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A gluon propagator is the inverse of the gauge-fixed quadratic Yang-Mills operator. It carries Lorentz and adjoint color indices and depends on the gauge-fixing condition, while gauge-invariant observables do not.

The proper two-point correction to the gluon propagator in QCD. At one loop it receives quark, gluon, Faddeev-Popov ghost field and four-gluon tadpole diagram contributions. The Slavnov-Taylor identity makes the total tensor transverse, $k_{\mu }{\Pi }^{\mu \nu ,ab}=0$, although the individual gluon and ghost loop contributions need not be transverse. The sign of the symbol must be tied to whether the amputated insertion is defined as $i\Pi$ or $-i\Pi$.

There is one four-gluon tadpole diagram. Temporarily replace its internal gluon propagator by $-i{g}_{\rho \sigma }{\delta }^{cd}/(l^2-m_g^2+i0)$. Contract the supplied four-gluon Feynman vertex with $g_{\rho \sigma }{\delta }^{cd}$. Its first colour factor vanishes because $f^{cce}=0$. Each of the other two terms contributes $C_A{\delta }^{ab}(D-1)g^{\mu \nu }$, soInclude the symmetry factor $1/2$ and the factor $-i$ from the internal gluon propagator:The exact regulated tadpole integral isIn particular,The entire expression is proportional to $(m_g^2{)}^{1-ϵ/2}$ and vanishes as the auxiliary mass tends to zero near four dimensions. Equivalently, its expanded terms are proportional to $m_g^2/ϵ$, $m_g^2\log m_g^2$ and $m_g^2$, all of which disappear when the regulator is removed. HenceThis is the massless scaleless integral result obtained directly in dimensional regularization. The artificial mass is only an intermediate device for this graph; retaining it in the full theory would modify the gauge symmetry and cannot be used when applying the massless transversality identity in part (iv).

For a bosonic quadratic action $S_2=\frac{1}{2}\int \varphi K\varphi$, completing the square in the generating functional gives $Z_0[J]/Z_0[0]=\exp (-iJ{K}^{-1}J/2)$. Two functional derivatives therefore give the Feynman propagator $\Delta =i{K}^{-1}$, with boundary conditions and a pole prescription included in the inverse. Zero modes must be removed or fixed before an inverse exists.

Take Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the Yang-Mills field strength $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$. The interaction terms do not contribute to the quadratic action. Integration by parts in $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-(\partial \cdot A^a{)}^2/(2\alpha )$ yieldsThe color kernel is diagonal. For $k^2\ne 0$, use the projectors $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$ and $P_T^{\mu \nu }={\eta }^{\mu \nu }-P_L^{\mu \nu }$. They obey $P_L^2=P_L$, $P_T^2=P_T$, $P_L{P}_T=0$, soMultiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the Feynman propagator prescription gives the gluon propagatorThe second term has the causal double-pole prescription. Away from the pole this is the usual $-i{\delta }^{ab}[{\eta }_{\mu \nu }-(1-\alpha )k_{\mu }{k}_{\nu }/k^2]/(k^2+i0)$. For $\alpha =1$ it becomes the Feynman-gauge adjoint propagator; $\alpha \to 0$ gives a transverse covariant Landau gauge propagator.

The limit $\alpha \to \infty$ removes gauge fixing and has no finite inverse on the full field space. The longitudinal eigenvalue $-k^2/\alpha$ tends to zero and its inverse diverges. A contraction with conserved external currents kills the longitudinal term, but that does not define a gauge-unfixed Gaussian integral. This massless limit is not the massive-vector unitary-gauge limit: here it exposes the unfixed gauge orbit degeneracy.

For $L[A^a]=n^{\mu }{A}_{\mu }^a$, eliminating $B^a$ gives the axial gauge term $(n\cdot A^a{)}^2/(2\xi )$ and ghost operatorIn the strict $\xi \to 0$ gauge, $n\cdot A=0$. The gauge-field-dependent part of $n\cdot D$ then vanishes, the Faddeev-Popov determinant becomes field independent, and the ghosts decouple. The one-loop gluon propagator therefore receives the gluon bubble but no ghost bubble. As in the covariant gauge, the four-gluon tadpole can occur with a hard cutoff and vanishes as a scaleless integral in dimensional regularization. Gauge-invariant observables agree between the two gauges even though their individual propagators and diagrammatic decompositions differ.

For $L[A^a]={\partial }^{\mu }{A}_{\mu }^a$, eliminating the Nakanishi-Lautrup field gives the covariant gauge LagrangianThe first term supplies the gluon kinetic term, three-gluon vertex, and four-gluon vertex. The second makes the quadratic gauge-field operator invertible. The last supplies the ghost propagator and ghost-antighost-gluon vertex.

The nonzero one-loop contributions to the gluon propagator are a gluon bubble with two three-gluon vertices and a closed ghost bubble with two ghost-antighost-gluon vertices. A four-gluon tadpole is also present with a cutoff regulator; for massless fields it is a scaleless integral and vanishes in dimensional regularization.