described by the last two solutions accelerate in the opposite direction. If the first two solutions describe electrons, then the last two solutions must describe positrons, the antimatter counterparts of electrons. Dirac stated that positrons cannot be real particles in the same sense that electrons are real particles, pointing out that if they were real particles, then the dynamics would go wrong, with particles having, for instance, negative kinetic energy, which is an absurdity. He then asserted that positrons must be holes poked into space by the creation of electrons. Richard Feynman hypothesized that positrons are electrons moving backward in time and used that hypothesis to make a significant contribution to the theory of quantum electrodynamics. In either case, when an electron meets a positron the two particles annihilate each other and their mass-energy gets re-manifested in a pair of 511-kev gamma photons.

Equations 15 give us two states for each of the particles, electron and positron, and we want an explanation for that fact. To get that explanation let’s put linear momentum into the system. The Dirac equation then gives us twelve sub-equations. With respect to the x-axis we have

(Eq’ns 16)

with respect to the y-axis we have

(Eq’ns 17)

and with respect to the z-axis we have

(Eq’ns 18)

We won’t be able to solve those equations by using the procedure that we used to solve Equations 14. The main difficulty lies in the fact that positron momentum states are mixed up with electron existence states and vice versa.

Let’s go back to the Dirac equation, only this time we express the gamma matrices as 2x2 arrays involving the Pauli spin matrices. This is quantum mechanics, so we want a plane-wave solution of the form

(Eq’n 19)

assuming that we can use Fourier’s theorem to create a real-world solution from the waves. In that equation we have

(Eq’ns 20)

representing the electron and positron states respectively. We thus get the Dirac equation as

(Eq’n 21)

The energy and mass terms implicitly involve the 2x2 identity matrix. Carrying out the indicated matrix multiplication gives us two equations,

(Eq’ns 22)

So now we know that

(Eq’ns 23)

We have explicitly

(Eq’n 24)

(Multiply that by itself and see what you get.) With four possible eigenvalue
sets for ϕ_{A}
and ϕ_{B},

(Eq’ns 25)

we have, by way of Equations 23, four solutions to the Dirac equation: Equation 19 becomes

(Eq’ns 26)

in which N represents a normalization factor.

The first two of those solutions describe particles with positive energy spinning through space. The first of those represents a particle traveling with spin up (oriented in the particle’s direction of motion) and the other represents a particle traveling with spin down (oriented against the direction of motion). The second pair of those solutions represent particles with negative energy; that is, particles of antimatter, interpreted as particles traveling backward in time. The first of those represents an antiparticle traveling with spin up and the other describes an antiparticle traveling with spin down.

The normalization factor must conform to the statement that

(Eq’n 27)

To work out the value of N we simply pick one of the solutions above and carry out the indicated multiplication, obtaining if we choose the first solution

(Eq’n 28)

That factor gives ψ*ψ units of linear momentum. Without it ψ*ψ is simply a number, as in the non-relativistic quantum theory.

We notice in Equations 26 the peculiar form of the spinors. In the spinors designating the electron spin states we see momentum-energy terms in the places that designate the positron spin states and in the spinors designating the positron spin states we see momentum-energy terms in the places that designate the electron spin states. That fact tells us that the spinor does not represent proper spin eigenstates unless and only unless the particle’s momentum equals zero. But the more important purpose of the momentum-energy terms comes clear when we use the spinors in calculating descriptions of particle - particle interactions by way of an expanded version of Born’s theorem.

So now we have the Dirac equation and its plane-wave solutions for a free particle.

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