We saw in a previous post that for a non-interacting theory (i.e. V(φ)=0V(\varphi) = 0) that the generating functional can be written as

Z[J]=e12JK1J. Z[J] = e^{\frac{1}{2} J \cdot K^{-1} \cdot J}.

We hinted that it is not always the case that KK can be naively inverted. The issue arises when we consider the Maxwell action for a U(1)U(1) gauge potential AμA_\mu:

S(A)=d4x[12Aμ(2gμνμν)Aν+AμJμ]. S(A) = \int d^4 x \left[ \frac{1}{2} A_\mu \left( \partial^2 g^{\mu \nu} - \partial^\mu \partial^\nu \right) A_\nu + A_\mu J^\mu \right].

Here the operator we are looking to invert is

Qμν2gμνμν \begin{equation} Q^{\mu \nu} \equiv \partial^2 g^{\mu \nu} - \partial^\mu \partial^\nu \end{equation}

which has zero eigenvalues for vectors of the form μθ(x)\partial_\mu \theta(x):

Qμνμθ(x)=(2gμννννμ)θ(x)=0 Q^{\mu \nu} \partial_\mu \theta(x) = \left(\partial^2 g^{\mu \nu} \partial_\nu - \partial^\nu \partial_\nu \partial^\mu \right) \theta(x) = 0

So, QμνQ^{\mu \nu} is not invertible and the issue is gauge redundancy - we need to fix a gauge.

the Faddeev-Popov procedure

An interpretation of the issue is that the redundancy in the gauge field, where acting with gU(1)g \in U(1), takes AμAμμθAgA_\mu \to A_\mu - \partial_\mu \theta \equiv A_g; which is not a physical symmetry of the system, but still gets naively integrated over in the generating functional:

Z=DA eiS(A) =DADg eiS(Ag)  \begin{align*} Z &= \int \mathcal{D}A \ e^{iS(A)} \\\ &= \int \mathcal{D}A \int \mathcal{D}g \ e^{iS(A_g)} \\\ \end{align*}

Which diverges.

The naive path integral effectively over counts and blows up. What we need to do is insert a constraint on the gauge (e.g. f(Ag)=0f(A_g) = 0), which can be done by inserting

1=Dg δ(f(Ag))det(f(Ag)g)=Δ(A)Dg δ(f(Ag)), \begin{equation} 1 = \int \mathcal{D}g \ \delta\left( f(A_g) \right) \det \left( \frac{\partial f(A_g)}{\partial{g}} \right) = \Delta(A)\int \mathcal{D}g \ \delta\left( f(A_g) \right), \end{equation}

where Δ(A)\Delta(A) is the Faddeev-Popov determinant which is gauge invariant. We then get

Z=DA eiS(A) =DA eiS(A) Δ(A)Dg δ(f(Ag)) =DgDA eiS(A) Δ(A) δ(f(Ag)) =(Dg)DA eiS(A) Δ(A) δ(f(A))  \begin{aligned} Z &= \int \mathcal{D}A \ e^{iS(A)} \\\ &= \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \int \mathcal{D}g \ \delta\left( f(A_g) \right) \\\ &= \int \mathcal{D}g \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \ \delta\left( f(A_g) \right) \\\ &= \left( \int \mathcal{D}g \right) \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \ \delta\left( f(A) \right) \\\ \end{aligned}

Where we arrived at the final result by noting that the path integral measure A\mathcal{A} is gauge invariant and shifting AAg1A \to A_{g^{-1}} removes the gauge dependence from δ(f(Ag))\delta(f(A_g)). The factor outside, (Dg)\left( \int \mathcal{D}g \right) is an infinite constant that we can drop, as it has no physical effect.

To see how this procedure works with QED check out Zee’s wonderful book Quantum Field Theory in a Nushell, which was my primary reference for this post.

ghosts in non-abelian gauge theories

We will now turn our attention towards quantum Yang-Mills theory, where the gauge group being considered is non-abelian, for all intents and purposes is SU(N)SU(N). Here things get very spooky.

In a pure non-abelian gauge theory, each component of the gauge potential AμA_\mu transforms under the adjoint representation of the gauge group, in such a way that the action remains gauge invariant–i.e.

Aμ(x)U(x)Aμ(x)U(x)+iU(x)μU(x) A_\mu(x) \to U(x)A_\mu(x)U^\dagger(x) + iU(x)\partial_\mu U^\dagger(x)

One can note that if UU does not vary throughout space, AA really does transform in the adjoint representation; the extra term comes from the requirement for gauge invariance in a theory where the gauge group can act differently at each spacetime coordinate.

Now, given a set of generators for the gauge group, Ta{T^a}, really a Lie algebra, we can see that under an infinitesimal gauge transformation (U(x)1+iθaTaU(x) \simeq 1 + i\theta^a T^a), and noting we can decompose Aμ=AμaTaA_\mu = A^a_\mu T^a , we see that

AμaAμafabcθbAμc+μθa. A^a_\mu \to A^a_\mu - f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a.

Where the structure constants fabcf^{abc} are defined by [Ta,Tb]=fabcTc[T^a, T^b] = f^{abc}T^c.

Let’s now choose a gauge fixing condition to be f(A)=μAμσf(A) = \partial^\mu A_\mu - \sigma, with σ\sigma arbitrary. By (2) we then have

Δ(A)={Dθ δ[f(A)]}1 ={Dθ δ[μAμaσa]}1 ={Dθ δ[μAμaσaμ(fabcθbAμc+μθa)]}1 =’’ {Dθ δ[μ(fabcθbAμc+μθa)]}1  \begin{align*} \Delta(A) &= \left\{\int \mathcal{D}\theta \ \delta\left[ f(A) \right] \right\}^{-1} \\\ &= \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu A^a_\mu - \sigma^a \right] \right\}^{-1} \\\ &= \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu A^a_\mu - \sigma^a - \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right)\right] \right\}^{-1} \\\ ``&= \text{’’} \ \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right)\right] \right\}^{-1} \\\ \end{align*}

Where there are quotes around the equality of the last line because we have yet to muliply by the gauge constraint δ[f(A)]\delta[f(A)].

Now lets formally define an operator Kab(x,y)K^{ab}(x,y) by

μ(fabcθbAμc+μθa)=d4y Kab(x,y)θb(y). \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right) = \int d^4y \ K^{ab}(x, y) \theta^b(y).

that is

Kab(x,y)μ(fabcAμcμδab)δ(4)(xy) K^{ab}(x,y) \equiv \partial^\mu \left( f^{abc} A^c_\mu - \partial_\mu \delta^{ab} \right) \delta^{(4)}(x - y)

A generalization of the delta function identity, dθ δ(Kθ)=1/K\int d\theta \ \delta(K\theta) = 1/K, to a vector θ\theta and matrix KK, gives us dθ δ(Kθ)=1/detK\int d\theta \ \delta(K\theta) = 1 / \det K. We also have that anti-commuting Grassmann variables allow us to write detK=dηdηexp[ηKη]\det K = \int d\eta d\eta^\dagger \exp\left[-\eta^\dagger \cdot K \cdot \eta\right]. Pulling all of this together we see that

Δ(A)=detKab(x,y)=DcDceiSghost(c,c) \Delta(A) = \det K^{ab}(x,y) = \int \mathcal{D} c \mathcal{D} c^\dagger e^{i S_{\text{ghost}}(c^\dagger, c)}

where

Sghost(c,c)=d4xd4y ca(x)Kab(x,y)cb(y) =d4x ca(x)[μfabcAμc(x)cb(x)2cb(x)] =d4x [μca(x)μcb(x)μcafabcAμc(x)cb(x)] =d4x μca(x)[μcb(x)fabcAμc(x)cb(x)] =d4x μca(x)Dμca(x)  \begin{aligned} S_{\text{ghost}}(c^\dagger, c) &= \int d^4x d^4y \ c_a^\dagger(x)K^{ab}(x, y)c_b(y) \\\ &= \int d^4x \ c_a^\dagger(x) \left[ \partial^\mu f^{abc} A^c_\mu(x) c_b(x) - \partial^2 c_b(x) \right] \\\ &= \int d^4x \ \left[ \partial^\mu c_a^\dagger(x) \partial_\mu c_b(x) - \partial^\mu c^\dagger_a f^{abc} A^c_\mu(x) c_b(x) \right] \\\ &= \int d^4x \ \partial^\mu c_a^\dagger(x) \left[ \partial_\mu c_b(x) - f^{abc} A^c_\mu(x) c_b(x) \right] \\\ &= \int d^4x \ \partial^\mu c_a^\dagger(x) D_\mu c_a(x) \\\ \end{aligned}

Here DμD_\mu is the covariant derivative of the adjoint representation; ca,ca,c_a, c_a^\dagger, and AμaA_\mu^a all transform in this way.

cac_a and cac_a^\dagger are ghost fields and seemingly violate the spin-statistics connection, because they are grassmann anti-commuting variables, but they are not physical, just a convenient representation of Δ(A)\Delta(A).

Zooming out, we now have

Z=DADcDc eiS(A)+iSghost(c,c) δ(Aσ) Z = \int \mathcal{D}A\mathcal{D}c\mathcal{D}c^\dagger \ e^{iS(A) + iS_{\text{ghost}}(c^\dagger, c)} \ \delta(\partial A - \sigma)

which after integrating over σ\sigma with a Gaussian weight, e(i/2ξ)d4xσa(x)2e^{ -(i/2\xi) \int d^4x \sigma^a(x)^2}, we arrive at a final gauge fixed (with gauge parameter ξ\xi) generating functional for a pure non-abelian Yang-Mills theory:

Z(J)=DADcDc exp{iS(A)+iSghost(c,c)i2ξd4x(μAμ)2+d4xJμAμ} \boxed{ Z(J) = \int \mathcal{D}A\mathcal{D}c\mathcal{D}c^\dagger \ \exp \left\{iS(A) + iS_{\text{ghost}}(c^\dagger, c) - \frac{i}{2\xi} \int d^4x (\partial^\mu A_\mu)^2 + \int d^4x J^\mu A_\mu\right\} }