QCD Feynman Rules The classical chromodynamics has a fairly simple Lagrangian 1 a a µν if = − + = F F + Ψ (iD + m )Ψ (1) L LYang Mills Lquarks − 4 µν if 6 f Xf where i denotes the color of a quark and f its flavor. In my notations, I follows the usual summation convention for the Lorentz or color indices, — and the Dirac indices are implicit altogether; but the sum over the quark flavors is explicit since the mass mf depends on the if if a a i jf flavor. OOH, the covariant derivatives Dµ are flavor-blind, DµΨ = ∂µΨ + igAµ(t ) jΨ where ta are matrices representing the gauge group generators in the quark representation; in a 1 QCD the quarks belong to the fundamental 3 representation of the SU(3)C so t are Gell- 2 × Mann matrices λa. The Quantum ChromoDynamics is more complicated, even at the Lagrangian level: including the gauge-fixing and the ghost terms as well as the counterterms, we have 1 a 2 1 µ 2 a µ a if = (F ) (∂µA ) + ∂µc¯ D c + Ψ (iD + m )Ψ L −4 µν − 2ξ if 6 f Xf 2 (4g) δ3 a a 2 (3g) abc b c aν g δ1 abc b c 2 (∂µA ∂ν A ) + gδ f A A ∂µA (f A A ) − 4 ν − µ 1 µ ν − 4 µ ν (2) (gh) a µ a (gh) abc a bµ c + δ ∂µc¯ ∂ c gδ f ∂µc¯ A c 2 − 1 (qf ) (qf ) (qf ) a a if + Ψif iδ ∂ + δm gδ A t Ψ . 2 6 − 1 6 Xf (qf ) (qf ) (qf ) On the last line here, the quark-related counterterms δ2 , δ1 , and δm could be flavor- dependent due to flavor-dependence of the quark mass. QCD Feynman rules follow from expanding the Lagrangian (2) into the free quadratic terms and the interaction terms (cubic, quartic, and all the counterterms). Thus we have: — Gluon propagator a b iδab kµkν = − gµν + (ξ 1) . (3) µ ν k2 + i0 − k2 + i0 1 — Quark propagator ′ i f f f iδjδf ′ = . (4) i j p m + i0 6 − f — Ghost propagator a b iδab = . (5) k2 + i0 Three-gluon vertex • β b k2 k1 a abc αβ γ βγ α γα β = gf g (k1 k2) + g (k2 k3) + g (k3 k1) . (6) α − h − − − i k3 γ c Four-gluon vertex • a β α b f abef cde(gαγgβδ gαδgβγ) − = ig2 + f acef bde(gαβgγδ gαδgγβ) . (7) − − + f adef bce(gαβgδγ gαγgδβ) − d γ δ c Quark-gluon vertex • i f a ′ = igγµ δf (ta)j . (8) µ − × f × i ′ j f 2 Ghost-gluon vertex • b p a abc ′µ µ = +gf p . (9) p′ c In addition, the renormalized theory has a whole bunch of the counterterm vertices: Two-gluon counterterm vertex ∗ a b = iδ δab k2gµν kµkν . (10) µ ν − 3 − Three-gluon counterterm vertex ∗ β b k2 k1 a (3g) abc αβ γ βγ α γα β = gδ1 f g (k1 k2) + g (k2 k3) + g (k3 k1) . α − × h − − − i k3 γ c (11) Four-gluon counterterm vertex • a β α b f abef cde(gαγgβδ gαδgβγ) − = ig2δ(4g) + f acef bde(gαβgγδ gαδgγβ) . (12) − 1 × − + f adef bce(gαβgδγ gαγgδβ) − d γ δ c Two-quark counterterm vertex ∗ ′ ′ f f f j (qf ) (qf ) = δf δi iδm iδ2 p . (13) i j × − ×6 3 Quark-gluon counterterm vertex ∗ i f a (q ) ′ = igδ f δf γµ (ta)j . (14) µ − 1 f × × i ′ j f Two ghost counterterm vertex ∗ a b = δab iδ(gh) k2. (15) × 2 × Ghost-gluon counterterm vertex ∗ b p a ′ = +gδ(gh) f abcp µ. (16) µ 1 × p′ c ⋆ Remember that the ghost fields are fermionic, so each closed loop of ghost propagators carries a minus sign. ⋆ The flavor f remains constant along any quark line, open or closed. For an open line, f matches both the incoming and the outgoing quarks (or antiquarks); for closed quark loops, we sum over all the flavors. ⋆ The color of a quark changes from propagator to propagator since the quark-quark-gluon a j a vertices carry the (t ) i factors. In matrix notations, the t generators should be multiplied 4 right-to-left in the order of arrows on the quark line, for example c b a j = tctbta other factors. ⇒ i × j i For the closed quark lines, one starts at an arbitrary vertex, multiplies all the genera- tors right-to-left in the order of the arrows, than takes the trace over the color indices, tr( tctbta). ··· Ward Identities QCD has weaker Ward identities than QED. In particular, consider the on-shell scattering amplitudes involving the longitudinally polarized gluons. When one gluon is longitudinal and all other gluons are transverse, the amplitude vanishes. But when two or more gluons are longitudi- nal, the amplitude does not vanish; instead, it is related to the amplitudes involving the external ghosts instead of the longitudinal gluons. As an example, consider the tree level annihilation of a quark and an antiquark into a pair of gluons, qq¯ gg. In QED there are two tree diagrams for the e−e+ γγ annihilation, but in → → QCD there are three diagrams: (k2,ν,b) (k1,µ,a) (k2,ν,b) (k1,µ,a) (k2,ν,b) (k1,µ,a) (k1 + k2,λ,c) q1 q2 (p2, j) (p1, i) (p2, j) (p1, i) (p2, j) (p1, i) (17) 5 According to the QCD Feynman rules, these diagrams evaluate to ∗ i ∗ j i =v ¯(p )( igγνe ) ( igγµe ) u(p ) tbta , (18) M1 2 − 2ν q m − 1µ 1 × i 6 1 − ∗ i ∗ j i =v ¯(p )( igγµe ) ( igγνe ) u(p ) tatb , (19) M2 2 − 1µ q m − 2ν 1 × i 6 2 − c j i i 3 =v ¯(p2)( igγλ) u(p1) (t ) i − 2 M − × × (k1 + k2) × ( g)f abc gµν( k + k )λ + gνλ( k (k + k ))µ + gλµ((k + k )+ k )ν × − − 1 2 − 2 − 1 2 1 2 1 the 3 gluon vertex; the unusual signs are due to momenta’s directions hh ii the k and k are outgoing while the k = k + k is incoming hh 1 2 3 1 2 ii ∗ ∗ e e , (20) × 1µ 2ν net = + + . (21) Mtree M1 M2 M3 Clearly, each term in the net amplitude is O(g2) and each term includes the polarization vectors for the two gluons, thus ∗ ∗ = e e µν . (22) M 1µ 2ν ×M µν ?? So let us check the Ward identity k µ M = 0. 1 × For the first diagram’s amplitude we have µν 2 b a j ν 1 k µ = g t t vγ¯ k u. (23) 1 ×M1 − i × q m 6 1 6 1 − In the second factor here, q = p k , hence 1 1 − 1 1 1 1 k = (p q ) = 1 + (p m), (24) q m 6 1 q m 6 1−6 1 − q m 6 1 − 6 1 − 6 1 − 6 1 − which for the on-shell quark gives 1 1 k u(p ) = u(p ) + (p m)u(p ) = u(p )+0 (25) q m 6 1 1 − 1 q m 6 1 − 1 − 1 6 1 − 6 1 − because (p m)u(p ) = 0. Thus, 6 1 − 1 µν 2 b a j ν k µ = +g t t v¯(p )γ u(p ). (26) 1 ×M1 i × 2 1 6 Likewise, for the second diagram µν 2 b a j 1 ν k µ = g t t v¯ k γ u. (27) 1 ×M2 − i × 6 1q m 6 2 − where in the second factor 1 1 1 q2 = k1 p2 = k1 = 1 (p2 + m) = v(p2) k1 = +v(p2) 0, − ⇒ 6 q2 m − 6 q2 m ⇒ 6 q2 m − 6 − 6 − 6 − (28) thus µν 2 a b j ν k µ = g t t v¯(p )γ u(p ). (29) 1 ×M2 − i × 2 1 µν µν In QED, k µ and k µ would have canceled each other, but in QCD eqs. (26) 1 ×M1 1 ×M2 and (29) carry different color-dependent factors. So instead of cancellation, we have µν 2 ν b a a b j 2 ν abc c j k µ = g vγ¯ u t t t t = g vγ¯ u if t . (30) 1 ×M1+2 × − i × − i But the net color-dependent factor is similar to the third amplitude, so there is a hope that the Ward identity might work when all three diagrams are put together. For the third diagram we have µν 2 abc c j λ 1 k µ = ig f t vγ¯ u 1 ×M3 − i × × (k + k )2 × 1 2 (31) µν λ νλ µ λµ ν k µ g (k k ) + g ( k 2k ) + g (2k + k ) , × 1 × 2 − 1 − 1 − 2 1 2 where on the second line ν λ νλ 2 λ ν k µ [ ] = k (k k ) + g ( k 2k k ) + k (2k + k ) 1 × ··· 1 2 − 1 − 1 − 1 2 1 1 2 = gλν (k + k )2 + k2 + (2 1)kλkν + kλkν + kλkν − 1 2 2 − 1 1 1 2 2 1 (32) on shell hh ii = gλν(k + k )2 + (k + k )λ(k + k )ν kλkν . − 1 2 1 2 1 2 − 2 2 Plugging the three terms here back into eq. (31), we obtain µν µν µν µν k µ = k µ + k µ + k µ (33) 1 ×M3 1 ×M3,a 1 ×M3.b 1 ×M3,c where µν 2 abc c j ν k µ = +ig f t v¯(p )γ u(p ), (34) 1 ×M3,a i × 2 1 7 ν µν 2 abc c j (k1 + k2) k1µ 3,b = ig f t i v¯(p2)(k1+k2)u(p1) 2 , (35) ×M − × 6 6 × (k1 + k2) ν µν 2 abc c j k2 k µ = +ig f t v¯(p )(k )u(p ) .