chapter  20
6 Pages

Gauge Fixing and Ghosts

The quantisation of gauge theories in the path integral formalism requires more discussion, since the gauge condition (like the Lorentz gauge ∂µ Aµ = 0) seems to remove only one degree of freedom of the two that are eliminated in the Hamiltonian formulation (see Chapter 16). From a simple example it is easily demonstrated what the effect of gauge fixing on a (path) integral is. For this we take f (x) to be a function on IR3, which is invariant under rotations around the origin, such that it is a function f (r ) of the radius r = |x| only. The symmetry group is hence SO(3), and we can attempt to compute the integral∫

d3x f (x) by introducing a ‘gauge’ fixing condition like x2 = x3 = 0. But it is clear that ∫

d3x f (x) = ∫

d3x δ(x2)δ(x3) f (x) = ∫

dx1 f (x1). (20.1)

We know very well that we need a Jacobian factor for the radial integral∫ d3x f (x) = 4π

r2dr f (r ). (20.2)

This Jacobian, arising in the change of variable to the invariant radial coordinates and the angular coordinates, can be properly incorporated following the method introduced by Faddeev and Popov. The starting point is a straightforward generalisation of the identity

∫ dx | f ′(x)|δ[ f (x)] = 1, assuming the

equation f (x) = 0 to have precisely one solution (in a sense the right-hand side of the equation counts the number of zeros). It reads

1 = ∫ Dg | det (M(g A))|δ(F(g A)), (20.3)

where F( A) ∈ LG is the gauge-fixing function [with the gauge condition F( A) = 0, e.g., F( A) = ∂µ Aµ = 0]. The gauge transformation g of the gauge field Ais indicated by g A; see Equation (18.34). Furthermore, M( A) : LG → LG plays the role of the Jacobian,

M(g A) = ∂F( g A)

∂g ≡ ∂F(

e Xg A) ∂ X

at X = 0. (20.4)

A in

Equivalently, with respect to the Lie algebra basis, where F( A) ≡ Fa ( A)Ta , one has

Mab( A) = dFa ( exp(tTb ) A) dt

at t = 0. (20.5)

To relate this to the previous equation, one makes use of the fact that

h(g A) = (hg) A, (20.6) which states that two successive gauge transformations, g and h, give the same result as a single gauge transformation with hg.