## A p 3 m a p 3 mf xp

= (p2 +m2)i> . Demanding that this relation hold for all ip gives

where {A, B} = AB + BA is the anticommutator of two operators A and B. This equation can hold only if

To construct such an equation therefore requires a vector space large enough to contain four anticommuting, Hermitian matrices.

It is easy to prove that such a space must have a minimum of four dimensions and that therefore the matrices and ¡3 must be at least 4x4. The proof follows in four steps:

Lemma 1: The matrices 0 and al are traceless.

To prove this for the matrices a„ note that the anticommutation relations imply

Making use of the fact that the trace of a product of matrices is unchanged by cyclic permutation of the matrices gives tr {/toi/3} = -tra,

Hence tr a, = 0. A similar argument shows that tr (3 = 0. Lemma 2: The eigenvalues of Qj and (3 must be ±1.

Since ai and 0 are Hermitian, they can be diagonalized, and because (at)2 = /32 = 1, their diagonal elements (eigenvalues) can only be ±1.

Lemma 3: The dimension of the matrices must be even.

We will use the non-covariant form of the Dirac equation in the next few sections and will return to the covariant form in Sec. 5.9 when we discuss the covariance of the equation.

5.2 CONSERVED NORM

The conserved norm is easily obtained from the equations. Note that, for any two solutions of the Dirac equation, i-Q^Pb = rpiHipb

Hence, if the electromagnetic interaction (or any other potential) is independent of energy, it will cancel when we subtract the above two equations, and subtracting the first from the second gives i^- (H>hpb) = -ii>l<Xi i>b where the arrow over the derivative tells us in which direction in acts, just as in Sec. 4.2. Hence the two terms on the right-hand side become a perfect divergence and i— (ipltpb) +iVi (^a,^6) =0 .

As in the KG case, we have a four-current which is conserved. The conservation law can be written

If we integrate this equation over all space and use the periodic boundary conditions to eliminate the spatial part, just as we did in our discussion of Eq. (4.13), we find that the following quantity is a constant of the motion:

J d3riplipb = constant .

Note that this expression is positive definite if a = b, and hence the states can be normalized as follows:

This norm is a constant of the motion, and the Dirac equation has no states with negative norm. This was first believed to be a great advantage of the Dirac equation, but as we will soon see, the Dirac equation suffers from the same

'The symbol p will be used to denote either the four-vector or the magnitude of the three-vector. They can be distinguished from each other by the context in which they are used.

ipp^s(x): is the wave function for a particle with positive energy, momentum p, and spin projection s.

^-pt-a{x)\ is the wave function for a negative energy state with momentum —p and spin projection — s, which is interpreted as an antiparticle state with positive energy, momentum p, and spin projection s.

In our study of the KG equation in Chapter 4, we also interpreted negative energy states as antiparticles. However, the way in which this interpretation is developed is significantly different for the two equations. First, KG particles (spin zero) obey Bose-Einstein statistics, and there is no limit to the number of negative energy particles which can occupy any negative energy state. Any positive energy KG state is therefore intrinsically unstable; there is no way to prevent it from decaying eventually to a negative energy state. On the other hand, Dirac particles (spin obey Fermi-Dirac statistics (which will be shown in Chapter 7). This means that no more than one particle can occupy any one state (the Pauli exclusion principle). If the physical vacuum is assumed to be the state in which all negative energy states are filled, a single positive energy state will be stable, since decay to negative energy states will be Pauli blocked by the filled negative energy sea, and we are able to "explain" why the lowest energy of a single particle is m (and not — oo as might be expected if the negative energy states were not already occupied). Furthermore, since the energy of the ground state can always be chosen to be zero [by choosing the appropriate constant Eo in Eq. (3.19)], this picture of the vacuum is physically sensible. In this picture, referred to as hole theory, an antiparticle is interpreted as a "hole" in the vacuum, i.e., as the absence of one of the particles from the otherwise filled negative energy sea. Being the absence of a negative energy state, the antiparticle has positive energy.

These ideas are illustrated in Fig. 5.1. In Fig. 5.1 A the vacuum has no particles, so a single particle with energy —E < -m, momentum -p, and spin projection -s can exist. In Fig. 5.IB the vacuum is assumed to be filled with negative energy states. The absence of a single negative energy state with quantum numbers -E < -m, —p, and -s then appears as a hole in this vacuum. Since the vacuum values for these quantum numbers must be zero (by definition), the hole therefore behaves just like a particle with energy 0 - (-E) = E, momentum 0 ~ (—p) — P< and spin projection 0 - (—s) = s, or a positive energy antiparticle with energy E > m, momentum p, and spin projection s. Thus hole theory provides a physical picture of how negative energy and antiparticle states are related. Mathematically, this relation is expressed through the charge conjugation transformation.

Hole theory played an important role in the development of relativistic quantum mechanics but is superseded by modern field theory. We no longer think of the vacuum as filled with negative energy particles. In Chapter 7 we will see that a quantum field is equally well suited to the description of either spin zero

This is further reduced by using p} ■ p{ — p2 cos 6, where 6 is the scattering angle

Hence, finally

0 0