Gauge-Fixed Yang–Mills Action and Ghost Sector

Covariant Yang–Mills perturbation theory requires more than adding $-(\partial\cdot A)^2/(2\xi)$ . The gauge condition has a field-dependent Jacobian, represented by interacting Grassmann ghosts, and a gauge fermion generates the combined gauge-fixing and ghost action. Keeping the auxiliary field makes the BRST differential nilpotent without equations of motion. This page constructs that action and its free propagators in a local perturbative gauge patch; physical-state cancellation additionally requires the quantum identities and suitable asymptotic states.

Required background. Yang–Mills equations, constraints, and observables supplies the gauge orbits and Gauss constraint being fixed. The BRST differential and gauge-fixed complex supplies the general cohomological construction specialized here.

Helpful background. Gribov copies and the limits of local gauge fixing explains why this perturbative gauge slice is not a global nonperturbative section of configuration space.

The Faddeev–Popov operator

Work in four-dimensional flat spacetime with a compact gauge group, an invariant orthonormal Lie-algebra basis and the Yang–Mills action conventions. Expand around the trivial background, with boundary conditions that justify integration by parts and with residual gauge zero modes separately removed. The local Faddeev–Popov construction requires an invertible orbit derivative on that domain. For an infinitesimal transformation with parameter $\alpha^a$ ,

\delta_\alpha A_\mu^a=(D_\mu\alpha)^a, \qquad (D_\mu\alpha)^a =\partial_\mu\alpha^a+g f^{abc}A_\mu^b\alpha^c.

Choose the linear covariant gauge function $G^a[A]=\partial^\mu A_\mu^a$ . Its derivative along a gauge orbit is

\delta_\alpha G^a =\partial^\mu(D_\mu\alpha)^a, \qquad \mathcal M^{ab}[A] =-\partial^\mu D_\mu^{ab}[A].

Here $\mathcal M$ is the negative of the orbit derivative $M_A$ used in the general BRST construction. This overall determinant sign contributes only a field-independent normalization in the regulated local construction. The determinant $\det\mathcal M[A]$ is field dependent because $D_\mu$ contains $A_\mu$ . Independent Grassmann fields $c^a$ and $\bar c^a$ exponentiate it. For a constant finite $\xi\neq0$ , the gauge average gives

\mathcal L_{\rm gf+gh} =-\frac{1}{2\xi}(\partial^\mu A_\mu^a)^2 -\bar c^a\partial^\mu(D_\mu c)^a.

After an integration by parts, the ghost term is

\mathcal L_{\rm gh} =(\partial^\mu\bar c^a)\partial_\mu c^a +g f^{abc}(\partial^\mu\bar c^a)A_\mu^b c^c.

The ghost–gauge interaction is absent in Abelian theory but unavoidable in non-Abelian loops. The derivative acts on the antighost; retaining that field order fixes the momentum and color signs in the ghost vertex. The auxiliary-field construction below also defines the Landau gauge $\xi=0$ , where division by $\xi$ is inappropriate.

Gauge fermion and nilpotent BRST symmetry

Introduce a commuting Nakanishi–Lautrup field $B^a$ . The differential $s$ acts from the left and obeys the graded product rule

s(XY)=(sX)Y+(-1)^{|X|}X(sY),

where $|X|$ is the Grassmann parity. The gauge and auxiliary fields are even, while $c$ and $\bar c$ are odd. Their ghost numbers are respectively zero, zero, $+1$ and $-1$ ; $s$ raises ghost number by one. The transformations are

\begin{aligned} sA_\mu^a&=(D_\mu c)^a,\\ sc^a&=-\frac g2 f^{abc}c^bc^c,\\ s\bar c^a&=B^a,\\ sB^a&=0,\\ s\psi&=igc^aT_R^a\psi. \end{aligned}

With gauge fermion density

\Psi=\bar c^a\left(\partial^\mu A_\mu^a+\frac\xi2B^a\right),

the BRST-exact term is

s\Psi =\frac\xi2B^aB^a+B^a\partial^\mu A_\mu^a -\bar c^a\partial^\mu(D_\mu c)^a,

For $\xi\neq0$ , the algebraic equation $B^a=-\partial\cdot A^a/\xi$ returns the gauge-fixing term above. At $\xi=0$ , keep $B$ : its integral imposes $\partial\cdot A=0$ . Integrating the ghost term by parts gives the displayed positive-kinetic convention. This is the gauge-fermion construction in Srednicki 2006, author manuscript, § 74, pp. 437–438, Eqs. (74.15)–(74.24), PDF, with $\bar c_{\rm here}=-\bar c_{\rm source}$ and $B_{\rm here}=-B_{\rm source}$ ; the gauge and ghost fields are unchanged.

To verify nilpotency, write $c=c^aT^a$ and $A_\mu=A_\mu^aT^a$ . Anticommutation of the coefficients gives $sc=igc^2$ , while $D_\mu c=\partial_\mu c-ig[A_\mu,c]$ . First, the three-ghost expression is

\begin{aligned} s^2c&=ig\bigl[(sc)c-c(sc)\bigr]\\ &=-g^2(c^2c-cc^2)=0. \end{aligned}

In components, the cancellation is the Jacobi identity

f^{abe}f^{ecd}+f^{bce}f^{ead}+f^{cae}f^{ebd}=0.

For the gauge field there are only two ghost factors. Since the covariant derivative is even, $D_\mu(c^2)=(D_\mu c)c+c(D_\mu c)$ . Applying the odd product rule to its connection term gives

\begin{aligned} s^2A_\mu &=D_\mu(sc)-ig\bigl[(sA_\mu)c+c(sA_\mu)\bigr]\\ &=igD_\mu(c^2)-ig\bigl[(D_\mu c)c+c(D_\mu c)\bigr]=0. \end{aligned}

This is the component cancellation displayed in Srednicki 2006, author manuscript, § 74, pp. 436–437, Eqs. (74.10)–(74.16), PDF. For matter in representation $R$ , use the same matrix $c=c^aT_R^a$ :

s^2\psi=ig(sc)\psi-igc(s\psi) =-g^2c^2\psi+g^2c^2\psi=0.

Finally, $s^2\bar c=sB=0$ and $s^2B=0$ . No field equation has entered. Gauge invariance gives $s\mathcal L_{\rm YM}=0$ , and nilpotency then proves $s(\mathcal L_{\rm YM}+s\Psi)=0$ . This is a classical algebraic statement; preservation by a regulator and renormalization, and the absence of a gauge anomaly, are further requirements.

Free propagators and field content

Use the inherited Fourier convention $\widetilde f(k)=\int\mathrm d^4x\,e^{+ik\cdot x}f(x)$ . Propagators below are time-ordered vacuum contractions with the overall momentum-conserving delta function suppressed. After eliminating $B$ at $\xi\neq0$ , the quadratic gauge kernel and its inverse can be checked away from $k^2=0$ with $P_{\rm L}^{\mu}{}_{\nu}=k^\mu k_\nu/k^2$ and $P_{\rm T}=1-P_{\rm L}$ :

\begin{aligned} K&=-k^2(P_{\rm T}+\xi^{-1}P_{\rm L}),\\ D&=-\frac{i}{k^2}(P_{\rm T}+\xi P_{\rm L}), \qquad KD=i\mathbf1. \end{aligned}

Thus the projector weights belong to the propagator before multiplying by $K$ . Restoring the Feynman boundary prescription gives

D_{\mu\nu}^{ab}(k) =\frac{-i\delta^{ab}}{k^2+i0} \left[ \eta_{\mu\nu} -(1-\xi)\frac{k_\mu k_\nu}{k^2+i0} \right],

The repeated Feynman prescription denotes a distributional boundary value, not an exact inverse at finite regulator; the canonical propagator prescription explains that distinction. The oriented ghost propagator is

\langle c^a(k)\bar c^b(-k)\rangle_0 =\frac{i\delta^{ab}}{k^2+i0}.

If $B$ is retained rather than integrated out, the first-order free rules are

\langle A_\mu^a(k)B^b(-k)\rangle_0 =-\frac{\delta^{ab}k_\mu}{k^2+i0}, \qquad \langle B^a(k)B^b(-k)\rangle_0=0,

for the stated Fourier and $s\Psi$ convention. These also follow by inverting the coupled quadratic form rather than using an auxiliary equation inside a time-ordered product. With color indices suppressed and columns $(A_\nu(k),B(k))$ , its kernel is

\mathbb K(k)= \begin{pmatrix} -k^2\eta^{\mu\nu}+k^\mu k^\nu & +ik^\mu\\ -ik^\nu & \xi \end{pmatrix}.

The full contraction matrix obeys $\mathbb K\mathbb D=i\mathbf1$ away from the pole. In particular, its $BA$ entry has the opposite momentum sign to the displayed $AB$ entry, as required by exchanging the fields and reversing $k$ . The auxiliary equation alone misses a local contact contraction; the exercise below derives its cancellation in $\langle BB\rangle_0$ . There is no independent $B$ asymptotic particle, including at $\xi=0$ .

The mass dimensions are

[A_\mu]=[c]=[\bar c]=1, \qquad [B]=2, \qquad [\xi]=0.

Every term therefore has dimension four. In Feynman gauge $\xi=1$ the gauge propagator is proportional to $\eta_{\mu\nu}$ ; in Landau gauge $\xi=0$ it is transverse. Neither choice changes a complete physical amplitude.

BRST cohomology and physical states

When the quantum BRST symmetry is preserved, the conserved charge $Q$ satisfies $Q^2=0$ . At ghost number zero, the perturbative physical state space is the cohomology of the indefinite-metric state space,

\mathcal H_{\rm phys}=\frac{\ker Q}{\operatorname{im}Q}.

Longitudinal and timelike gauge modes, ghosts, antighosts, and the auxiliary sector assemble into BRST-trivial combinations; the free one-particle cohomology contains the two transverse gauge polarizations for each gauge generator. A BRST-invariant time evolution preserves $\ker Q$ and descends to equivalence classes. Establishing positive physical unitarity additionally requires the Slavnov identity, absence of gauge anomalies, and suitable asymptotic states; the argument and its hypotheses continue on Consistency of perturbative Yang–Mills theory.

This construction is perturbative around a gauge slice near a chosen background. Gribov copies can obstruct a global gauge condition, and BRST cohomology in a confining nonperturbative theory need not be represented by colored one-gauge-boson states. Covariant ghosts are indispensable internal fields, but they never appear as physical external particles.

Independent checks and limits

Determinant check: differentiating $G^a[A^\alpha]$ must give $\partial^\mu D_\mu^{ab}$ with the same $g f^{abc}$ sign as the ghost vertex.
Nilpotency check: calculate $s^2c$ ; the three-ghost coefficient must vanish by Jacobi, not by treating the ghosts as commuting.
Free inverse check: away from the pole, multiplying the gauge kinetic operator by the Feynman contraction gives $KD=i\mathbf1$ ; its transverse and longitudinal pieces both give $i$ times their projector.
Abelian limit: $f^{abc}\to0$ makes $\mathcal M=-\partial^2$ field independent, so the ghost determinant is an irrelevant constant and ghost interactions disappear.

Common pitfalls

Dropping ghosts because they are unphysical. “Unphysical” means absent from external cohomology classes, not absent from loops. Ghost loops are required by the Slavnov identities and the one-loop beta function.

Checking BRST invariance only after eliminating $B$ . Nilpotency on $\bar c$ is off shell with $B$ present. For $\xi\neq0$ , the reduced rule gives $s_{\rm red}^2\bar c=-\partial^\mu(D_\mu c)/\xi$ , which vanishes using the ghost equation obtained by varying the antighost. The eliminated auxiliary equation alone does not set this expression to zero.

Interpreting $\xi$ as a coupling. It changes the representation of off-shell Green functions. A residual $\xi$ dependence in a complete physical amplitude is a failed gauge-consistency check.

Exercise: the auxiliary contact term

For $\xi\neq0$ , complete the square in $B$ . Derive the mixed contraction and explain why substituting the auxiliary equation directly into both factors of $\langle BB\rangle_0$ gives the wrong answer. Suppress color deltas and work off the momentum pole before imposing the same Feynman boundary prescription.

Solution

Set $G=\partial\cdot A$ and $B'=B+G/\xi$ . The quadratic terms become

\frac\xi2B^2+BG=\frac\xi2(B')^2-\frac{G^2}{2\xi}.

The shifted field is independent of $A$ in the free Gaussian integral. Its contraction is the local term $\langle B'(x)B'(y)\rangle_0=(i/\xi)\delta^{(4)}(x-y)$ . Since $G(k)=-ik^\mu A_\mu(k)$ and $G(-k)=+ik^\mu A_\mu(-k)$ ,

\begin{aligned} \langle G(k)G(-k)\rangle_0 &=k^\mu k^\nu D_{\mu\nu}(k)=-i\xi,\\ \langle B(k)B(-k)\rangle_0 &=\frac{i}{\xi}+\frac{1}{\xi^2}(-i\xi)=0,\\ \langle A_\mu(k)B(-k)\rangle_0 &=-\frac{ik^\nu}{\xi}D_{\mu\nu}(k) =-\frac{k_\mu}{k^2}. \end{aligned}

The first term in the second line is lost if $B=-G/\xi$ is substituted into a product without the Gaussian contact term. Restoring the boundary prescription gives the mixed propagator above. At $\xi=0$ , the completion of the square is unavailable; invert the coupled $A,B$ kernel or take the limit of the complete contractions, which remain finite.

References

Srednicki, Mark. Quantum Field Theory. Author’s prepublication manuscript, copyright 2006, § 74, pp. 435–438. Open PDF. Author page identifying the draft and published-edition errata. The printed labels here refer to this manuscript, not the 2007 book.