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

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 ∗

p a ′ = +gδ(gh) f abcp µ. (16) µ 1 × p′

⋆ 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 generators 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 longitudinal, 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 ) . (36) 1 ×M3,c i × 2 6 2 1 × (k + k )2 1 2

By inspection of eqs. (34) and (30), the ﬁrst term in eq. (33) precisely cancels the contributions of the ﬁrst two diagrams,

µ µν k µ + k µ = 0. (37) 1 M1+2 1 ×M3,a

The second term’s contribution (35) vanishes for the on-shell quarks. Indeed, by momentum conservation k1 + k2 = p1 + p2, hence v¯(p )(k +k )u(p )=v ¯(p )(p +p )u(p )=v ¯(p )(p + m)u(p )+v ¯(p )(p m)u(p )=0+0 2 6 1 6 2 1 2 6 1 6 2 2 2 6 2 1 2 6 1 − 1 (38) µν and therefore k µ = 0. 1 ×M3,b But the third term’s contribution (35) does not vanish, and this breaks the Ward identity for the net QCD amplitude:

µν µν 2 abc c j 1 ν k µ = k µ = +ig f t vk u k = 0. (39) 1 ×Mnet 1 ×M3,c i × 6 2 × (k + k )2 × 2 6 1 2

ν However, the net violation of the Ward identity is proportional to the k2 factor. Therefore, when we contract the amplitude µν with the polarization vector of the second gluon, we obtain Mnet

µν ∗ ∗ k µ e = [ ] (k e ), (40) 1 ×Mnet 2ν ··· × 2 2 which vanishes when the second gluon is transversely polarized! This agrees with the weakened Ward identity of QCD: Amplitudes involving one longitudinal gluon vanish if all the other gluons are transverse, but if two (or more) gluons are longitudinal, the amplitude does not have to vanish. Instead, such amplitudes are related to the amplitudes involving ghosts and antighosts.

Indeed, consider the annihilation amplitude of two quarks into two longitudinal gluons,

(qq¯ gLgL). In Minkowski space, there are two distinct longitudinal polarizations for a M →

8 µ gluon moving in the direction n, namely e± = (1, n)/√2. In light of eq. (40), the ampli- ± tude for producing two gluons with polarizations L+ (i.e., eµ kµ for each gluon) vanishes, ∝ (qq¯ gL gL ) = 0. The amplitude for producing two gluons with longitudinal polarizations M → + + L also vanishes, (qq¯ gL−gL−) = 0, although I am not going to prove it in these notes. − M → Instead, let me focus on the non-zero amplitude for producing one gluon with the longitudinal L+ polarization and the other gluons with the longitudinal L polarization. − In light of eq. (40), we get

k1µ µν 1 ν (qq¯ gL+gLi) = e1µ(L+) = e2ν (L ) = [ ] e2ν (L )k , M → ω √2 ×Mnet × − ω √2 × ··· × − 2 1 1 (41) where [ ] stands for the factors from eq. (39) which I did not write down explicitly in eq. (40), ··· namely j 1 [ ] = +ig2f abc tc vk u , (42) ··· i × 6 2 × (k + k )2 1 2 while n µ µ ν (1, ) ν ω2 e (L )gµνk = − gµν ω2(1, +n) = 2. (43) 2 − 2 √2 × × √2 ×

1 2 2 Thus, in the center of mass frame where ω1 = ω2 = 2 Ecm while (k1 + k2) = s = Ecm, we have

2 ig abc c j (q +¯q gL + gL−) = f t v(p )k u(p ). (44) M → + s i × 2 6 2 1

Let’s compare this amplitude to the annihilation of the same quark and the same antiquark into a ghost and antighost. At the tree level, there is only one diagram for the later process,

9 which yields the amplitude

j igλν i (q +¯q gh + gh) =v ¯(p )( igγλ)u(p ) (tc − gf abckν Mtree → 2 − 1 × i × s × 2 2 (45) g j = f abc tc v(p )k u(p ). − s i × 2 6 2 1 By inspection, this amplitude is equal to the amplitude (44) for q +q ¯ annihilating into two longitudinal gluons instead of a ghost and an antighost,

(q +¯q gh + gh) = (q +¯q gL + gL−). (46) M → M → + In the next set of notes we shall learn that such relations stem from the BRST symmetry, but right now we may use eq. (46) to understand how the physical cross-sections work in QCD.

The ghosts violate the spin-statistics theorem, so we must give up one one of its assumptions: relativity, positive particle energies, or the positive norm in the Hilbert space. The correct choice is to give up on the norm positivity in the extended Hilbert space including both the physical and the unphysical quanta: While the physical (anti)quarks and transverse gluons must have positive norm, the norm for the unphysical longitudinal gluons is ghost has mixed signature — positive for the longitudinal gluons but negative for the ghosts and antighosts. And because of the negative norm for the (anti)ghosts states, the cross-section for the annihilation-into-ghosts process comes out negative, dσ 2 = |M| . (47) dΩ −64π2s By themselves, the negative cross-sections are impossible, but they make sense in the context of net unpolarized cross-section where the ﬁnal states could be either gluons or ghosts,

dσ(q +¯q ) dσ(q +¯q gT + gT ) dσ(q +¯q gL + gL) dσ(q +¯q gh + gh) →··· = → + → + → . dΩ dΩ dΩ dΩ (48) Thanks to eq. (46), the negative cross-section for the annihilation into ghosts precisely cancels the positive cross-section for the annihilation into longitudinal gluons,

dσ(q +¯q gL + gL) dσ(q +¯q gh + gh) → + → = 0. (49) dΩ dΩ Thus, the un-physical processes cancel each other, and the net annihilation cross-section is just

10 the cross-section for producing the physical states only. At the O(g4) level, this means annihilation into a pair of transverse gluons only,

dσ(q +¯q g + g orgh+ gh) dσ(q +¯q gT + gT only) → = → . (50) dΩ dΩ

Note: this relation is important for the unitarity of QCD.