QCD β Function. ǫ C. multiplet
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- Catherine Cole
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1 QCD β Function In these notes, I shall calculate to 1-loop order the δ counterterm for the gluons and hence the β functions of a non-abelian gauge theory such as QCD. For simplicity, I am going to refer to the gaugefields as gluons and to the fermions as quarks, but I am going to allow for any simple gauge group G and for the fermions in any complete multiplet of G. Once I finish the calculation, I ll spell out the group factors for G = SUN c ) and N f fundamental multiplets of quarks, but until then I ll keep all the group factors generic. As explained in class see also my notes on minimal subtraction the β function for the gauge coupling can be obtained from the infinite parts of the counterterms as dgµ) dlogµ = βg) = g N loops Residue of 1 ǫ pole of[ δ 1 δ 1 δ ]. 1) In class, I have also calculated the δ and the δ 1 counterterms for the quarks to one-loop order: in the Feynman gauge ξ = 1) for the gluons and the MS regularization scheme, δ quark) = g 16π 1 ) quark ǫ C, ) multiplet δ 1 quark) = g 16π 1 [ ) quark ǫ C multiplet ] + Cadjoint). ) Note: in theories where the fermions form several different multiplets of the gauge group, each multiplet would have its own δ and δ 1 counterterms, and they would differ from multiplet to multiplet. However, all the multiplets would have the same difference δ 1 δ = g 16π 1 Cadjoint). 4) ǫ In light of eq. 1), this means that all the fermions couple to the gauge fields with exactly the same renormalized coupling gµ). In these notes, I shall calculate the δ counterterm to the one-loop order; once I have it, eqs. 1) and 4) would give me the one-loop) βg). 1
2 At the one-loop order, the self-energy corrections to the gluons come from 5 diagrams: 1) ) ) 4) 5) where the fifth diagram s contribution Σ µν 5 k) = δ k g µν k µ k ν) 5) cancels the UV divergences of the first 4 diagrams. So let s calculate those divergences. The first diagram the quark loop gives us iσ µν d 4 ) 1 k) = p π) 4 tr igγ µ i i ) p m+i igγν ) p+ k m i ) tr t a q) tb q) 6) 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 t a m) tb m) = δ ab Rm) 7) where Rm) is the index of the multiplet m). For several quark multiplets, their contributions add up, thus tr ) t a q) tb q) = δ ab R net = δ ab quark multiplets Rmultiplet). 8) In particular, in QCD the quarks comprise N f copies of a fundamental N multiplet of the SUN) gauge group one fundamental multiplet for each flavor hence R net = N f Rfundamental) = N f 1. 9) Apart from this group factor, the rest of the quark loop 6) looks exactly like the electron loop in QED. We have calculated that loop back in February cf. my notes so let me
3 simply recycle the result in the present context: Σ µν 1 k) = k g µν k µ k ν) ) g 1 1π R net ǫ + finite. 1) Consequently, the counterterm needed to cancel this divergence is for a general gauge theory; for QCD δ 1 st ) = g 16π 1 ǫ 4 R net 11) δ 1 st ) = g 16π 1 ǫ N f. 1) Now consider the second diagram the gluon loop α,c) µ ν a k k b p 1 p β,d) 1) Evaluating this diagram in the Feynman gauge, we get iσ µν k) = 1 d 4 p 1 π) 4 p 1 i i +i p V µαβ k,p 1,p ) gf bcd V ν +i gfacd αβ k, p 1, p ) where 1 is the symmetry factor due to similar gluon propagators, and the V s are the momentum- and Lorentz-index-dependent parts of the gluon vertices, V µαβ k,p 1,p ) = g αβ p 1 p ) µ + g βµ p k) α + g µα k p 1 ) β, 14) V ναβ k, p 1, p ) = V ναβ k,p 1,p ) 15) = [ g αβ p 1 p ) ν + g βν p k) α + g να k p 1 ) β].
4 Let s start with the group factor in eq. 14). As explained in class, ) cd ) dc it a adj +it a adj = tr T a adj Tadj b ) = δ ab Radjoint); 16) cd f acd f bcd = cd for an SUN) gauge group, Radjoint) = N. For a general gauge group, Radjoint) = Cadjoint), often denoted CG). 17) Plugging the group factor into eq. 14) and assembling all the constant factors, we obtain where the numerator is Σ µν g d 4 k) = i CG) p 1 N µν π) 4 p 1 +i) p +i) 18) N µν = V µαβ k,p 1,p ) V ν αβ k, p 1, p ) = +V µαβ k,p 1,p ) V ν αβ k,p 1,p ) = D p 1 p ) µ p 1 p ) ν + g µν p k) + g µν k p 1 ) + p 1 p ) µ p k) ν) + p k) µ k p 1 ) ν) + k p 1 ) µ p 1 p ) ν). 19) As usual, the first step in evaluating the momentum integral like 18) is to simplify the denominator using the Feynman parameters. By momentum conservation p k p 1, hence where 1 1 p 1 +i)p +i) = Plugging this denominator int eq. 18) we get 1 dx [ 1 x)p 1 +xp 1 +k) +i ] = dx [l +i] ) l = p 1 + xk and = x1 x)k. 1) Σ µν g k) = i CG) d 4 l N µν dx π) 4 [l +i], ) and now we need to re-express the numerator in terms of the shifted momentum l. Using 4
5 p 1 = +l xk and p = l 1 x)k, we obtain p 1 p = l x 1)k, p k = l + x )k, k p 1 = l + x+1)k, ) and hence N µν = D l x 1)k ) µ l x 1)k ) ν + g µν [ l+x )k) + l+x+1)k) ] + l x 1)k) µ l+x )k) ν) + l+x )k) µ l+x+1)k) ν) + l+x+1)k) µ l x 1)k) ν) 4) This whole big mess is a quadratic polynomial in l and k, but the mixed terms like lk) or l µ k ν are odd with respect to l l and hence cancel out from the momentum integral ). Thus, keeping only the terms carrying two or zero l s, we arrive at N µν [ = D 4l µ l ν + x 1) k µ k ν] + g µν [ l + x ) +x+1) )k ] 4l µ l ν x 1)x )k µ k ν + l µ l ν + x )x+1)k µ k ν 4l µ l ν x 1)x+1)k µ k ν = g µν [ l + x x+5)k ] + 4D 6)l µ l ν + [ D 6) 4D 6)x1 x) ] k µ k ν. 5) Moreover, in the context of the momentum integral ) l µ l ν = l D gµν, 6) hence N µν = k g µν k µ k ν) [ 6 D) + 4D 6)x1 x) ] + g µν 6 6 ) ) l + D 1)1 4x+4x ) k D 7) where I have re-arranged the k terms so that the first line has the right tensor structure for the gluon s self-energy corrections. The second line has wrong tensor structure, and it does not integrate to zero, but we shall see that it cancels against similar bad terms from the two remaining diagrams. 5
6 To see how the cancellation works, let us postpone taking the momentum integral ) until we have evaluated the sideways gluon loop and the ghost loop diagrams and brought them to a similar form Σ µν g k) = i CG) Σ µν g 4 k) = i CG) d 4 l N µν dx π) 4 [l +i], 8) d 4 l N µν 4 dx π) 4 [l +i], 9) for some numerators N µν and N µν 4, then we are going to add up the numerators, N µν = N µν + N µν + N µν 4, ) and only then take the momentum integral. For the sideways gluon loop a,µ) c,γ) p b,ν) d,δ) 1) we have iσ µν k) = 1 d 4 p ig γδ δ cd π) 4 p +i ig f abe f cde g µγ g νδ g µδ g βγ ) +f ace f bde g µν g γδ g µδ g γν ) +f ade f bce g µν g δγ g µγ g δν ) ) where the overall factor 1 comes from the symmetry of the propagator. The group factors 6
7 in this amplitude evaluate to δ cd f abe f cde =, δ cd f ace f bde = f ace f bce = CG) δ ab, ) δ cd f ade f bce = f ace f bce = CG) δ ab, cf. eq. 16) hence g γδ δ cd [ ] = CG)δ ab g γδ g µν g γδ g µδ g γν + g µν g δγ g µγ g δν) = CG)δ ab Dg µν g µν + Dg µν g µν) 4) = CG)δ ab D 1)g µν. Plugging this result into eq. ), we obtain Σ µν k) = +ig CG) D 1)g µν d 4 p π) 4 1 p +i 5) Instead of directly evaluate 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 1 or p = p 1 k 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 p +i = 1/ 1/ + +i p +i = p 1 +p p 1 +i)p +i) = p 1 Consequently, the amplitude 5) takes form 8) for the numerator dx l xk) + l 1 x)k) [l +i] 6) N µν = D 1)g µν [ l xk) + l 1 x)k) ] = D 1)g µν [ l +1 x+x )k ]. 7) 7
8 Finally, there the ghost loop diagram c p 1 µ ν a k k b p d 8) which evaluates to iσ µν 4 k) = d 4 p 1 π) 4 i p 1 +i i p +i gfacd p µ gfbdc p ν 1. 9) Note: the ghost propagators are oriented and go in opposite directions, so this diagram does not have the symmetry factor 1. Instead, it carries an overall minis sign for the fermionic loop. Another minus sign hides in the group factor: f acd f bdc = f acd f bcd = CG) δ bc 4) Consequently, Σ µν d 4 4 k) = ig CG) p 1 p µ pν 1 π) 4 p 1 +i)p +i) 41) for p = +p 1 + k. Combining the two denominator factors via the Feynman parameter integral, this amplitude takes for 9) form the numerator N µν 4 = p µ pν 1 = l xk +k) µ l xk) ν = l µ l ν + x1 x)k µ k ν = D l g µν + x1 x)k g µν x1 x) k g µν k µ k ν). 4) 8
9 Now let s add up the numerators of the three diagrams: N µν = N µν + N µν + N µν 4 = k g µν k µ k ν) [ 6 D) + 4D 6)x1 x) + x1 x) ] [ + g µν l 6 6 D D 1) ] D + g µν k [ D 1)1 4x+4x ) D 1)1 x+x ) + x1 x) ] = k g µν k µ k ν) [ 6 D) + 4D )x1 x) ] + g µν l D ) D + g µν k D)x1 x) k g µν k µ k ν) N good + g µν N bad 4) where N good x) = 6 D) + 4D )x1 x), D N bad x,l) = D ) D l + x1 x)k = D ) D l, 44) for the same = x1 x)k as in the denominator of the momentum integral. The 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 in any dimension D for which the integral converges which takes D < ). Indeed, d D l π) D N bad l) [l +i] = d D D l E D = D )i l E + π) D l E + ) d D ) l E D )/D /D) = D )i π) D l + e + l + ) = D )i d D l E D π) D = D )i D dt D ) + t ) exp tl E + )) dt D ) + t ) e t = D )i D4π) D/ D ) Γ 1 D d D ) l E E π) D = 4πt) D/ e tl ) D 1 + Γ D ) D ) 9
10 = 4D )i D 1 D4π) D/ D 1 ) Γ 1 D ) ) ) + Γ D 45) where the expression inside ) on the last line has form yγy) + Γy +1) for y = 1 D 46) and vanishes y = D. Thus, the net vacuum polarization tensor for the gluons does have the right k dependence, Σ µν ++4 k) = k g µν k µ k ν) Π ++4 k ) 47) where Π ++4 = ig CG) dx d 4 l π) 4 N good l +i). 48) Since the numerator N good does not depend on the loop momentum l but only on the Feynman parameter x, the momentum integral here becomes the familiar d 4 l 1 d 4 π) 4 l +i) = l E +i π) 4 l E + ) DR ) i 1 16π ǫ + finite. 49) Consequently, Π ++4 = + g CG) π dxn good x) ) 1 ǫ + finite, 5) and to get just the coefficient of the 1ǫ divergence, we may replace the N good x) for general dimension D with its value for D = 4. Thus N good x) + 8x1 x) = dxn good x) 1, 51) hence Π ++4 = + g 16π 5CG) ) 1 ǫ + finite, 5) 1
11 and the δ counterterm which cancels this divergence is is δ nd + rd +4 th ) = + g 16π 5CG) 1 ǫ. 5) Altogether includingthequarkloops contribution11) theone-loopδ counterterm δ oneloop = g 5 16π CG) 4 ) R netquarks) 1 ǫ. 54) Consequently, applying eqs. 1) and 4), we obtain δ 1 δ 1 δ )1loop = g 16π CG) 5 6 CG) + ) R netquarks) 1 ǫ and therefore 55) βg) 1loop = g 16π 11 CG) + 4 ) R netquarks). 56) For QCD and QCD-like theories with SUN c ) gauge group and N f flavors of fundamental multiplets of quarks, βg) 1loop = g 16π 11 N c + ) N f. 57) For completeness sake, let me give you without proof the 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, Weyl fermions, and complex scalars can be in any multiplets of the gauge group G, while the Majorana fermions and real scalars must be in real multiplets of G. Altogether, for the gauge fields, βg) 1loop = g 16π all physical multiplets for Dirac fermions, + for Majorana fermions, Rmultiplet) + for chiral Weyl fermions, + 1 for complex scalar fields, for real scalar fields. Note: the gauge fields contribution here includes both the vector fields A a µ themselves and 58) 11
12 the ghosts c a and c a, so do not count the ghosts as separate multiplets. Also 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, QCD-like theories are asymptotically free only for N f < 11 N c. Finally, a note on theories with product gauge groups G = G 1 G. In such theories, each component group G i abelian or non-abelian has its own gauge coupling g i. At the one-loop level, the beta functions of each g i are independent from each other and also from the other couplings like Yukawa or λφ 4 ), although at higher loops that s no longer true, i, β i = g i 16π const + g i 4π) 4 Og i, otherg j, yukawa,λ) 59) Moreover, the constant factors in the one-loop terms are obtained exactly as in eq. 58) where we count multiplets of each G i without paying attention to the other gauge groups G j. For example, a m,n) multiplet of SUm) SUn) counts as m fundamental multiplets of SUn) when you calculate the β n or as n fundamental multiplets of SUm) when you calculate the β m. 1
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