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 Hamil-tonian can again be derived through a Legendre transform of the Lagrangian
S=
d4xL=
d4x�(x)(iγµ∂µ−m)�(x). (14.1) The canonical momentum is hence
πa(x)= δS δ�˙a(x) =
�(x)iγ0
a=i�a∗(x), (14.2) such that
H=
d3�x
πa(x)�˙a(x)−L
=
d3x��(x)(−iγj∂j+m)�(x)
=
d3�x�†(x)(−iαj∂j+mβ)�(x)
=
d3k k� 0(k)�
α
b†α(k)b� α(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 85
86 A Course in Field Theory operators
{bα(k), b β(p)} =0,{bα(k), b β†(p)} =δαβδ3(k− p), {dα(k), d β(p)} =0,{dα(k), d β†(p)} =δαβδ3(k− p),
{dα(k), b β(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> (suppress-ing the spinor indices), the anticommutation relations imply that|k, p >=
−| p,k>.
A hole in the Dirac sea is by definition the state that is obtained by anni-hilating a negative energy state in the Dirac sea. As annihilation lowers the total energy by the energy of the annihilated state, which in this case is nega-tive, the net energy is raised. The wave function for the negative energy state is given by exp(ikx)v(α)(k)/ √
k0 and has momentum−k [k0 ≡ (k2+m2)12];
see Equation (13.8). The reason to associate its Fourier coefficient in Equa-tion (13.17) with a creaEqua-tion operator dα†(k), is that conservation of energy and momentum implies that it creates an antiparticle as a hole in the Dirac sea, with momentumk and helicity 12 forα = 2, whereas forα =1 the helicity is−12(hence the helicity and momentum are opposite to the negative energy state it annihilates). The wave function of an antiparticle with momentumk is hence given by exp(−ikx)v(α)(k)†/√
k0. If we now use the anticommutation relations of creation and annihilation operators (as a consequence of the Pauli exclusion principle), we see that with the present interpretation the Dirac Hamiltonian is bounded from below
H=
d3k k 0(k)
α
b†α(k)b α(k) +dα†(k)d α(k) −1 . (14.5)
As in the case for scalar field theory, we can normalise the energy of the vacuum state to zero by adding a (infinite) constant. Note that this constant has its sign opposite to the scalar case (and is in magnitude four times as large).
In so-called supersymmetric field theories, this is no longer an accident as the Dirac fields will be related to the scalar fields by a symmetry, which is, however, outside the scope of these lectures.
It is also important to note that the anticommuting relations are crucial to guarantee locality of the Dirac field. In this case, Dirac fields specified in differ-ent regions of space-time that are space-like separated should anticommute.
And indeed, in Problem 24 you are asked to prove that
{a(x),b†(x)} =0 for (x−x)2<0. (14.6) Also, as in the scalar theory, we can couple a source to the free Dirac field (we again introduce ¯εas the expansion parameter for taking the interactions
The Dirac Hamiltonian 87 due to the source into account in Hamiltonian perturbation theory)
L(x)=(x)(iγµ∂µ−m)(x)−ε¯J¯(x)(x)−ε(x)J¯ (x),
H(x)=(x)(−iγk∂k+m)(x)+ε¯J¯(x)(x)+ε(x)J¯ (x), (14.7) where, as before, the Hamiltonian H is the spatial integral over the Hamiltonian densityH(x), H =
d3xH(x). Note that the sourcesJ(x) and J¯(x) are independent, as for a complex scalar field, see Problem 17. In Prob-lem 24 you are asked to prove that in Hamiltonian perturbation theory one obtains
<0|Texp
−i T
0
H(t)dt
|0>ei E0T =1−i ¯ε2
d4xd4y ¯J(x)GF(x−y)
×J( y)+O( ¯ε3)
=1−i × > ×
εJ¯ ε¯J¯ +O( ¯ε3) (14.8) where
GF(x−y)=
d4p (2π)4
e−i p(x−y) p/−m+iε =
d4p (2π)4
( p/+m)e−i p(x−y)
p2−m2+iε . (14.9) Hence, the Green’s function for fermions is in Fourier space given by GF( p)= ( p/−m+iε)−1, which is the inverse of the quadratic part of the Lagrangian, as for the scalar fields. In Problem 24 it will be evident that, nevertheless, the anticommuting properties of the Dirac field play a crucial role (compare this to chapter 5). It becomes, however, plausible that there is also for the fermions a path integral formulation, as splitting of a square [compare this to Equations (7.11) and (7.12)] is independent of the details of the path integrals.
This will be the subject of the next chapter.
To conclude, we note that the coupling to an electromagnetic field Aµ(x) should be achieved through the current jµ(x), defined in Equation (12.27)
d4x jµ(x) Aµ(x)=
d4x(x)γµAµ(x)(x). (14.10) Since this current is conserved, which can also be seen as a consequence of the Noether theorem applied to the invariance of Equation (14.1) (the free Dirac Lagrangian) under global phase transformations, the coupling is gauge invariant. Using the minimal coupling defined as in Equation (3.35),
Dµ(x)≡
∂µ−ie Aµ(x)
(x), (14.11)
one can fix the normalisation of the electromagnetic current to be Jµ(x) =
−e jµ(x), since [compare this to Equation (3.31)]
(x)(iγµDµ−m)(x)=(x)(iγµ∂µ−m)(x)+e jµAµ(x). (14.12)
88 A Course in Field Theory The current jµ(x) is the charge density current, whose time componentρ(x) at the quantum level is no longer positive definite. Using the anticommuting properties of Equation (14.4) one finds
d3x ˆρ(x) =
d3x†(x)(x)
=
d3k
α
b†α(k)b α(k) +dα(k)d α†(k)
=
d3k
α
b†α(k)b α(k) −dα†(k)d α(k) +1
=Q0/e+
d3k
α
b†α(k)b α(k) −dα†(k)d α(k) . (14.13) The vacuum value of this operator is indicated by the (generally infinite) constant Q0/e, which can be normalised to zero. After all, we want the state with all negative energy states occupied to have zero charge. With the above normalisation of the electric current, Jµ(x)= −e jµ(x), we see that the b modes can be identified with the electrons with charge−e and the d modes with their antiparticles, the positrons, with opposite electric charge+e. To summarise, bα†(k) corresponds with the creation operator of a spin-up (α =1) or a spin-down (α=2) electron of momentumk, whereas d α†(k) corresponds with the creation operator of a spin-up (α = 2) or a spin-down (α = 1) positron of momentumk.
DOI: 10.1201/b15364-15