QCD Beta Function

In these notes I shall calculate the one-loop β(g) functions for QCD and other non- abelian gauge theories. To keep my formulae generic, I allow for any simple gauge group G — although I call the gauge bosons ‘the gluons’ — with Dirac fermions — which I call ‘the quarks’ in some multiplet (Q) of G.

As I explained in class, the non-abelian gauge theories do not obey the QED-like Ward identity Z1 = Z2. Instead, they have weaker identities

1/2 Z(q) Z(gh) Z(3g) Z4g g√Z 1 = 1 = 1 = 1 = 3 , (1) (q) (gh) Z3 Z3 gbare Z2 Z2 !

which nevertheless assure the universality of renormalized gauge coupling g. That is, we have the same g in all couplings of the non-abelian gauge theory: the quark-gluon coupling, the ghost-gluon coupling, the 3-gluon coupling, and the square of the same g in the 4-gluon coupling. In the MS renormalization schemes, the identities (1) translate to linear relations between the counterterms, or rather the simple 1/ǫ poles of the counterterm. Let Res[δ] stand for the residue of the pole at ǫ 0 of the counterterm δ, that is, the coeﬃcient of the → simple 1/ǫ pole regardless of the higher-order poles 1/ǫ2, 1/ǫ3, etc. Then eqs. (1) translate to

Res δ(q) δ(q) = Res δ(gh) δ(gh) = Res δ(3g) δ = 1 Res δ(4g) δ . (2) 1 − 2 1 − 2 1 − 3 2 1 − 3 Moreover, each one of these diﬀerences in combinations with the δ3 counterterm may be used to calculate the β function of the gauge theory:

dg(µ) = β(g) = gLˆ Res 2δ(q) 2δ(q) δ (3.1) d log µ 1 − 2 − 3 = gLˆ Res2δ(gh) 2δ(gh) δ (3.2) 1 − 2 − 3 = gLˆ Res2δ(3g) 3δ (3.4) 1 − 3 g = Lˆ Res 2δ(4g) 4δ , (3.4) 2 1 − 3 where Lˆ = g2(∂/∂g2) is the number-of-loops operator.

1 (q) (q) In these notes, I shall calculate the δ2 , the δ1 , and the δ3 counterterms to the one-loop order, and then use eq. (3.1) to calculate the one-loop beta-function.

(q) The δ2 Counterterm

At the one-loop level, the quark propagator renormalization comes from a single diagram

j j j i = iΣ(p) = hopefully = iΣ(p) δ − 6 i − 6 × i

(4) which evaluates to

4 µν d k i ig j ( igγ ) ( igγ ) − tata . (5) (2π)4 − µ p+k m + i0 − ν × k2 + i0 × i Z 6 6 − (For simplicity, I use the Feynman gauge with ξ = 1 for the gluon’s propagator.) Apart from a a j the group-theoretical factor t t i, the Dirac indexology and the momentum integral here are exactly as in the electron’s propagator correction in QED, so instead of recalculating it from scratch, let me simply copy it from the solutions to homework#17 (problem 3). Focusing on the UV divergence of the integral and disregarding the gory details of the ﬁnite part, we have

d4k i igµν iΣ (p) = ( igγ ) ( igγ ) − − 0 6 (2π)4 − µ p+k m + i0 − ν × k2 + i0 Z 6 6 − (6) g2 1 = i ( p +4m) + ﬁnite . − 16π2 −6 × ǫ

Now consider the group theoretical factor. The ta matrices in the quark-gluon vertices represent the gauge group generators Tâ in the quark multiplet (Q), so the matrix combination ˆ â â a tata represent the Casimir operator C2 = s T T . In any irreducible multiplet (Q), P P 2 this Casimir operators becomes a multiplet-dependent number C(Q) times a unit matrix, thus

j tata = C(Q) δj . (7) i × i In QCD, the gauge group is SU(3)color, the quarks belong to the fundamental 3 multiplets 4 (one such multiple for each quark ﬂavor), the Casimir of this multiple is C(3) = 3 , and we may just as well plug in this number into eq. (7) and subsequent formulae. However, in order to generalize from QCD to other gauge theories — with other gauge groups, and with fermions belonging to other kinds of multiplets — I am going to keep a generic C(Q) in these notes.

Altogether, the quark’s propagator correction is

2 j g C(Q) 1 Σ(p) = δj C(Q) Σ (p) = δj ( p +4m) + ﬁnite , (8) 6 i i × × 0 6 i × 16π2 × −6 × ǫ and to cancel the UV divergence here, we need the counterterms

2 2 (q) g C(Q) 1 (q) g C(Q) 1 δ = and δm = m . (9) 2 − 16π2 × ǫ − × 4π2 × ǫ

(q) The δm counterterm depends on the quark’s mass m, so it’s different for different quark (q) flavors, but the δ2 counterterm is the same for all flavors, at least in the MS renormalization scheme.

(q) The δ1 counterterm (q) The δ1 counterterm cancels the momentum-independent UV divergence of the quark- antiquark-gluon vertex µ a

j j j i = igΓµ,a = hopefully = igΓµ (ta) . − i − × i (10)

3 At the one-loop level, there are two diagrams correcting this vertex, namely

µ a µ a (1) (2)

(λ,c) (ν, b)

j i j i (ν, b) (11) The left diagram here looks just like the electron-photon vertex correction in QED, but the right diagram is new — it appears only in the non-abelian gauge theories like QCD.

Since we are interested only in the UV divergences of the diagrams, and by power- counting such divergences should be independent, let’s simplify the calculations by setting all the external momenta to zero. This also avoids the IR divergences — since the zero quark momenta are oﬀ-shell — and lets us use the same loop momentum kµ for all the propagators in the loop. Thus, for the left diagram (11.a) we have

4 j d k i i i igΓµ,a(0, 0) = ( igγν) ( igγµ) ( igγ ) − − 1 i (2π)4 − k m + i0 − k m + i0 − ν × k2 + i0 Z j 6 − 6 − tbtatb . × i (12) The Dirac indexology and the momentum integral on the top line looks exactly like its QED analogue, which was discussed in painful detail in my notes on the QED vertex corrections. For p′ = p = 0, the algebra becomes much simpler:

4 µ ′ d k Γµ (p = p =0) = ig2 N (13) QED − (2π)4 (k2 m2 + i0)2(k2 + i0) Z −

4 where µ = γν(k + m)γµ(k + m)γ N 6 6 ν = 2kγµ k + 8mkµ 2m2γµ + O(ǫ) − 6 6 − after averaging over the directions of kν (14) hh ii 2k2 = γνγµγ + 0 2m2γµ + O(ǫ) ∼ − D × ν − = (k2 2m2) γµ + O(ǫ). − × Consequently,

4 2 2 ′ d k i(k 2m ) + O(ǫ) Γµ (p = p =0) = g2γµ − − QED × (2π)4 (k2 m2 + i0)2(k2 + i0) Z − d4k (k2 +2m2) + O(ǫ) = g2γµ E E (15) × (2π)4 (k2 + m2)2 k2 Z E × E g2γµ 1 = + a ﬁnite constant . 16π2 ǫ

For the ﬁnite p and p′, we would get the same UV divergence but a much more complicated ﬁnite part (cf. my notes for the gory details),

2 µ ′ g γ ′ Γµ (p ,p) = + ﬁnite function of(p ,p) . (16) QED 16π2 ǫ

b a b j Now consider the group-theoretical factor t t t i (implicit sum over b) on the second a b b a a b b a abc c line of eq. (12). First, using t t = t t + [t , t ]= t t + if t , we get

tbtatb = tb tatb = tbtb ta + tb if abctc = C(Q) ta + if abctbtc. (17) × × × ×

Next, in the second term on the RHS we use the antisymmetry f abc = f acb to rewrite −

if abctbtc = if abc 1 [tb, tc] = if abc 1 if bcdtd = 1 if abc if bcd td. (18) × 2 × 2 × − 2 × − × Finally, we relate the structure constants f abc of the Lie algebra to the matrices representing bc cb the generators in the adjoint multiplet, T a = if abc and T d = if dcb = if bcd. adj adj − 5 Consequently,

bc cb if abc if bcd = + T a T d = tr T a T d tr (T aT b = R(adj) δad ×− adj adj adj adj ≡ adj × Xb,c Xb,c (19) where R(adj) is the index of the adjoint multiplet. As you should see in the homework set#22 (problem 1), for the adjoint multiplet the index is equal to the Casimir, R(adj) = C(adj); it is commonly denoted C(G) where G stands for the gauge group itself. Thus,

if abctbtc = 1 C(G) ta (20) − 2 × and therefore

tbtatb = C(Q) 1 C(G) ta. (21) − 2 × b X Altogether, the ﬁrst diagram (11.1) yields

′ j ′ igΓµ,a(p ,p) = igΓµ(p ,p) (ta)j − 1 i − 1 × i 2 µ (22) ′ g γ for Γµ(p ,p ) = + C(Q) 1 C(G) + ﬁnite. 1 − 2 × 16π2 × ǫ ⋆ ⋆ ⋆

Now consider the second diagram (11.2) with two gluon propagators and one three-gluon vertex. At zero external momenta, this diagram evaluates to

4 2 j d k i i igΓµ,a(0, 0) = ( igγ ) ( igγ ) − − 2 i (2π)4 − λ k m + i0 − ν × k2 + i0 × Z 6 − (23) gλµkν + gµνkλ 2gνλkµ × − j tbtc gf cab . × i × − The second line here (and also the gf cabfactor on the third line) stems from the three- − gluon vertex. Note that if we treat all gluons’ momenta as ﬂowing in to the vertex, then the left gluon has momentum k = +k, the right gluon has momentum k = k, and the top 1 3 − gluon has k2 = 0, hence

gλµ(k k )ν + gµν(k k )λ + gνλ(k k )µ = gλµkν + gµνkλ 2gνλkµ (24) 1 − 2 2 − 3 3 − 1 − on the second line of eq. (23).

6 The group factor on the third line of eq. (23) is an expression we have already evaluated. Indeed, multiplying both sides of eq. (20) by ig, we have

j C(G) j g f cab = f abc tbtc = ig ta . (25) − × i − × 2 × i Consequently,

j igΓµ,a = igΓµ (ta)j (26) − 2 i − × i for

ig2C(G) d4k γ (k + m)γ Γµ(0, 0) = λ6 ν kλgµν + kνgµλ 2kµgνλ . (27) 2 2 × (2π)4 (k2 m2 + i0)(k2 + i0)2 × − Z − (The color-index dependence (26) is valid for all external momenta, but the integral (27) is specialized for p′ = p = 0 only.)

Combining all the factors in the numerator inside the integral (27), we obtain

µ = γ (k + m)γ kλgµν + kνgµλ 2kµgνλ N λ6 ν × − =k(k + m)γµ + γµ(k + m)k 2kµ γν(k + m)γ 6 6 6 6 − × 6 ν = 2k2 γµ + 2mkµ + 2kµ (D 2)k Dm × × − 6 − = (2k2gµν + 2(D 2)kµkν) γ 2(D 1)mkµ − × ν − − (28) after averaging over the directions of kν hh ii 2(D 2) = 2k2 + − k2 gµν γ ∼ D × ν = (3+ O(ǫ)) k2 γµ, × × and therefore

3+ O(ǫ) d4k k2 Γµ(0, 0) = g2C(G) γµ i 2 2 × × × − (2π)4 (k2 m2 + i0)(k2 + i0)2 Z − 3+ O(ǫ) d4k k2 = g2C(G) γµ E E 2 × × × (2π)4 (k2 + m2) (k2 )2 Z E × E (29) 3+ O(ǫ) 1 1 = g2C(G) γµ + ﬁnite 2 × × × 16π2 × ǫ g2 3C(G) γµ = + ﬁnite. 16π2 × 2 × ǫ

7 ⋆ ⋆ ⋆

Combining the two diagrams’ (11) contribution, we obtain

′ j ′ igΓµ,a (p ,p) = igΓµ (p ,p) (ta)j , (30) − 1+2 i − 1+2 × i where at zero momenta p′ = p = 0 we have

g2 1 Γµ (0, 0) = γµ C(Q) 1 C(G)+ 3 C(G) + finite constant , (31) 1+2 × 16π2 × − 2 2 × ǫ while at finite momenta p and p′ we expect the same UV divergence but a very different finite piece

2 µ ′ g γ ′ Γµ (p ,p) = + (C(Q)+ C(G)) + ﬁnite f µ(p ,p) . (32) 1+2 16π2 ǫ To cancel the UV divergence here, we need the one-loop counterterm

g2 1 δ(q) = (C(Q)+ C(G)) . (33) 1 −16π2 × ǫ

(q) (q) Note that in QCD — or in any other non-abelian gauge theory — the δ1 and the δ2 counterterms are not equal to each other — they have diﬀerent group-theoretical factors: (q) (q) C(Q)+C(G) for the δ1 versus C(Q) for the δ2 . However, in gauge theories where fermions belong to several inequivalent irreducible multiplets (r) of the gauge group G, the diﬀerence δ(r) δ(r) is universal: 1 − 2 g2C(G) 1 same δ(r) δ(r) = (r). (34) 1 − 2 − 16π2 × ǫ ∀

Even at the higher loop orders, the diﬀerences δ(r) δ(r) — or at least the 1/ǫ parts of these 1 − 2 diﬀerences — are the same for fermions in any multiplet (r) of the gauge group,

same Res δ(r) δ(r) (r). (35) 1 − 2 ∀ This universality is the special case of the relations (2), which assure that all the coupling of the gauge theory have the same renormalized coupling g(µ).

8 The δ3 Counterterm At the one-loop order, the self-energy corrections to the gluons come from 5 diagrams:

(1) (2) (3) (4) (5)

where the ﬁfth diagram’s contribution

ab Σµν(k) = δ k2gµν kµkν δab (37) 5 − 3 × − × cancels the UV divergences of the ﬁrst 4 diagrams. So let’s calculate those divergences.

The ﬁrst diagram — the quark loop — gives us

ab iΣµν (k) = hopefully = δab iΣµν (k) 1 × d4p i i = tr ( igγµ) ( igγν) tr ta tb , − (2π)4 − p m + i0 − p+k m 0 × (q) (q) i Z 6 − 6 6 − (38) where the ﬁrst trace is over the Dirac indices while the second trace is over the quarks’ colors and ﬂavors. For a single quark multiplet (m) of the gauge group G

tr ta tb = δab R(m) (39) (m) (m) × where R(m) is the index of the multiplet (m). For several quark multiplets, their contributions add up, thus

tr ta tb = δab R = δab R(multiplet). (40) (q) (q) × net × quark multipletsX

In particular, in QCD the quarks comprise Nf copies of a fundamental N multiplet of the

9 SU(N) gauge group — one fundamental multiplet for each ﬂavor — hence 1 R = N R(fundamental) = N . (41) net f × f × 2 Apart from this group factor, the rest of the quark loop (38) looks exactly like the electron loop in QED. We have calculated that loop back in February — cf. my notes — so let me simply recycle the result in the present context:

2 ab g 1 Σµν(k) = δab k2gµν kµkν − R + ﬁnite . (42) 1 × − × 12π2 × net × ǫ Consequently, the counterterm needed to cancel this divergence is

g2 1 4 δ (1st) = R (43) 3 −16π2 × ǫ × 3 net for a general gauge theory; for QCD

g2 1 2 δ (1st) = N . (44) 3 −16π2 × ǫ × 3 f

⋆ ⋆ ⋆

Now consider the second diagram — the gluon loop (α,c)

µ ν a k k b −→ −→ p2

(β,d) (45)

Evaluating this diagram in the Feynman gauge, we get

4 µν ab 1 d p1 i i acd µαβ iΣ (k) = − − gf V (k,p1,p2) 2 2 (2π)4 p2 + i0 × p2 + i0 × − × Z 1 2 (46) gf bcdV ν ( k, p , p ) × − αβ − − 1 − 2 1 where 2 is the symmetry factor due to 2 similar gluon propagators, their momenta add to p + p = k, and the V ’s are the momentum- and Lorentz-index-dependent parts of the 1 2 −

10 3–gluon vertices,

µαβ αβ µ βµ α µα β V (k,p1,p2) = g (p1 p2) + g (p2 k) + g (k p1) , − − − (47) V ναβ( k, p , p ) = V ναβ(k,p ,p ). − − 1 − 2 − 1 2 Let’s start with the group factor in eq. (46). As we saw a few pages above in eq. (19),

cd dc f acd f bcd = iT a +iT b = +tr T a T b = δab R(adjoint) (48) × − adj × adj adj adj × bc bc X X where

R(adjoint) = C(adjoint), often denoted C(G); (49) for an SU(N) gauge group, C(G)= N.

Plugging the group factor into eq. (46) and assembling all the constant factors, we obtain

2 4 µν ab g d p i Σµν (k) = δab C(G) 1 − N2 (50) 2 × 2 × (2π)4 (p2 + i0) (p2 + i0) Z 1 2 × where the numerator is µν = V µαβ(k,p ,p ) V ν ( k, p , p ) = +V µαβ(k,p ,p ) V ν (k,p ,p ) N2 − 1 2 × αβ − − 1 − 2 1 2 × αβ 1 2 = D (p p )µ(p p )ν + gµν (p k)2 + gµν (k p )2 × 1 − 2 1 − 2 × 2 − × − 1 + (p p )(µ(p k)ν) + (p k)(µ(k p )ν) + (k p )(µ(p p )ν) 1 − 2 2 − 2 − − 1 − 1 1 − 2 (51) The second line here has form

A(µBν) + B(µCν) + C(µAν) = (A+B+C)µ(A+B+C)ν AµAν BµBν CµCν; (52) − − − moreover,

A + B + C = (p p ) +(p k)+(k p ) = 0. (53) 1 − 2 2 − − 1 Consequently, the numerator (51) simpliﬁes to

µν = D (p p )µ(p p )ν + gµν (p k)2 + gµν (k p )2 N2 × 1 − 2 1 − 2 × 2 − × − 1 µ ν µ ν µ ν (p1 p2) (p1 p2) (p2 k) (p2 k) (k p1) (k p1) − − − − − − − − − (54) = gµν (p k)2 +(k p )2 + (D 1) (p p )µ(p p )ν × 2 − − 1 − × 1 − 2 1 − 2 (p k)µ(p k)ν (k p )µ(k p )ν. − 2 − 2 − − − 1 − 1 As usual, the ﬁrst step in evaluating the momentum integral like (50) is to simplify the

11 denominator using the Feynman parameters. By momentum conservation p k p , 2 ≡ − − 1 hence

1 1 1 dx dx = = (55) 2 2 2 2 2 2 2 (p1 + i0)(p2 + i0) (1 x)p + x(p1 + k) + i0 [ℓ ∆+ i0] Z0 − 1 Z0 − where

ℓ = p + xk and ∆ = x(1 x)k2. (56) 1 − − Plugging this denominator int eq. (50) we get

1 g2 d4ℓ µν Σµν(k) = i C(G) dx N2 , (57) 2 − 2 × (2π)4 [ℓ2 ∆+ i0]2 Z0 Z − and now we need to re-express the numerator in terms of the shifted momentum ℓ. Using p =+ℓ xk and p = ℓ (1 x)k, we obtain 1 − 2 − − −

p p = 2ℓ (2x 1)k, p k = ℓ +(x 2)k, k p = ℓ +(x+1)k, (58) 1 − 2 − − 2 − − − − 1 − and hence

µν = gµν ( ℓ +(x 2)k)2 + ( ℓ +(x + 1)k)2 N2 × − − − µ ν + (D 1) 2ℓ (2x 1)k 2ℓ (2x 1)k − × − − − − ( ℓ +(x 2)k)µ( ℓ +(x 2)k)ν ( ℓ +( x + 1)k)µ( ℓ +(x 2)k)ν. − − − − − − − − − (59) This whole big mess is a quadratic polynomial in ℓ and k, but the mixed terms like (ℓk) or ℓµkν are odd with respect to ℓ ℓ and hence cancel out from the momentum integral (57). → − Thus, keeping only the terms carrying two or zero ℓ’s, we arrive at

µν = gµν 2ℓ2 + A k2 + B ℓµℓν + C kµkν (60) N2 × × × × where

A = (x 2)2 + (x + 1)2 = 5 2x(1 x), (61) − − −

12 B = (D 1) 4 1 1 = 4D 6, (62) − × − − − C = (D 1) (2x 1)2 (x 2)2 (x + 1)2 − × − − − − = (D 6) (4D 6) x(1 x). (63) − − − × −

Moreover, in the context of the momentum integral (57),

ℓ2 ℓµℓν = gµν, (64) ∼ D ×

hence

B µν = gµνℓ2 2 + + (A + C) gµν k2 C (gµνk2 kµkν N2 ∼ × D × − × − (65) = k2gµν kµkν good + gµν bad, − ×N2 ×N2 where

good = C = (6 D) + (4D 6)x(1 x), (66) N2 − − − − B bad = ℓ2 2 + + k2 (A + C) N2 × D × 6 = ℓ2 6 + k2 (D 1)(1 4x +4x2). (67) × − D × − −

Obviously, the good has the right tensor structure for the gluon’s self-energy corrections, N2 while the bad has the wrong tensor structure. In the context of the momentum integral (57), N2 the bad term bad does not integrate to zero. However, its integral cancels against integrals N2 of the similar bad terms stemming from the two two remaining diagrams.

To see how the cancellation works, let us postpone taking the momentum integral (57) until we have evaluated the sideways gluon loop and the ghost loop diagrams and brought them to a similar form

1 2 4 µν ab g d ℓ i Σµν (k) = δab C(G) dx − N3 , (68) 3 × 2 × (2π)4 [ℓ2 ∆+ i0]2 Z Z − 0 1 2 4 µν ab g d ℓ i Σµν (k) = δab C(G) dx − N4 , (69) 4 × 2 × (2π)4 [ℓ2 ∆+ i0]2 Z Z − 0

13 for some numerators µν and µν, then we are going to add up the numerators, N3 N4

µν = µν + µν + µν, (70) N234 N2 N3 N4 split the net numerator into the ‘good’ and the ‘bad’ tensor structures,

µν = good gµνk2 kµkν + bad gµν, (71) N234 N234 × − N234 × and only then take the momentum integral.

⋆ ⋆ ⋆

For the sideways gluon loop

(c,α) (a, µ)

(b, ν) (72) (d, β)

we have

f abef cde(gµαgνβ gµβgβα) 4 cd ab 1 d p ig δ − iΣµν (k) = − αβ ig2 + f acef bde(gµνgαβ gµβgαν) (73) 3 2 (2π)4 p2 + i0 × −  −  Z ade bce µν βα µα βν  + f f (g g g g )   −  1 where the overall factor 2 comes from the symmetry of the propagator. The group factors in this amplitude evaluate to

δcd f abef cde = 0, × δcd f acef bde = f acef bce = C(G) δab, (74) × × δcd f adef bce = f acef bce = C(G) δab, × ×

14 — cf. eq. (48), — and consequently

δcd [ ] = C(G)δab 2gµνgαβ gµ(αgβ)ν (75) × ··· × − and

g δcd [ ] = C(G)δab (2D 2)gµν. (76) αβ × ··· × −

Plugging this result into eq. (73), we obtain

4 ab d p 1 Σµν(k) = δab g2C(G) i(D 1)gµν (77) 3 × × − (2π)4 p2 + i0 Z Instead of directly evaluating the momentum integral here, we are going to combine the integrand with the other one-loop diagrams. Since this diagram has only one propagator rather than two, we may identify the loop momentum p here as either p or p = p k — 1 2 − 1 − as long as our UV regulator allows constant shifts of the integration variable, both choices are equivalent. For symmetry’s sake, let’s take the average between the two choices and identify

1 2 2 2 2 1 1/2 1/2 p1 + p2 (ℓ xk) + ( ℓ (1 x)k) 2 2 + 2 = 2 2 = dx − 2 − − 2− p + i0 → p1 + i0 p2 + i0 2(p1 + i0)(p2 + i0) 2[ℓ ∆+ i0] Z0 − (78) Consequently, the amplitude (77) takes form (68) for the numerator

µν = (1 D)gµν (ℓ xk)2 +( ℓ (1 x)k)2 = (1 D)gµν 2ℓ2 + (1 2x+2x2)k2 . N3 − × − − − − ∼ − × − (79) In terms of the ‘good’ and the ‘bad’ tensor structures along the lines of eq. (65), this whole numerator is ‘bad’, thus

good = 0, (80) N3 bad = (1 D) 2ℓ2 + (1 2x +2x2)k2 . (81) N3 − × −

15 ⋆ ⋆ ⋆

Finally, there is the ghost loop diagram c p1

µ ν a k k b −→ −→

p2 d (82)

which evaluates to 4 ab d p i i iΣµν (k) = 1 gf acdpµ gf bdcpν. (83) 4 − (2π)4 p2 + i0 × p2 + i0 × − 2 × − 1 Z 1 2 Note: the ghost propagators are oriented and go in opposite directions, so this diagram does 1 not have the symmetry factor 2 . Instead, it carries an overall minis sign for the fermionic loop. Another minus sign hides in the group factor:

f acdf bdc = f acdf bcd = C(G) δbc (84) − − × Consequently,

2 4 µ ν ab g d p +2ip p iΣµν (k) = δab C(G) 1 2 1 (85) 4 × 2 × (2π)4 (p2 + i0)(p2 + i0) Z 1 2 for p2 = +p1 + k. Combining the two denominator factors via the Feynman parameter integral, this amplitude takes form (69) for the numerator

µν = 2pµpν N4 − 2 1 = 2(ℓ xk + k)µ(ℓ xk)ν − − − (86) = 2ℓµℓν + 2x(1 x)kµkν ∼ − − 2 = ℓ2 gµν + 2x(1 x)kµkν , ∼ −D × − or in terms of ‘good’ and ‘bad’ tensor structures,

good = 2x(1 x), (87) N4 − −

16 2 bad = ℓ2 + 2x(1 x) k2. (88) N4 −D × − ×

⋆ ⋆ ⋆

Now let’s total up the numerators of the three diagrams according to the ‘good’ and ‘bad’ tensor structures:

= good k2gµν kµkν + bad gµν, (89) N234 N234 × − N234 ×

where

good = good + good + good N234 N2 N3 N4 = (6 D) + (4D 6) x(1 x) + 0+ 2x(1 x) − − × − − − = (6 D) + 4(D 2) x(1 x), (90) − − × −

bad = bad + bad + bad N234 N2 N3 N4 6 = 6 ℓ2 + (D 1)(1 4x +4x2) k2 − D × − − × + 2(D 1) ℓ2 (D 1)(1 2x +2x2) k2 − − × − − − × h 2 i + ℓ2 + 2x(1 x) k2 −D × − × 8 = ℓ2 8 2D + k2 ( 2D + 4) x(1 x) × − D − × − − 2(D 2) = − D ∆ (D 2) ℓ2 , (91) D × × − − ×

where ∆ = x(1 x)k2, exactly as in the denominator of the momentum integral. − − The net bad-tensor-structure term in the net numerator does not vanish, but it integrates to zero. Or rather, the dimensionally regularized integral of the bad term integrates to zero

17 in any dimension D for which the integral converges (which takes D < 2). Indeed,

dDℓ i bad(ℓ) − N234 = (92) (2π)D [ℓ2 ∆+ i0]2 Z − 2(D 2) dDℓ i D∆ (D 2)ℓ2 = − − − − D × (2π)D [ℓ2 ∆+ i0]2 Z − 2(D 2) dDℓ (D∆+(D 2)ℓ2 = − E − E D × (2π)D (ℓ2 + ∆)2 Z E 2(D 2) dDℓ D 2 2∆ = − E − + D × (2π)D ℓ2 + ∆ (ℓ2 + E + ∆)2 Z E ∞ 2(D 2) dDℓ = − E dt (D 2) + 2∆ t exp t(ℓ2 + ∆) D (2π)D − × × − E Z Z0 ∞ D 2(D 2) − d ℓ − 2 − = − dt (D 2) + 2∆ t e t∆ E e tℓE = (4πt) D/2 D − × × × (2π)D Z Z 0 2(D 2) D − D − = − (D 2) Γ 1 D ∆ 2 1 + 2∆ Γ 2 D ∆ 2 2 D(4π)D/2 × − × − 2 × − 2 4(D 2) D − = − ∆ 2 2 D 1 Γ 1 D + Γ 2 D (92) D(4π)D/2 × × 2 − − 2 − 2 because

D 1 Γ 1 D + Γ 2 D = 0. (93) 2 − × − 2 − 2 Thus, the net vacuum polarization tensor for the gluons does have the right k dependence,

ab Σµν (k) = δab k2gµν kµkν Π (k2) (94) 234 × − × 234 where 1 g2 C(G) d4ℓ i good Π = dx − N234 . (95) 234 2 × (2π)4 (ℓ2 ∆+ i0)2 Z0 Z −

Since the numerator good does not depend on the loop momentum ℓ but only on the N234

18 Feynman parameter x, the momentum integral here becomes the familiar

d4ℓ i d4ℓ 1 1 1 − = E + ﬁnite(x) . (96) (2π)4 (ℓ2 ∆+ i0)2 (2π)4 (ℓ2 + ∆)2 −−→DR 16π2 ǫ Z − Z E

Consequently

1 g2C(G) 1 Π (k2) = + dx (x, D) + ﬁnite(x, k2) 2+3+4 32π2 × Ngood × ǫ Z 0 (97) 1 g2C(G) 1 = + dx (x, D = 4) + ﬁnite(k2) . 32π2 ×  ǫ × Ngood  Z0  

Note that the UV divergence here — the pole at ǫ 0 — obtains from the good at D =4 → N234 (since the diﬀerence between the at D =4 2ǫ and at D = 4 is O(ǫ)), thus N −

1 10 good(x) = (6 D)+4(D 2) x(1 x) 2+8x(1 x) = dx good(x) . N234 − − × − → − ⇒ N234 → 3 Z0 (98) Therefore, g2 5C(G) 1 Π = + + ﬁnite(k2) , (99) 234 16π2 × 3 × ǫ and the δ3 counterterm which cancels this divergence is

g2 5C(G) 1 δ (2nd +3rd +4th) = + . (100) 3 16π2 × 3 × ǫ

Finally, adding the quark loops’ contribution (43), we arrive at the complete one-loop δ3 counterterm, g2 5 4 1 δ = C(G) R (quarks) . (101) 3 16π2 × 3 − 3 net × ǫ

19 The Beta Function

(q) (q) Now that we have the one-loop counterterms δ2 , δ1 , and δ3, we may use them to obtain the one-loop β-function for the gauge coupling. According to eq. (3.1),

1 loop β1 loop = g Res 2δ(q) 2δ(q) δ × 1 − 2 − 3 2 2 2 5 4 g C(Q)+ C(G) g C(Q) g 3 C(G) 3 Rnet(Q) = g 2 − 2 2 − 2 − 2 × " × 16π − × 16π − 16π # g3 11 4 = C(G) + R (Q) . 16π2 × − 3 3 net (102)

For the QCD and for QCD-like theories with an SU(Nc) gauge group and Nf ﬂavors of fundamental multiplets of quarks, C(G)= N , R (Q)= N 1 , hence c net f × 2

g3 11 2 β1 loop(g) = N + N . (103) 16π2 − 3 c 3 f

11 Note the negative coeﬃcient of the Nc-dependent term. Consequently, for Nf < 2 Nc, the whole β-function is negative — or at least it’s negative for a weak enough coupling g — which makes QCD or a QCD-like gauge theory asymptotically free. For more general gauge theories with fermions, the asymptotic freedom requires

11 R (all the fermions) < C(G). (104) net 4

More General Gauge Theories

For completeness sake, let me give you a formula for the one-loop beta function for any gauge theory coupled to several kinds of ‘matter’ ﬁelds: Dirac fermions like the quarks, but also chiral Weyl fermions (left-handed or right-handed only), Majorana fermions, complex scalars, or real scalars. In general, the Dirac fermions, the Weyl fermions, and the complex scalars can be in any multiplets of the gauge group G, while the Majorana fermions and the real scalars must be in real multiplets of G.

20 As required by the gauge coupling universality, for any kind of the matter multiplets —fermionic or scalar — coupled to G we should have the same diﬀerence δ δ as for the 1 − 2 quarks: To all loop orders

same Res δ(m) δ(m) matter multiplet (m), (105) 1 − 2 ∀ and speciﬁcally at one loop,

g2C(G) 1 δ(m) δ(m) = matter multiplet (m), (106) 1 − 2 − 16π2 × ǫ ∀ cf. eq. (34) for the multiplets of Dirac fermions. But let me skip the proof of these formulae and focus on the δ3 counterterm.

At the one loop level, multiplets of Majorana or Weyl fermions aﬀect the δ3 counterterm similarly to the Dirac fermions, via the loop

(36.1)

However, for the Majorana fermions, the solid lines have no arrows, which gives the diagram 1 an extra symmetry factor 1/2. Consequently, their contribution to the δ3 counterterm is 2 of what the Dirac fermions in the same multiplets would contribute,

g2 4/3 1 δ (Majorana) = R (Majorana) . (107) 3 −16π2 × 2 net × ǫ

For the Weyl fermions, the solid lines in the diagram (36.1) do have arrow — which avoids the 1 symmetry factor, — but the vertices have extra chiral projection factors 1 (1 γ5) onto 2 2 ∓ left or right chiralities, which changes the trace over the Dirac indices to

1 γ5 i 1 γ5 i tr ( igγµ) ∓ ( igγν) ∓ . (108) − 2 ×p + i0 × − 2 ×p+k + i0 6 6 6 To evaluate this trace we use the anticommutativity of the γ5 matrix with the γµ and hence

21 with the massless fermion propagators, and also (1 γ5)2 = 2(1 γ5). Consequently, ∓ ∓ 1 i i tr Weyl) = tr (1 γ5)2 ( igγµ) ( igγν) 4 × ± × − × Dp + i0 × − ×p+k + i0 6 6 1 i i = tr ( igγµ) ( igγν) 2 × − × Dp + i0 × − ×p+k + i0 6 6 (109) 1 i i tr γ5 ( igγµ) ( igγν) ± 2 × × − × Dp + i0 × − ×p+k + i0 6 6 1 1 = tr Dirac tr extra , 2 × ± 2 × where the ‘extra’ trace on the bottom line vanishes by the Lorentz symmetry after integrating over the loop momentum p. Indeed,

1 1 tr extra = g2 tr γ5 γµ γν × × p + i0 p+k + i0 6 6 6 tr γ5γµ pγν(p+k) = 4iǫµλνρp (p + k) 6 6 6 λ ρ g2 =  = 4iǫµλνρp k  (p2 + i0)((p + k)2 + i0) × λ ρ    µλνρ   by the antisymmetry of ǫ    p = 4ig2ǫµνρλ k λ . ρ × (p2 + i0)((p + k)2 + i0) (110) By Lorentz, symmetry, when we integrate the p-dependent factor here over p, we obtain

dDp p λ = (scalar) k , (111) (2π)D (p2 + i0)((p + k)2 + i0) × λ Z hence dDp tr extra = 4ig2ǫµνρλ k (scalar) k =0 (112) (2π)D ρ × × λ Z by the antisymmetry of the ǫµνρλ tensor. Consequently, eq. (109) for the trace over a loop of Weyl fermions simpliﬁes to

1 tr Weyl) = tr Dirac), (113) ∼ 2 × 1 so the whole loop diagram for a Weyl fermion multiplet is precisely 2 of what we would get

22 for a similar multiplet of Dirac fermions. Thus, just like for the Majorana fermions,

g2 4/3 1 δ (Weyl) = R (Weyl) . (114) 3 −16π2 × 2 net × ǫ

For the complex scalars, we have two one-loop diagrams

(115)

These diagrams look exactly similar to the diagrams renormalizing the gauge coupling in the scalar QED, which you should have evaluated in homework set#16. The only diﬀerence from the scalar QED is that the scalar’s electric charge2 (in units of e2) is replaced by the trace over the color indices

tr tatb = δab R(scalar multiplet). (116) colors × All other aspects of the two diagrams — the Lorentz indexology and the momentum integrals

— work exactly as for the scalar QED, so copying the formula for the scalar QED’s δ3 from the solutions to homework#16 and multiplying by the group factor (116), we obtain

g2 1 1 δ (complex scalars) = R (complex scalars) . (117) 3 −16π2 × 3 × net × ǫ

Finally, for the real scalars we have two one-loop diagrams similar to (115), but without the arrows on the dotted lines. Consequently, both diagrams get an overall symmetry factor 1 1/2, hence the δ3 counterterm as in eq. (117) times 2 ,

g2 1/3 1 δ (real scalars) = R (real scalars) . (118) 3 −16π2 × 2 × net × ǫ

23 Altogether, for a gauge theory with a simple gauge group G and matter (fermionic and/or scalar) in some multiplets (m), we have

11 for the gauge fields, − 3 4  + 3 for Dirac fermions,  2 g3  + for Majorana fermions, β1 loop(g) = R(multiplet)  3 (119) 16π2 × ×  2 all physical  + 3 for chiral Weyl fermions, X  multiplets + 1 for complex scalar fields,  3  1  + 6 for real scalar fields.   a In this formula, the gauge fields’ contribution includes both the vector fields Aµ themselves as well as the ghosts ca andc ¯a, so please do not count the ghosts as separate multiplets.

Note that only the non-abelian gauge fields give negative contributions to the β function, all other fields’ contributions are positive. Consequently, only the non-abelian gauge theories can be asymptotically free, and only when there are not too many fermionic or scalar fields coupled to the gauge fields. For example, the QCD-like theories are asymptotically free only 11 for Nf < 2 Nc. In a theory with a product gauge group G = G G , each component group G 1 ⊗ 2 ⊗··· i — abelian or non-abelian — has its own gauge coupling gi. At the one-loop level, the beta

functions of each gi are independent from each other, and also from the other couplings like Yukawa of λφ4, thus g3 g3 i, β = i b + i O(g2, other g2, yukawa2,λ) (120) ∀ i 16π2 × i (4π)4 × i j

3 2 where bi are the numerical factors which obtain exactly as the factor multiplying g /16π in eq. (119). However, for each bi you should count multiplets of the appropriate Gi without paying attention to the other gauge groups Gj. For example, a bi-fundamental (m, n) multiplet of an SU(m) SU(n) gauge group counts as m fundamental multiplets of SU(n) ⊗ when you calculate the βn, — oras n fundamental multiplets of SU(m) when you calculate

the βm, — thus

R ((m, n)) = n R((m)) = m 1 , R ((m, n)) = m R((n)) = n 1 . (121) SU(m) × × 2 SU(n) × × 2 And for the abelian U(1) factors, the index of a charged singlet is simply its charge squared, while the index of a complete multiplet WRT all the other G factors is R = (multiplet size) j × 24 (abelian charge)2. For example, in the Standard Model SU(3) SU(2) U(1), a multiplet × × (m of SU(3); n of SU(2); hypercharge = y) has

R ((m, n,y)) = mn y2. (122) U(1) ×