chapter  14
4 Pages

The Dirac Hamiltonian

In the Dirac equation we encountered an additional difficulty, namely that the negative energy solutions even arise at the level of the classical theory. In field theory the negative energy solutions had an interpretation in terms of antiparticles, and the field theory Hamiltonian was still positive, and most importantly, bounded from below (see Chapter 2 and Problem 5). The field theory Hamiltonian for the Dirac field no longer has this property. The Hamiltonian can again be derived through a Legendre transform of the Lagrangian

S = ∫

d4x L = ∫

d4x (x)(iγ µ∂µ − m)(x). (14.1)

The canonical momentum is hence

πa (x) = δS δ˙a (x)

= ((x)iγ 0)a = i∗a (x), (14.2) such that

H = ∫

d3x ( πa (x)˙a (x) − L

) = ∫ d3x (x)(−iγ j∂ j + m)(x) =

∫ d3x †(x)(−iα j∂ j + mβ)(x)

= ∫

d3 k k0(k) ∑

( b†α(k)bα(k) − dα(k)d†α(k)

) . (14.3)

Note the resemblance with Equation (12.1) for the middle term. We used Equation (13.17) for the expansion of the Dirac field in plane waves. From this result it is clear that the Hamiltonian is not bounded from below, and this would make the vacuum unstable, as the negative energy states, described by the dα(k), can lower the energy by an arbitrary amount. It is well known how Dirac repaired this problem. He postulated that all negative energy states are occupied, and that the states satisfy the Pauli principle, i.e., two particles cannot occupy the same quantum state. (It is only in that case that we can make sense of what is meant with filling all negative energy states.) This implies that one should use anticommuting relations for the creation and annihilation

A in

operators

{bα(k), bβ( p)} = 0, {bα(k), b†β( p)} = δαβδ3(k − p), {dα(k), dβ( p)} = 0, {dα(k), d†β( p)} = δαβδ3(k − p), {dα(k), bβ( p)} = 0, {dα(k), b†β( p)} = 0. (14.4)

Indeed, if we define a two-particle state as |k, p >≡ b†(k)b†( p)|0 > (suppressing the spinor indices), the anticommutation relations imply that |k, p >= −| p, k >.