86
CHAPTER 12. PROBLEM SET SOLUTIONS
12.3
Problem Set 3 Solutions
Feynman rules:
−i(gµ0 − α
pµ p0 δab
)
p2 p2
µ,a
(12.99)
ν,b
p
Figure 12.1: Feynman Rules (Equation 12.99).
Fermion:
ip/
p2
(12.100)
p
Figure 12.2: Fermion (Equation 12.100).
Ghost:
iδab
p2
a
(12.101)
b
p
Figure 12.3: Ghost (Equation 12.101).
Scalar:
i
δik
p2
(12.102)
gf abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
(12.103)
−ig 2 [f abc f cde (g µν g νσ − g µν g νρ ) +
f ace f bde (g µν g ρσ − g µσ g νρ ) +
f cde f bce (g µν g ρσ − g µρ g νσ )]
(12.104)
12.3. PROBLEM SET 3 SOLUTIONS
1.
87
igγ µ tra
(12.105)
gf abc pµ
(12.106)
ig(p + p� )µ (tra )ij
(12.107)
ig 2 (tra trb + trb tra )ij
(12.108)
Λ
i Ω3
4
log( )2
−→ −g 2 C2 (ΓF )δab (gµν p2 − pµ pν )
4
3
2 (2π )
µ
log part
(12.109)
where Λ is cut-off scale, µ is renormalization scale, and Ω3 is 3D surface area.
log div.part
−→
1
Λ
i Ω3
−g 2 C2 (Γs )δab (gµν p2 − pµ pν )
log( )2
4
4
2 (2π )
µ
log part
−→ 0
(12.110)
(12.111)
Similarly all tadpoles have no log div. part.
i Ω3
Λ
1 1
log( )2
−→ −g 2 C2 (G)δab ( gµν − pµ pν )
4
6 2
2 (2π )
µ
log div.
(12.112)
11 α i Ω3
Λ 2
1
19 α
log(
) (12.113)
−→ −g 2 C2 (G)δab (p2 gµν ( + ) − pµ pν ( + ))
6
12 2
6
2 2 (2π )4
µ
log div.
log div.
−→ 0
(12.114)
i Ω3
Λ 2 2 2
log(
) g (p gµν − pµ pν )
2 (2π)4
µ
5 α
4
1
{( + )C2 (G) − C2 (ΓF ) − C2 (Γs )}
3 2
3
2
⇒ −δ3 |µ2
(12.115)
Add up all contributions to get:
(12.116)
88
CHAPTER 12. PROBLEM SET SOLUTIONS
i
j
p
Figure 12.4: Scalar (Equation 12.102).
µ,a
k
p
q
ν,b
ρ, c
Figure 12.5: Equation 12.103.
ν,b
µ,a
ρ,c
σ,d
Figure 12.6: Equation 12.104.
µ,a
Figure 12.7: Equation 12.105.
µ,a
p
b
c
Figure 12.8: Equation 12.106.
µ,a
p’,j
p,i
Figure 12.9: Equation 12.107.
12.3. PROBLEM SET 3 SOLUTIONS
89
ν, b
µ,a
j
i
Figure 12.10: Equation 12.108.
µ,a
ν, b
p
Figure 12.11: Equation 12.109.
µ,a
ν, b
p
Figure 12.12: Equation 12.110.
µ,a
ν, b
p
Figure 12.13: Equation 12.111.
µ,a
ν, b
p
Figure 12.14: Equation 12.112.
µ,a
ν, b
p
Figure 12.15: Equation 12.113.
µ,a
ν, b
p
Figure 12.16: Equation 12.114.
90
CHAPTER 12. PROBLEM SET SOLUTIONS
• Note that inclusion of ghost contribution was crucial to get right tenser –
structure (p2 gµν − pµ pν ).
• We only evaluate log div. parts, since linear and quadratic divergence of
the term aΛ + bλ2 will not yield and contribution to β–function as β gets
∂
∂
δ1 and ∂µ
δ3 .
contributions from ∂µ
2. (a)
log div.part
−→
−ig 3 f abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
1
i Ω3
Λ 2
(13 − 9α)C2 (G)
log(
)
(12.117)
8
2 (2π )4
µ
µ,a
k
p
q
ν,b
ρ,c
+ permutations
Figure 12.17: Equation 12.117.
log part
−→
ig 3 f abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
i Ω3
Λ
1
3
(9 − α)C2 (G)
log( )2
(12.118)
4
4
2
2 (2π )
µ
µ,a
k
p
q
ν,b
ρ, c
+ permutations
Figure 12.18: Equation 12.118.
log div.part
−→
ig 3 f abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
i Ω3
Λ
1
C2 (G)
log( )2
(12.119)
4
24
2 (2π )
µ
12.3. PROBLEM SET 3 SOLUTIONS
log div.part
−ig 3 f abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
4
i Ω3
Λ
C2 (ΓF )
log( )2
(12.120)
4
3
2 (2π )
µ
−→
log part
−→
91
−ig 3 f abc [(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
i Ω3
Λ
1
C2 (Γs )
log( )2
(12.121)
4
3
2 (2π )
µ
log part
−→ 0
(12.122)
Adding all up we get
2 3
4
1
−ig 3 f abc {C2 (G)( + α) − C2 (ΓF ) − C2 (Γs )}
3 4
3
3
[(k − p)ρ g µ0 + (p − q)µ g νρ + (q − k)ν g ρµ ]
i Ω3
Λ
δ1
log( )2 ⇒ 2
(12.123)
4
2 (2π )
µ
µ
Therefore,
β=µ
∂
3
(−δ1 + δ3 )
∂µ
2
(12.124)
Therefore,
β3g =
ρ3
11
4
1
[C2 (G)(− ) + C2 (ΓF ) + C2 (Γs ) ]
2
(4π)
3
3
3
(12.125)
Note that α–dependence has cancelled out.
(b) Courtesy of Guide Festucchia
4 Gluons vertex: We have to evaluate the following diagrams
Using mathematica we obtain:
d4 K 1 K µ K ν K ρ K σ
(
(34 − 12α + 2α2 ) +
(2π)4 K 4
K4
K µ K ν K ρ K σ + K µ K σ g ρν + K ρ K σ g νµ + K ν K ρ g µσ
(3 + 2α − α2 ) +
2
K
4 f al ebg gch hdf
g f
f
f
f
�
92
CHAPTER 12. PROBLEM SET SOLUTIONS
µ,a
k
p
q
+ permutations
ν,b
ρ, c
Figure 12.19: Equation 12.119.
µ,a
k
p
q
+ permutations
ν,b
ρ, c
Figure 12.20: Equation 12.120.
µ,a
k
p
ν,b
q
+ permutations
ρ, c
Figure 12.21: Equation 12.121.
+ permutations
Figure 12.22: Equation 12.122.
12.3. PROBLEM SET 3 SOLUTIONS
93
(g µν g ρσ + g µσ g νρ )(1 + 2α + α2 ))
b c d 1
= g 4 tr(taG tG
tG tG )
24
µν ρσ
µσ νρ
((g g + g g )(94 + 60α + 14α2 ) + g µρ g σν (34 − 12α + 2α2 ))
�
d4 K 1
(12.126)
(2π)4 K 4
b, ν
a, µ
c, ρ
d, σ
Figure 12.23: 4 Gluons Vertex (Equation 12.126).
The different permutations of the gluon vertices will be accounted for later.
d4 K 1 K µ K ν K ρ K σ
(
(−3α + α2 ) +
4
4
4
(2π) K
K
µ ν ρσ
2
ν ρ µσ
K K g (−1 − 2α − α ) + (K K g + K µ K σ g νρ )(3 + 3α) +
K ρ K σ g µν (6 − 3α − α2 ) + g µν g ρσ (2 + 3α + α2 ) +
g µρ g νσ (−2 − 2α) + g µσ g νρ (1 + α))(c.ρ ↔ d, σ)
c d 1
= −g 4 tr(taG tbG tG
tG ) ((g µν g ρσ + g µσ g νρ ))
(12.127)
24
1
= − g 4 (1 + 3α)[(f ade f bce + f ace f bde )gµν gρσ +
3
(f ade f cbe + f abe f cde )gµρ gνσ + (f abe f dce + f ace f dbe )gµσ gνρ ]
�
d4 K 1
(12.128)
(2π)4 K 4
1
= − g 4 (1 + 3α)[f abe f cde (gµρ gνσ − gµσ gνρ ) +
3
ace bde
f f (gµν gρσ − gµσ gνρ ) + f ade f bce (gµν gρσ − gµσ gνρ )]
�
d4 K 1
(12.129)
(2π)4 K 4
(12.130)
−g
4
b c d
tr(taG tG
tG tG )
�
This computation has been done with α → −α
94
CHAPTER 12. PROBLEM SET SOLUTIONS
1
− g 4 (1 − 3α)[f abe f cde (gµρ gνσ − gµσ gνρ ) +
3
f ace f bde (gµν gρσ − gµσ gνρ ) + f ade f bce (gµν gρσ − gµσ gνρ )]
�
d4 K 1
(12.131)
(2π)4 K 4
(12.132)
d4 K Kα Kβ Kγ Kλ
tr[γ α γ µ γ β γ ν γ δ γ ρ γ λ γ σ ] +
(2π)4
K4
(color trace reversed)
1
= −g 4 tr(ta tb tc td ) (tr[γ α γ µ γα γ ν γ β γ ρ γβ γ σ ] + tr[γ α γ µ γ β γ ν γα γ ρ γβ γ σ ] +
24
�
d4 K 1
tr[γ α γ µ γ β γ ν γβ γ ρ γα γ σ ])
(12.133)
(2π)4 K 4
8
= − g 4 tr(ta tb tc td )(g µγ g ρσ − 2g σγ g µρ + g µσ g ργ ) +
3
(µ, a ↔ γ, b) + (µ, a ↔ σ, d) + (color trace reversed)
(12.134)
4
a b c d
−g tr(t t t t )
�
This has exactly the same structure as before so the result is
4
− g 4 [f abe f cde (gµρ gνσ − gµσ gνρ ) +
3
ace bde
f f (gµν gρσ − gµσ gνρ ) + f ade f bce (gµν gρσ − gµσ gνρ )]
�
d4 K
C2 (ΓF )
(12.135)
(2π)4
(12.136)
= 32g 4 tr(ta tb tc td )
=
�
d4 K Kµ Kν Kρ Kσ
(2π)4
K8
4 4
g tr(ta tb tc td )(g µν g ρσ + g µρ g νσ + g µσ g νρ )
3
�
d4 K 1
+ permutations
(2π)4 K 4
(12.137)
12.3. PROBLEM SET 3 SOLUTIONS
95
b, ν
a, µ
c, ρ
d, σ
Figure 12.24: Equation 12.127.
b, ν
a, µ
K
c, ρ
d, σ
Figure 12.25: Equation 12.133.
b, ν
a, µ
K
c, ρ
d,σ
Figure 12.26: Equation 12.137.
96
CHAPTER 12. PROBLEM SET SOLUTIONS
d4 K Kρ Kσ
(2π)4 K 6
�
d4 K 1
4
a b c d µν ρσ
= −2g tr({t t }t t )g g
+ permutations(12.138)
(2π)4 K 4
= −8g 4 tr({ta tb }tc td )g µν
b, ν
�
a, µ
c, ρ
d, σ
Figure 12.27: Equation 12.138.
= g 4 tr({ta tb }{tc td })g µν g ρσ
b, ν
c, ρ
�
d4 K 1
+ permutations
(2π)4 K 4
(12.139)
a, µ
d, σ
Figure 12.28: Equation 12.139.
Summing the last three graphs together and permuting the vertices we
obtain:
1 4 µν ρσ
g [g g tr(−2ta tb tc td + 4td tb tc ta − 2tb ta tc td ) +
3
g µρ g νσ tr(−2ta tc tb td + 4td tc tb ta − 2tc ta tb td ) +
�
d4 K 1
g µσ g νρ tr(−2ta tb td tc + 4tc tb td ta − 2tb ta td tc )]
(12.140)
(2π)4 K 4
Now using
12.3. PROBLEM SET 3 SOLUTIONS
tr(−2ta tb tc td + 4td tb tc ta − 2tb ta tc td )
= tr([a, d][b, c] + [a, c][b, d])
= −C2 (G)[f ade f bce + f ace f bde ]
97
(12.141)
(12.142)
and similar relocations we obtain:
1
− g 4 [f abe f cde (gµρ gνσ − gµσ gνρ ) + f ace f bde (gµν gρσ − gµσ gνρ )
3
�
d4 K 1
ade bce
C2 (ΓB )
(12.143)
+f f (gµν gρσ − gµσ gνρ )]
(2π)4 K 4
All the calculations sum to:
g 4 [f abe f cde (gµρ gνσ − gµσ gνρ ) + f ace f bde (gµν gρσ − gµσ gνρ )
�
d4 K 1
ade bce
+f f (gµν gρσ − gµσ gνρ )]
(2π)4 K 4
1
4
1
{− (1 − 3α)C2 (G) − C2 (ΓF ) − (ΓB )}
(12.144)
3
3
3
1 2g 3
β = ( )∗
2 (4π)2
1
4 5 α
4
{[− (1 − 3α) − ( + )]C2 (G) + C2 (ΓF )(−1 + 2) +
3
2 3 2
3
1
C2 (ΓB )(−1 + 2)}
(12.145)
3
11
4
1
2g 3
= (− C2 (G) + C2 (ΓF ) + C2 (ΓB ))
(12.146)
3
3
3
(4π)2
* 21 comes from the fact that the vertex appears at color g 2 2 from the log
div. of ln( π12 ) with respect to π.
3. (a)
i Ω3
1
Λ
log
log( )2
−→ g 3 γ µ ta (1 − α)[(2(ΓF ) − C2 (G))]
4
2
2 (2π )
µ
(12.147)
i Ω3
3 3
1
Λ
log
log( )2
−→ g 3 C2 (G)γ µ ta ( − α)[(2(ΓF ) − C2 (G))]
4
2 4
2
2 (2π )
µ
(12.148)
98
CHAPTER 12. PROBLEM SET SOLUTIONS
µ,a
p’
p
Figure 12.29: Equation 12.147.
µ,a
p
p’
Figure 12.30: Equation 12.148.
12.3. PROBLEM SET 3 SOLUTIONS
99
From these read off δ1f
Λ
i Ω3
1
log
log( )2
−→ C2 (ΓF )g 3 p(1
/ − α)[(2(ΓF ) − C2 (G))]
4
2
2 (2π )
µ
p
(12.149)
p
Figure 12.31: Equation 12.149.
Read off δ2f
βF = µ
∂
δ3
(−δ1f + δ2f + )
∂µ
2
(12.150)
Same as β3g .
(b)
log
−→ −g 2 C2 (ΓF )(2 + α)p2
i Ω3
Λ
log( )2
4
2 (2π)
µ
(12.151)
p
Figure 12.32: Equation 12.151.
log
−→ 0
(12.152)
Figure 12.33: Equation 12.152.
Read off δ2s
i Ω3
Λ
1
log
log( )2
−→ g 3 ta (C2 (Γs ) − C2 (G))(p + p� )µ (1 − α)
4
2 (2π)
µ
2
(12.153)
i Ω3
3
Λ
log
−→ g 3 ta C2 (G) (1 − α)(p + p� )µ
log( )2
4
2 (2π)
4
µ
(12.154)
100
CHAPTER 12. PROBLEM SET SOLUTIONS
µ,a
p
p’
Figure 12.34: Equation 12.153.
µ,a
p
p’
Figure 12.35: Equation 12.154.
µ,a
p
p’
+ permutations
Figure 12.36: Equation 12.155.
12.3. PROBLEM SET 3 SOLUTIONS
101
3
1
i Ω3
Λ
log
log( )2
−→ −g 3 ta (p + p� )µ (2C2 (Γs ) − C2 (G))
4
2
2
2 (2π)
µ
(12.155)
From last 4 graphs read off δ1s
log
−→ 0
(12.156)
Figure 12.37: Equation 12.156.
Therefore,
βs = µ
δ3
∂
(−δ1s + δ2s + )
∂µ
2
(12.157)
Same as β3g .
(c)
1
Λ
i Ω3
log
−→ ig 2 C2 (G)(1 − α)(p + p� )µ f abc
log( )2
4
8
µ
2 (2π)
(12.158)
µ,a
p
b
p’
+ perturbations
c
Figure 12.38: Equation 12.158.
i Ω3
3
Λ 2
log(
)
ig 3 C2 (G)(1 − α)(p + p� )µ f abc
8
2 (2π )4
µ
(12.159)
102
CHAPTER 12. PROBLEM SET SOLUTIONS
Read off δ1g from there.
i Ω3
1 α
Λ
−g 2 ( + )C2 (G)δbc p2
log( )2
4
2 4
2 (2π )
µ
(12.160)
Read off δ2g from this.
Therefore,
βg = µ
δ3
∂
(−δ1g + δ2g + )
∂µ
2
(12.161)
Same as β3g .
Looking back we see that
1
δ1s − δ2s = δ1f − δ2f = δ1g − δ2g == δ13g − δ3 = (δ14g − δ3 )
2
(12.162)
These relations show the universality at the β–function coefficients which
is a direct consequence of gauge invariance. One could have proven these
relations directly from the renormalized Lagrangian.
4. In light of problem 3, the easiest way to compute the β–function is
(a) Set α = 0 (since β is independent of α).
(b) Compute δ1g − δ2g (the ghost contribution).
(c) Compute δ3 .
Therefore,
β=
∂
δ3
(−δ1g + δ2g + )
∂µ
2
(12.163)
12.3. PROBLEM SET 3 SOLUTIONS
103
µ,a
p
p’
+ permutations
Figure 12.39: Equation 12.159.
b
c
p
Figure 12.40: Equation 12.160.