Introduction to Renormalization QCD (Dated: May 21, 2007) Abstract We will show the renormalization procedures for Quantum Chromodynamics , and derive the renormalization group equation of strong coupling, αs . As a non-Abelian gauge theory, the ghost elds have to be introduced. Because the contributions from the ghost elds and 3-gluon interaction, more Feynman diagrams of fermion self-energy, gauge boson self-energy and quark-quark-gluon interaction are needed to get Z1,2,3. And the simple relation of Z1 = Z2 is not valid. We use the dimensional regularization to regularize the UV divergences in our calculation.
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1 I. INTRODUCTION
In the previuos two lectures, we have introduced the procedures to regularize the di- vergences as calculating the one-loop Feynman diagrams of the Quantum Electrodynamics (QED). We also derived the running coupling which has a powerful prediction on the cou- pling strength at the dierent energy scales. In this lecture, we will use the same techniques to deal with another successful theory, Quantum Chromodynamics (QCD). QCD is the theory describing the strong interaction of quarks and gluons at all energy scales. Unlike QED, QCD is a non-Abelian gauge theory described by SU(3) gauge theory, the quanta, gluons, carry the color quantum numbers, and could interact with each other. This causes a signi cant results, asymptotic freedom, which means the coupling strength becomes weaker as energy scale becomes higher, we will see this features later as we derive the renormalization group equation. And also because of the asymptotic freedom, the in- teractions between quarks and gluons are so strong at low energy scale that the quarks will form bound states to be hadrons, either mesons (qq¯ bound state) or baryons (qqq bound state). All the hadrons seen in Nature are color singlet objects, and no free quark is ob- served. Since we can not observe the colored object, it is interesting to know how the 3-color model be con rmed. The rst suggestive experiment was π → γ γ decay, described by the Fig. 1 (a). Another experiment to con rm the number of colors was the ratio R, de ned as R ≡ σ(e+e− → quarks)/σ(e+e− → µ+µ−), shown in Fig. 1 (b). Since the coupling strength is strong at low energy, the perturbation calculations are not reliable anymore. The tech- niques to do the calculations at the low energy scale is Lattice QCD, we are not interested in this aspect in this notes. This note is organized as follows, we give an introduction of colors in Sec. II. The Lagrangian and renormalization of QCD are give in Sec. III. The Sec. IV is the summary of this notes. We also give the Feynman rules in Appendix. Some useful formulas can be found in the QED renormalization notes.
II. SU(3) AND COLORS
In the QCD, new quantum numbers are introduced, called colors. Colored Quark trans- forms as a fundamental representation under SU(3), Each quark carries one of the three colors which are red, blue and green. The gluon, quanta of QCD, transforms as an adjoint
2 e γ − q 0 q π γ∗
γ + q¯ (a) e (b)
Figure 1: (a) π0decays to γγ, (b) e+e− annihilation to quarks. representation under SU(3). The eight generators, ta, of SU(3) should satisfy the Lie algebra,
[tA, tA] = if ABC tC , (A, B, C) = 1 ∼ 8,
where f ABC is the structure function. One of the representations for the generators is provided by Gell-Mann matrices, λa, which are hermitian and traceless, 1 tA = λA 2 0 1 0 0 −i 0 1 0 0 1 2 3 λ = 1 0 0 , λ = i 0 0 , λ = 0 −1 0 , 0 0 0 0 0 0 0 0 0 0 0 1 0 0 −i 0 0 0 4 5 6 λ = 0 0 0 , λ = 0 0 0 , λ = 0 0 1 , 1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 1 7 8 λ = 0 0 −i , λ = √ 0 1 0 . 3 0 i 0 0 0 −2 By convention, the normalization of the SU(N) matrices is chosen to be 1 T r{tAtB} = δAB. 2 With this choice, we can have the color matrices obey the following relations which are needed in calculations,
X 4 tA tA = C δ ,C = ; ab bc F ac F 3 A C D X ABC ABD CD C CAB T r{T T } = f f = CAδ , where (T )AB ≡ −if ,CA = 3. A,B
3 III. RENORMALIZATION OF QCD
In this section, we will proceed what we did in the QED and see how it becomes more complicated because of the self-interaction of gluons. First, we begin with the QCD La- grangian,
1 L = − F F 0µν + Ψ¯ (i 6∂ + m )Ψ − c¯a∂2ca + g Ψ¯ γ taΨ Aaµ − g f (∂ Aa )AbµAcν QCD 4 0µν 0 0 0 0 0 0 0 µ 0 0 0 abc µ 0ν 0 0 2 a b cµ dν a µ b c (1) −g0fabcfdecA0µA0νA0 A0 − g0fabcc¯0(∂ A0µc0),
where Ψ is the quark eld, Fµν ≡ ∂µAν − ∂νAµ, A is the gluon eld and c is the ghost eld. As we did in QED, the subscript 0 denotes the bare quantity. Now we introduce Z0s to de ne the normalized elds, therefore, we rewrite the Eq. (1) in terms of renormalized (without the subscript 0) quantities as
1 L = − Z F F µν + Z Ψ(¯ i 6∂ + m)Ψ − Z c¯a∂2ca + Z gΨ¯ γ taΨAaµ − Z3ggf (∂ Aa)AbµAcν QCD 4 3 µν 2 c 1 µ 1 abc µ ν 4g 2 a b cµ dν c a µ b c (2) −Z1 g fabcfdecAµAνA A − Z1gfabcc¯ (∂ Aµc ),
1 = − F F µν + Ψ(¯ i 6∂ + m)Ψ − c¯a∂2ca + gΨ¯ γ taΨAaµ − gf (∂ Aa)AbµAcν 4 µν µ abc µ ν 2 a b cµ dν a µ b c −g fabcfdecAµAνA A − gfabcc¯ (∂ Aµc ) + Lcounter.
The Feyman rules are given in the Appendix. By comparing Eqs. (1) and (2), we can easily have
p aµ p aµ a p a (3) Ψ0 = Z2Ψ,A0 = Z3A , c0 = Zcc and
p Z1g = Z2 Z3g0 (4)
3g 3/2 (5) Z1 g = Z3 g0 4g 2 2 2 (6) Z1 g = Z3 g0 c p (7) Z1g = Zc Z3g0 .
4 k
pk + p p
Figure 2: The quark self-energy Feynman diagram.
We could choose one relation among the Eqs. (4), (5), (6) and (7) to derive the running coupling , or ( g2 ). Here, we choose Eq. (4), since we already have the results for , g αs ≡ 4π Z1
Z2 and Z3 from QED. The extra works for QCD is that we need to take care about the colors and add in more diagrams. From Eq. (4), we have
Z2 2 2 2 (8) g = 2 Z3g0 . Z1
As we learn in QED case, g, Z1,2,3 are scale dependent, therefore, we can take the deriva- tive and get
2 2 2 dg 2Z2 dZ2 2Z2 dZ1 Z2 dZ3 2 2 = 2 2 Z3 − 3 2 Z3 + 2 2 g0 dµ Z1 dµ Z1 dµ Z1 dµ 2 dZ 2 dZ 1 dZ 2 1 3 2 (9) = 2 − 2 + 2 g Z2 dµ Z1 dµ Z3 dµ
The Feynman diagram related to Z2 is shown in Fig. 2. Compared to the diagram in QED, the only dierence is that the color factor is needed. So the result is the result of QED multiplied by color factor, tAtA. From Eq. (26) in the QED notes, taking n → 4 and 2 2 p = −µ , we got the Z2 in QCD is
∂ P α πn/2−2Γ(2 − n/2) Z = 1 + 2 = 1 − s C (10) 2 ∂ 6p 4π (µ2)2−n/2 F where 2 . The Feynman diagrams related to are shown in Fig. 3. Form the αs ≡ gs /(4π) Z1 µ A vertex Feynman rule, we write the corrections of vertex from one loop diagrams as −igsΛ t . The Fig. 3 (a) can be easily got from the QED result, and multiply it by the factor factor, tBtAtB, which we need to calculate.
5 p′ C B t A t ABC t q f
tB k tB (a) (b) p
Figure 3: The Feynman diagrams of one-loop correction to vertex of quark-quark-gluon.
tBtAtB = tB([tA, tB] + tBtA) = tB(if ABC tC ) + tBtBtA f ABC [tB, tC ] = i + C tA 2 F ABC BCD D A = if if t /2 + CF t tD = −T r{T AT D} + C tA 2 F −C = ( A + C )tA (11) 2 F Therefore,
−C Λµ(Fig.3a) = ( A + C )Λµ(QED, e → g ) 2 F s −C α πn/2−2Γ(2 − n/2) = ( A + C ) s . (12) 2 F 4π (µ2)2−n/2
Now we will work on the diagram Fig. 3 (b), Λ0µ(p, p0).
f ABC tC tB Z N µ Fig.3(b) = −ig3 d4k , (13) s (2π)4 D where
µ µν λ νλ µ µλ ν N = γλ 6kγν(g k − 2g k + g k ), (14) D = (k2 + i)(k2 + 2p · k + i)(k2 − 2q · k + q2 + i). (15)
Here, we already drop the linear term in the N µ, since the integral is zero for linear term. The color factor
6 1 f ABC tC tB = f ABC [tC , tB] 2 1 = f ABC if CBDtD 2 i = − (T A) (T D) tD 2 BC CB i = − C tA. (16) 2 A
The Λ(Fig.3b) then is C 1 Z N µ Λ(Fig.3b) = −i A g2 d4k . (17) 2 s (2π)4 D By using the Feynman tricks to evaluate the denominator,
1 Z 1 Z y 1 = 2 dx dy 2 3 D 0 0 {[k + px − (q + p)y] − ∆} Z 1 Z y 1 = 2 dx dy , (18) 2 3 0 0 [k˜ − ∆] where k˜ = k + px − (q + p)y and ∆ = −q2y(1 − x). The numerator now is
µ 2 µ ν µ N = 2k γ − 2γ 6kγνk = 2(k˜2 + qxy)γµ − 2(2 − n)(6k˜ − x 6p + y 6p0)(k˜ − xp + yp0)µ.
After taking n → 4 and p = p0, i.e. q → 0, for non-divergent term, and let −q2 → µ2, we have
2 Z 1 Z x CA 2 π 6Γ(2 − n/2) µ Λ(Fig.3b) = gs 4 dx dy 2 2−n/2 γ 2 (2π) 0 0 [−q y(1 − x) + i] 3C α πn/2−2Γ(2 − n/2) = A s . (19) 2 4π (µ2)2−n/2 Combine Eq. (12) and Eq. (19), We shall have
α πn/2−2Γ(2 − n/2) −C 3C Z = 1 − s ( A + C + A ) 1 4π (µ2)2−n/2 2 F 2 α πn/2−2Γ(2 − n/2) = 1 − s (C + C ). (20) 4π (µ2)2−n/2 F A
Now the only piece left to derive the running αs is Z3. The relevant Feynman diagrams are shown in Fig. 4.
7 q k k A, µ D, ν
(a) (b)
(c) (d)
Figure 4: The one-loop contributions to the gluon self-energy.
The tensor of Fig. 4 (a) can be calculated easily by multiplying color factor to the QED result which we already have, i.e. Eq. (52) in lecture note II. The color factor is
1 T r{tAtA} = . 2 Therefore, the contribution from Fig. 4 (a) is
n/2−2 Z 1 2 2 1 8αs π Γ(2 − n/2) x(1 − x) Π(a)(k = −µ ) = − 2 2−n/2 dx 2−n/2 2 4π (µ ) 0 [x(1 − x)] 1 8α πn/2−2Γ(2 − n/2) Γ(2)Γ(2) = − s 2 4π (µ2)2−n/2 Γ(4) 2 α πn/2−2Γ(2 − n/2) = − s (21) 3 4π (µ2)2−n/2 The tensor of Fig. 4 (b) is
g2f ABC f DCB Z N µν Πµν (k) = − dnq (b) 2(2π)4i (q2 + i)((q + k)2 + i) g2C δAD Z N µν = A dnq , (22) 2(2π)4i (q2 + i)((q + k)2 + i) where
N µν = [(q + 2k)2 + (q − k)2]gµν + n(2q + k)µ(2q + k)ν +[−(q + 2k)µ(2q + k)ν + (q − k)µ(q + 2k)ν − (q − k)µ(2q + k)ν + µ ↔ ν]
Because R dnqq−2 vanishes in dimensional regularization, any terms of the form Z q2 dnq q2(q + k)2
8 should be consistently eliminated. The non-vanishing terms of N µν are
(2kq + 5k2)gµν + n(2q + k)µ(2q + k)ν +[−(q + 2k)µ(2q + k)ν + (q − k)µ(q + 2k)ν − (q − k)µ(2q + k)ν+ µ ↔ ν].
Using Feynman trick, the denominator becomes Z 1 Z 1 1 1 (23) dx 2 2 2 q → q − xk dx 2 2 , 0 [(q + xk) + k x(1 − x)] −−−−−−−→ 0 [q − ∆] where ∆ = −k2x(1 − x), and the non-vanishing terms of N µν after symmetric integration are 3 n 4(n − )qµν + [4k2 + (1 − 2x)k2]gµν + 2[( − 1)(1 − 2x)2 − (1 + x)(2 − x)]kµkν. 2 2 After combining the denominators and using the useful integral Z 1 dx(1 − 2x)F (x(1 − x)) = 0, 0 the result for µν is Π(b)(k) 2 AD n/2 Z 1 µν g CAδ π dx 3 2 µν Π(b)(k) = 4 2−n/2 {[(n − )x(1 − x)Γ(1 − n/2) + 2Γ(2 − n/2)]k g (2π) 0 (∆ − i) 2 n −[(1 − )(1 − 2x)2 + (1 + x)(2 − x)]Γ(2 − n/2)kµkν. (24) 2 The contribution from Fig. 4 (c) is g2f BAC f CDB Z (q + k)µqν Πµν (k) = − dnq (c) (2π)4i (q2 + i)((q + k)2 + i) 2 AD Z 1 Z µ ν µ ν g CAδ n q q − x(1 − x)k k = 4 dx d q 2 2 (2π) i 0 (q − ∆) g2C δADπn/2 = − A (2π)4 Z 1 dx 1 2 µν µ ν (25) 2−n/2 x(1 − x)[ Γ(1 − n/2)k g − Γ(2 − n/2)k k ]. 0 (∆ − i) 2 Adding Fig. 4 (b) and (c) we have
2 AD n/2 Z 1 µν g CAδ π dx 2 µν Π(b)+(c)(k) = 4 2−n/2 {[(n − 2)x(1 − x)Γ(1 − n/2) + 2Γ(2 − n/2)]k g (2π) 0 (∆ − i) n −[2 + (1 − )(1 − 2x)2]Γ(2 − n/2)kµkν} 2 g2C δADπn/2Γ(2 − n/2) = A (2π)4 Z 1 dx n 2 µν 2 µ ν (26) 2−n/2 {[2 − 2x(1 − x)]k g − [2 + (1 − )(1 − 2x) ]k k }. 0 (∆ − i) 2
9 Now, we got to show that, the k2gµν term and kµkν term are equal, then we can rewrite Eq. (26) as
µν 2 µν µ ν Π(b)+(c)(k) = (k g − k k )Π(k).
We will start from the second term of the coe cient of kµkν, and show it is indeed equal to the second term of the coe cient of k2gµν.
n Z 1 dx(1 − 2x)2 n Z 1 dx(1 − 4x + 4x2) (1 − ) 2−n/2 = (1 − ) 2−n/2 2 0 (∆ − i) 2 0 (∆ − i) Z 1 n 1 n/2−2 n/2−1 n/2 n/2−2 = (1 − ) 2 2−n/2 dx(x − 4x + 4x )(1 − x) 2 (−k ) 0 n 1 Γ(n/2 − 1)Γ(n/2 − 1) = (1 − ) { 2 (−k2)2−n/2 Γ(n − 2) Γ(n/2)Γ(n/2 − 1) Γ(1 + n/2)Γ(n/2 − 1) −4 + 4 } Γ(n − 1) Γ(n) 1 Γ(n/2)Γ(n/2) (n − 1)(n − 2) n = [ + 4(n − 1) − 4 ] (−k2)2−n/2 Γ(n) 1 − n/2 2 −2 Γ(n/2)Γ(n/2) = (−k2)2−n/2 Γ(n) Z 1 dx = −2 2−n/2 x(1 − x). 0 (∆ − i) The The contribution from Fig. 4 (d) is proportional to the integral
Z dnq , q2 which vanishes. Finally, we have
µν µν µν µν ΠGauge(k) = Π(b)(k) + Π(c)(k) + Π(d)(k) 2 µν µ ν = (k g − k k )ΠGauge(k),
where
2 AD n/2 Z 1 g CAδ π Γ(2 − n/2) dx ΠGauge(k) = 4 2−n/2 [2 − 2x(1 − x)] (2π) 0 (∆ − i) α C πn/2−2Γ(2 − n/2) 5 = s A 4π(−k2)2−n/2 3 5C α πn/2−2Γ(2 − n/2) = A s , (k2 → −µ2) (27) 3 4π(µ2)2−n/2
Combining Eq. (21) , (27) and considering nf light fermions with masses lighter than
10 scale µ, we have
2 2 Z3 = 1 + Π(k = −µ ) α πn/2−2Γ(2 − n/2) 5C 2n = 1 + s ( A − f ). (28) 4π(µ2)2−n/2 3 3
Let's recall the Z1 and Z2, which are α πn/2−2Γ(2 − n/2) Z = 1 − s (C + C ), 1 4π (µ2)2−n/2 F A α πn/2−2Γ(2 − n/2) Z = 1 − s C . 2 4π (µ2)2−n/2 F Therefore, the derivatives with respective to µ2 are ∂Z α α 1 = s Γ(3 − n/2)(C + C ) = s (C + C ), (29) ∂µ2 4πµ2 F A 4πµ2 F A ∂Z α α 2 = s Γ(3 − n/2)C = s C , (30) ∂µ2 4πµ2 F 4πµ2 F ∂Z α 5C 2n α 5C 2n 3 = − s Γ(3 − n/2)( A − f ) = − s ( A − f ). (31) ∂µ2 4πµ2 3 3 4πµ2 3 3 We can see that the pole for n → 4 are all eliminated. From the Eq. (9), the running of QCD coupling at one-loop is then
2 dg 2 dZ2 2 dZ1 1 dZ3 2 2 = 2 − 2 + 2 g dµ Z2 dµ Z1 dµ Z3 dµ α α α 5C 2n = 2 s C − 2 s (C + C ) − s ( A − f ) g2 4πµ2 F 4πµ2 F A 4πµ2 3 3 α 11 2 = − s ( C − n )g2 4πµ2 3 A 3 f dg2 α = − s (11C − 2n )g2, (32) dµ2 12πµ2 A f , take and g2 , CA = 3 αs = 4π dα α2 s = − s (33 − 2n ). (33) dµ2 12πµ2 f 0 2 Since from Eqs. (10), (20) and (28), the Z s = 1 + O(αs), i.e. ∂Z/∂µ ∼ O(αs), therefore,
in the second step of Eq. (32), we just keep the leading term which is at order O(αs) inside
the ( ). Solving Eq. (33), we have the relation of αs at two dierent scales, α (µ2) 2 s 1 (34) αs(µ2) = 2 2 . αs(µ1) µ2 1 + (33 − 2nf )ln( 2 ) 12π µ1 As we can see the asymptotic freedom that is, as energy scale becomes higher and higher the coupling αs becomes weaker and weaker, because (33 − 2nf ) is a positive value.
11 IV. CONCLUSION
In this notes, we have proceeded the renormalization of QCD. As a non-Abelian gauge theory, unlike QED, we needed to calculate more diagrams related to various renormalization
Z factors, and take care of the color factors. The relations, Z1 = Z2, in QED case is no longer satis ed in QCD because the gluon will interact themselves and the ghost elds will also contribute. After carefully dealing with the poles for dimension n → 4 in dimensional
regularization, we see that all the singularities are eliminated in nal results of αs. And we 2 can derive the β function and the running coupling αs(µ ) to see the most important feature of QCD, the asymptotic freedom.
V. APPENDIX
Feynman rules of QCD: Propagators:
A, µ B,ν AB µν pµpν 2 Gauge choice. −iδ [g − (1 − λ) p2+i ]/(p + i), λ :
p i (p6 −m+i)
p B A AB i δ p2+i
Vertex: b
A, µ
a µ A −igsγ (t )ba
12 C,λ
p3 p2 B, ν
p1
A, µ ABC λ µν µ νλ ν λµ −gsf (p1 − p2) g + (p2 − p3) g + (p3 − p1) g
A p B,µ
ABC µ C gsf p
A, α B, β
C,γ D, δ
2 XAC XBD αβ γδ αδ βγ 2 XAD XBC αβ γδ αγ βδ −igs f f (g g − g g ) − igs f f (g g − g g )
2 XAB XCD αγ βδ αδ βγ −igs f f (g g − g g )
13