# Solution 9

```Physics 523, Quantum Field Theory II
Homework 8
Due Wednesday, 10th March 2004
Jacob Lewis Bourjaily
Renormalization of Pseudo-Scalar Yukawa Theory
Let us consider the theory generated by the Lagrangian
1
1
L = (∂&micro; φo )2 − m2φo φ2o + ψ o (i 6∂ − meo )ψo − igo ψ o γ 5 ψo φo .
2
2
Superficially, this theory will diverge very similarly to quantum electrodynamics because the fields
and the coupling constant have the same dimensions as in quantum electrodynamics. Therefore, we see
that the superficial divergence is given by D = 4L − 2Pφ − Pe where L represents the number of loops
and Pφ and Pe represent the number of pseudo-scalar and fermion propagator particles, respectively.
Furthermore, we see that this can be reduced to
3
D = 4 − Nφ − Ne ,
2
(a.1)
where Nφ and Ne represent the number of external pseudo-scalar and fermion lines, respectively.
We see that this implies that the following diagrams are superficially divergent:
a)
D=4
b)
d)
D=1
e)
g)
D=3
c)
D=0
f)
D=0
D=2
D=1
Although vacuum energy is an extraordinarily interesting problem of physics, we will largely ignore
diagram (a) which is quite divergent. We note that because the Lagrangian is invariant under parity
transformations φ(t, x) → −φ(t, −x) any diagram with an odd number of external φ’s will give zero. In
particular, the divergent diagrams (b) and (d) will be zero.
The first divergent diagram we will consider, (c), is clearly ∼ a0 Λ2 + a1 p2 log Λ where we note that the
term proportional to p in the expansion vanishes by parity symmetry. Similarly, we naı̈vely suspect that
the divergence of diagram (f) would be ∼ a0 Λ+ 6 p log Λ but the term linear in Λ is reduced to me log Λ
by the symmetry of the Lagrangian of chirality inversion of ψ together with φ → −φ. The diagrams (e)
and (g) are both ∼ log Λ. All together, there are six divergent constants in this theory.
We note that because the diagram (e) diverges, we must introduce a counterterm δλ which implies
λ 4
that our original Lagrangian should have included a term 4!
φ .
1/2
1/2
We define renormalized fields, φo ≡ Zφ φ and ψo ≡ Z2 ψ, where Zφ and Z2 are as would be
defined canonically. Using these our Lagrangian can be written as
1
λ
1
1/2
Zφ (∂&micro; φ)2 − Zφ m2φo φ2 − Z2 ψ(i 6∂ − meo )ψ − −igo Z2 Zφ ψγ 5 ψφ − Zφ2 φ4 .
2
2
4!
Let us define the counterterms,
go
1/2
δmφ ≡ Zφ m2φo −m2φ , δme ≡ Z2 meo −me , δφ ≡ Zφ −1, δλ ≡ λo Zφ2 −λ, δ1 ≡ Z2 Zφ −1,
g
L =
δ2 ≡ Z2 −1.
Therefore, we may write our renormalized Lagrangian
1
λ
1
(∂&micro; φ)2 − m2φ φ2 + ψ(i 6∂ − me )ψ − igψγ 5 ψφ − φ4
2
2
4!
1
δλ
1
+ δφ (∂&micro; φ)2 − δmφ φ2 + ψ(iδ2 6∂ − δme )ψ − igδ1 ψγ 5 ψφ − φ4 .
2
2
4!
L =
1
(a.4)
2
JACOB LEWIS BOURJAILY
Let us compute the pseudo-scalar self-energy diagrams to the one-loop order, keeping only the divergent pieces. This corresponds to:
k
−iM 2 (p2 ) =
p
k
p
+
p
p
+
p
&times;
p
k+p
Using the ‘canonical procedure’ and dropping all but divergent pieces (linear in &sup2;−1 ) we see that
&cedil;
&middot; 5
Z
Z d
d k
λ dd k
i
γ i(6k + 6p + me )iγ 5 (6k + me )
2
−iM 2 (p2 ) = −i
−
g
+ i(p2 δφ − δme ),
Tr
2 (2π)d k 2 − m2φ
(2π)d
((k + p)2 − m2e )(k 2 − m2e )
&cent;
&iexcl;
Z 1 Z d
Γ 1 − d2
λ
1
d k `2 − x(1 − x)p2 − m2e
2
= −i
+ i(p2 δφ − δm2 ),
−
4g
dx
2 (4π)d/2 (m2φ )1−d/2
(2π)d
(`2 − ∆)2
0
#
&iexcl;
&cent;
&iexcl;
&cent;
&iexcl;
&cent;
Z 1 &quot;
m2φ
&cent;
Γ 2 − d2
Γ 2 − d2 &iexcl;
d Γ 1 − d2
λ
1
i
i
2
2
2
= −i
− 4g
dx −
+
x(1 − x)p + me
2 (4π)d/2 (1 − d/2) (m2 )2−d/2
(4π)d/2 2 ∆1−d/2
(4π)d/2 ∆2−d/2
0
+ i(p2 δφ − δm2 ),
Z
Z
&cent;
&cent;
λm2φ 2
i 2 1 &iexcl; 2
i 2 1 &iexcl; 2
2
2
2
− 8g
dx me − x(1 − x)p + 4g
dx me + x(1 − x)p2 + i(p2 δφ − δm2 ),
∼i
32π 2 &sup2;
(4π)2 &sup2; 0
(4π)2 &sup2; 0
&micro;
&para;
λm2φ 1
g2 2
2 2 1 2
2
2
=i
+i 2
−2me + p + p + me + i(p2 δφ − δm2 ),
16π 2 &sup2;
4π &sup2;
6
6
&Atilde;
!
2
2
2
2
2
λmφ
g p
g me 1
=i
+
−
+ i(p2 δφ − δm2 ).
16π 2
4π 2
2π 2
&sup2;
Therefore, applying our renormalization conditions, we see that1
&Atilde;
!
&micro; 2 &para;
λm2φ
g 2 m2e 1
g
1
∴ δmφ =
−
,
δ
=
−
.
φ
2
2
2
16π
2π
&sup2;
4π
&sup2;
(b.1)
Similarly, let us compute the fermion self-energy diagrams to one-loop order, keeping only divergent
parts. This corresponds to:
p−k
−iΣ22 (6p) =
p
k
p
+
p
&times;
p
Again, using the ‘canonical procedure’ and dropping all but divergent pieces (linear in &sup2;−1 ) we see that
#
Z d &quot;
d k
i
i(6k + me ) 5
2
5
−iΣ(6p) = g
γ
γ + i(6pδ2 − δme ),
(2π)d
((p − k)2 − m2φ ) (k 2 − m2e )
Z d
d k
6k − me
2
= −g
+ i(6pδ2 − δme ),
d
2
2
(2π) (k − me )((p − k)2 − m2φ )
Z 1 Z d
d ` 6pz − me
2
dz
= −g
+ i(6pδ2 − δm2 ),
d (`2 − ∆)2
(2π)
0
Z
g2 2 1
∼ −i
dz (6pz − me ) + i(6pδ2 − δme ),
(4π)2 &sup2; 0
&micro; 2
&para;
g 6p
g 2 me 1
=i
−
+ i 6pδ2 − iδme .
16π 2
8π 2
&sup2;
Therefore, applying our renormalization conditions, we see that
&micro; 2 &para;
&micro; 2 &para;
g me 1
g
1
∴ δme = −
(b.2)
,
δ2 = −
.
8π 2
&sup2;
16π 2 &sup2;
1For renormalization conditions and Feynman rules please see the Appendix.
PHYSICS 523: QUANTUM FIELD THEORY II HOMEWORK 8
3
Let us now compute the δ1 counterterm by computing δΓ5 (q = 0) given by:
p0
k+q
5
δΓ (q = 0) = p − k
k
←
−
q ∼0
+
&times;
p
Again, using the ‘canonical procedure’ and dropping all but divergent pieces (linear in &sup2;−1 ) we see that
Z d
d k
γ 5 (6k + me )γ 5 (6k + me )γ 5
+ δ1 γ 5 ,
δΓ5 (q = 0) = −ig 2
(2π)d ((p − k)2 − m2φ )(k 2 − m2e )(k 2 − m2 )
Z d
d k
(6k + me )(6k − me )
2 5
+ δ1 γ 5 ,
= ig γ
(2π)d ((p − k)2 − m2φ )(k 2 − m2e )(k 2 − m2 )
Z 1 Z d
d ` `2 + (z 2 − 1)m2e
= ig 2 γ 5 dz
+ δ1 γ 5 ,
d
2 − ∆)3
(2π)
(`
0
&cedil;
&middot;
Z 1
i d2
2 5
= ig γ
+ δ1 γ 5 ,
dz(1 − z)
2 2 &sup2;
(4π)
0
2
g
1
= −γ 5 2 + δ1 γ 5 .
8π &sup2;
Therefore, applying our renormalization conditions, we see that
&micro; 2 &para;
g
1
∴ δ1 =
(b.3)
.
8π 2 &sup2;
Let us now compute the δλ counterterm by computing the one-loop correction to the standard φ4
vertex. The five contributing diagrams are:
iM =
+
+
+
+
&times;
We may save a bit of sweat by noting that the sum of the first four diagrams is identical to the analogous
diagrams in φ4 -theory. The sum was computed fully both in class and in the text and give a divergent
3λ2 1
contribution of 16π
2 &sup2; to δλ . Therefore, we are only burdened with the calculation of the remaining two.
We see that, (note the combinatorial factor of 6)
&middot;
&cedil;
Z d
3λ2 1
d k
γ 5 (6k + me )γ 5 (6k − 6p1 + me )γ 5 (6k − 6p1 − 6p2 + me )γ 5 (6k − 6p1 − 6p2 + 6p3 + me )
4
iM = i
−
6g
Tr
− iδλ ,
16π 2 &sup2;
(2π)d
(k 2 − m2e )((k − p1 )2 − m2e )((k − p1 − p2 )2 − m2e )((k − p1 − p2 + p3 )2 − m2e )
&pound;
&curren;
Z d
3λ2 1
d k Tr γ 5 6k γ 5 6k γ 5 6k γ 5 6k
4
∼ i
− 6g
− iδλ ,
k→∞
4
16π 2 &sup2;
(2π)d
(k 2 − m2e )
Z d
d k
4k 4
3λ2 1
4
−
6g
− iδλ ,
=i
2
d
2
16π &sup2;
(2π) (k − m2e )4
&iexcl;
&cent;
3λ2 1
i
d(d + 2) Γ 2 − d2
2
=i
− 24g
− iδλ ,
16π 2 &sup2;
4
(4π)d/2
6∆2−d/2
3g 4 1
3λ2 1
−
i
− iδλ .
=i
16π 2 &sup2;
π2 &sup2;
Therefore, applying our renormalization conditions, we see that
&micro; 2
&para;
3λ
3g 4 1
∴ δλ =
− 2
.
(b.4)
16π 2
π
&sup2;
4
JACOB LEWIS BOURJAILY
Appendix
Feynman Rules and Renormalization Conditions
Given the Lagrangian for pseudo-scalar Yukawa theory,
1
1
λ
L = (∂&micro; φ)2 − m2φ φ2 + ψ(i 6∂ − me )ψ − igψγ 5 ψφ − φ4
2
2
4!
1
1
δλ
+ δφ (∂&micro; φ)2 − δmφ φ2 + ψ(iδ2 6∂ − δme )ψ − igδ1 ψγ 5 ψφ − φ4 ,
2
2
4!
we can derive the renormalized Feynman rules.
=
i
p2 −mφ2 +i&sup2;
= −iλ
&times;
= i(p2 δφ − δmφ )
&times;
= −iδλ
p
&times;
p
&times;
= 6 p−mie +i&sup2;
= gγ 5
= i(6pδ2 − δme )
= gδ1 γ 5
To derive the counter terms explicitly, it is necessary to offer a convention of renormalization conditions. Above, we have used the conditions:
=
i
p2 −mφ2 +i&sup2;
with pole = 1.
= −iλ at s = 4m2 , t = u = 0.
Σ(6p = m) = 0.
&macr;
dΣ(6p) &macr;&macr;
= 0.
d 6p &macr;6 p=m
gΓ5 (q = 0) = gγ 5 .
ˆ
∆ ≡ m2
n=1
ˆ
#
\$
1
i (−1)n Γ n − d2
!
&quot;n =
,
d
d/2
n−d/2
Γ (n) ∆
(2π) ⃗ℓ2 − ∆
(4π)
dd ⃗ℓ
1
%
&amp;
d
1
=−
Γ 1−
,
d ⃗2
d/2
1−d/2
2
∆
(2π) ℓ − ∆
(4π)
dd ⃗
ℓ
i
1
iM = −
d≡4−ε
&amp;
%
1
d
.
Γ
1
−
2−d
d/2
2 m
2 (4π)
i δλ
ε→0
!ε
&quot; 1
Γ
−
1
2−ε/2
2
mε−2
2 (4π)
&quot; % 4π &amp;ε/2
i δ λ m2 ! ε
Γ
.
−
1
=−
2
2
m2
2 (4π)
i δλ
iM = −
Γ
!ε
&quot;
2
− 1 = − + (γ − 1) + O (ε) ,
2
ε
ε
Aε/2 ≡ e 2 log A = 1 −
# \$
ε
log A + O ε2 ,
2
%
&amp;
&amp;%
# 2\$
2
ε
4π
iM = −
− + (γ − 1) + O (ε)
1 − log 2 + O ε
2
ε
2
m
2 (4π)
%
&amp;
2
2
i δλ m
− − log m2 + (γ − 1 + log 4π) ,
=−
2
ε
2 (4π)
i δ λ m2
m→0
ε
m2 log m2 → 0
\$
#
i p⃗2 δZ − δm
iM = i⃗
p2 &middot;
i p⃗2 δZ
λ2
1
+
12 (4π) ε
,
4
δZ ≡ −
λ2
1
,
12 (4π) ε
4
!
φ4
S
φ4
s→∞
i M (s, t) ∼ − i λ − i
t
λ2
2
(4π)
log s − i
3λ3
4
2 (4π)
log2 s + &middot; &middot; &middot; .
p⃗1 , p⃗2
⃗p3 , p
⃗4
m
2
2
2
2
2
2
s ≡ (⃗
p1 + ⃗p2 ) = (⃗
p3 + ⃗p4 ) ,
t ≡ (⃗p1 − p⃗3 ) = (⃗
p2 − p⃗4 ) ,
u ≡ (⃗p1 − p⃗4 ) = (⃗p2 − p⃗3 ) ,
s + t + u = 4m2
iM
S
•
#
\$−1
i p⃗2 − m2
⃗p
•
−iλ
•
− i δλ
•
⃗k
&acute;
•
d
dd⃗k/ (2π)
−iλ
iM
⃗p1 , p
⃗2
(− i λ)
2
p⃗ ≡ ⃗
p1 + p⃗2
2
dd⃗k
ˆ
i
i
d
(2π) ⃗k 2 − m2 (⃗k + ⃗p)2 − m2
p⃗2 = s
# \$
2
≡ (− i λ) &middot; i V p⃗2 ,
2
s
− i δλ
u
i M = − i λ − i λ2 (V (s) + V (t) + V (u)) − i δλ ,
# 2\$
V p
⃗
# \$
i
V p⃗2 ≡
2
# 2\$
V ⃗
p =−
1
2 (4π)
2
%
ˆ
dd⃗k
1
1
.
2
2
⃗
⃗
(2π) k − m (k + p⃗)2 − m2
d
2
− γ + log (4π) −
ε
ˆ
0
1
&amp;
# 2
\$
2
,
dx log m − x (1 − x) ⃗p
t
ε ≡ 4−d → 0
m→0
# 2\$
V p
⃗ =−
=−
1
2
2 (4π)
1
2
%
%
2
− γ + log (4π) −
ε
ˆ
1
0
&amp;
#
\$
dx log −x (1 − x) p⃗2
2
− γ + log (4π) − log p⃗2 − i π −
ε
ˆ
1
2 (4π)
0
&amp;
%
1
2
− γ + log (4π) − log p⃗2 − i π + 2
=−
2
ε
2 (4π)
%
&amp;
1
2
2
≡−
− log p⃗ + C ,
2
ε
2 (4π)
&amp;
dx log (x (1 − x))
C ≡ −γ + log (4π) − i π + 2.
m→0
i M = − i λ − i λ2 (V (s) + V (t) + V (u)) − i δλ
% %
&amp;
&amp;
2
i λ2
3
+
C
−
log
s
−
log
t
−
log
u
− i δλ .
= −iλ +
ε
2 (4π)2
δλ ≡
iM = −iλ −
log s
s→∞
s→∞
3λ2
2 (4π)
2
i λ2
2
2 (4π)
%
&amp;
2
+C ,
ε
(log s + log t + log u) .
t
s≫m
log t
m
s = 2⃗
p1 &middot; p⃗2 = 2⃗
p3 &middot; p⃗4 ,
p⃗1 + ⃗
p2 = p⃗3 + ⃗p4
u = −2⃗
p1 &middot; ⃗p4 = −2⃗
p2 &middot; p⃗3 ,
=⇒
⃗p2 − p⃗4 = p⃗3 − p⃗1 ,
s + u = (⃗
p1 &middot; p⃗2 + ⃗p3 &middot; ⃗p4 ) − (⃗p1 &middot; ⃗p4 + p⃗2 &middot; p⃗3 )
= (⃗
p1 − ⃗p3 ) &middot; (⃗
p2 − ⃗p4 )
= − (⃗
p1 − p⃗3 )2
= −t → 0,
u ≈ −s
log u ≈ log (−s) = i π + log s ≈ log s,
iM = −iλ − i
λ2
2
(4π)
iM
log s,
s
t
u
s t
u
s
3
(− i λ)
i M1 ≡
4
# 2\$
V p
⃗
ˆ
dd⃗k1
i
i
d
(2π) ⃗k12 − m2 (⃗k1 + p⃗)2 − m2
ˆ
dd⃗k2
# \$\$2
i
i
3 #
,
≡ (− i λ) &middot; i V p⃗2
2
2
2
2
⃗
⃗
(2π) k2 − m (k2 + p⃗) − m
d
2! &middot; 2! = 4
m→0
%
%
&amp;
&amp;
# 2\$
2
2
1
1
2
2
V p⃗ = −
− log p⃗ + C → −
− log p⃗ ,
2
2
ε
ε
2 (4π)
2 (4π)
ε→0
C
3
i M1 = − i λ
'
−
1
2
2 (4π)
%
%
2
− log p⃗2
ε
&amp;(2
&amp;
1
1
1
2 2
2
=−
− log p⃗ + log p⃗
4
ε2
ε
4
(4π)
%
&amp;
1
1
1
i λ3
2
−
log
s
+
log
s
.
=−
4
ε2
ε
4
(4π)
i λ3
s
t
−⃗k
V ((⃗k + p⃗3 )2 )
⃗p3 , p
⃗4
⃗p1 , p
⃗2
⃗k + ⃗p
2
⟨φ1 φ2 φ3 φ4 φ4a φ4b φ4c ⟩
p⃗1,2,3,4
a
φ1
φa 4
φ1,2,3,4
c
φa 3
φ3
φb 4
φb 3
φ4
φa
φc
φc 4
φc 3
b
φ2
φa 2
φb
2
3!
3!
3
(4!)
3
S=
(− i λ)3
2
dd⃗k1
dd⃗k2
i
i
d ⃗2
2
2
2
2
2
⃗
⃗
⃗
⃗
(2π) k1 − m (k1 + p⃗) − m
(2π) k2 − m (k2 + k1 + p⃗3 )2 − m2
ˆ
i
i
dd⃗k
&middot; i V ((⃗k + ⃗p3 )2 ).
≡ (− i λ)3
d ⃗2
2
⃗
(2π) k − m (k + p⃗)2 − m2
i M2 ≡
ˆ
d
i
(3!) (4!)
= 2.
4 &middot; 3 &middot; 4 &middot; 4 &middot; 2 &middot; 3 &middot; 3 &middot; 2 &middot; 3!
i
ˆ
m→0
i λ3
2
i M2 =
dd⃗k1 1
1
d
(2π) ⃗k 2 (⃗k1 + p⃗)2
ˆ
1
dd⃗k2 1
=
d ⃗2 ⃗
(2π) k (k2 + ⃗q )2
ˆ
ˆ
1
dd⃗k2 1
1
.
d
(2π) ⃗k 2 (⃗k2 + ⃗k1 + p⃗3 )2
2
⃗q ≡ ⃗k1 +⃗
p3
ˆ
2
1
dx
0
ˆ
dd ⃗ℓ2 1
,
d
(2π) D2
D ≡ (1 − x) ⃗k22 + x(⃗k2 + ⃗q )2
= ⃗k 2 + 2x⃗k2 &middot; ⃗q + x⃗q 2
2
≡ ⃗ℓ22 − ∆,
⃗ℓ2 ≡ ⃗k2 + x⃗q
ˆ
ˆ
ˆ
0
1
dx
dd ⃗ℓ2
⃗ℓ2 = ⃗k 2 + 2x⃗k2 &middot; ⃗q + x2 ⃗q 2 ,
2
2
∆ ≡ −x (1 − x) ⃗q 2 .
1
d
(2π)
=⇒
!
&quot;2 =
⃗
ℓ22 − ∆
#
\$
i Γ 2 − d2
d/2
(4π)
1
,
∆2−d/2
#
\$ˆ 1
2−d/2
(−1)
i Γ 2 − d2
1
dd⃗k2 1
1
=
dx
d ⃗2 ⃗
d/2
2
q)
(2π) k2 (k2 + ⃗
(4π)
0
(x (1 − x) ⃗q 2 )2−d/2
\$' #
\$2 (
#
(−1)2−d/2 i Γ 2 − d2
Γ −1 + d2
1
=
,
d/2
2−d/2
2
Γ (−2 + d)
(4π)
(⃗q )
1
n =
(x (1 − x))
ˆ
2
1
dx x−n (1 − x)
0
B (a, b) ≡
ˆ
0
1
−n
≡ B (1 − n, 1 − n) =
xa−1 (1 − x)b−1 dx =
Γ (1 − n)
,
Γ (2 − 2n)
Re n &lt; 1,
Γ (a) Γ (b)
Γ (a + b)
d→4
#
\$2
Γ −1 + d2
= 1,
Γ (−2 + d)
ˆ
#
\$
2−d/2
(−1)
i Γ 2 − d2
dd⃗k2 1
1
1
.
=
d ⃗2 ⃗
d/2
2−d/2
2
2
(2π) k2 (k2 + ⃗q )
(4π)
(⃗q )
'
(
#
\$
2−d/2
(−1)
i Γ 2 − d2
1
dd⃗k1 1
1
d
d/2
2−d/2
(2π) ⃗k12 (⃗k1 + p⃗)2
(4π)
(⃗q 2 )
#
\$ˆ
λ3 (−1)2−d/2 Γ 2 − d2
1
dd⃗k 1
1
.
=−
!
&quot;
d ⃗2 ⃗
d/2
2
(2π) k (k + p⃗) (⃗k + p⃗ )2 2−d/2
2 (4π)
3
i λ3
i M2 =
2
ˆ
1
) αi =
Ai
1
=
ABC α
ˆ
1
1
=
ˆ
1
ˆ
0
!*
&quot; !
+ &quot; ) xαi −1 Γ (, αi )
xi , i ! α )
dxi δ 1 −
,
i
Γ (αi )
( xi Ai )
dx dy dz δ (1 − x − y − z)
0
n = 3 α1 = α2 = 1 α3 ≡ α
i = 1, . . . , n
dy dz
0
z
z α−1
2+α
(xA + yB + zC)
α−1
2+α
((1 − y − z) A + yB + zC)
α ≡ 2 − d/2
1
1
1
=
!
&quot;
2
⃗k 2 (⃗k + ⃗
p) (⃗k + p⃗ )2 2−d/2
3
ˆ
1
0
Γ (2 + α)
Γ (1) Γ (1) Γ (α)
Γ (2 + α)
.
Γ (α)
#
\$
z 1−d/2 Γ 4 − d2
\$,
dy dz 4−d/2 #
D
Γ 2 − d2
D ≡ (1 − y − z) ⃗k 2 + y(⃗k + p⃗)2 + z(⃗k + ⃗p3 )2
= ⃗k 2 − y⃗k 2 − z⃗k 2 + y⃗k 2 + ys + 2y⃗k &middot; p⃗ + z⃗k 2 + zm2 + 2z⃗k &middot; p⃗3
= ⃗k 2 + 2⃗k &middot; (y⃗
p + z⃗p3 ) + ys + zm2
2
≡ ℓ⃗2 − (y⃗
p + z⃗
p3 ) + ys + zm2
≡⃗
ℓ2 − ∆,
p⃗2 = s
⃗23 = m2
p
⃗ℓ ≡ ⃗k + (y⃗
p + z⃗
p3 )
=⇒
2
⃗ℓ2 = ⃗k 2 + 2⃗k &middot; (y⃗
p + z⃗p3 ) + (y⃗
p + z⃗p3 ) ,
∆ ≡ (y⃗
p + z⃗p3 )2 − ys − zm2
= 2yz⃗
p &middot; ⃗p3 − y (1 − y) s − z (1 − z) m2 .
s→∞
s = 2⃗
p1 &middot; ⃗
p2 = 2⃗
p3 &middot; ⃗
p4 ,
m2
t
t = −2⃗
p1 &middot; p⃗3 = −2⃗
p2 &middot; p⃗4 ,
∆
m2 ≪ s
u = −2⃗
p1 &middot; p⃗4 = −2⃗
p2 &middot; p⃗3 ,
2⃗
p &middot; p⃗3 = 2 (⃗p1 + p⃗2 ) &middot; p⃗3 = −t − u,
u ≈ −s
t
∆ = y (y + z − 1) s.
#
\$ˆ
Γ 2 − d2
\$
#
z 1−d/2 Γ 4 − d2
\$
i M2 = −
dy dz 4−d/2 #
d
d/2
D
Γ 2 − d2
(2π) 0
2 (4π)
\$ˆ 1
ˆ
2−d/2 #
λ3 (−1)
Γ 4 − d2
1
dd ⃗ℓ
1−d/2
.
=−
dy
dz
z
&quot;
d !
d/2
(2π) ⃗ℓ2 − ∆ 4−d/2
2 (4π)
0
2−d/2
λ3 (−1)
ˆ
dd ⃗ℓ
ˆ
1
#
\$
n
i (−1) Γ n − d2
1
&quot;n =
!
,
d
n−d/2
d/2
Γ
(n)
∆
(2π) ⃗ℓ2 − ∆
(4π)
dd ⃗ℓ
1
n ≡ 4 − d/2
4−d/2
dd ⃗ℓ
2−d/2
λ3 (−1)
i M2 = −
#
\$ˆ
Γ 4 − d2
d/2
2 (4π)
i λ3 Γ (4 − d) 1
=−
2 (4π)d s4−d
d≡4−ε
1
4−d/2
dy dz z 1−d/2
1
1
0
0
(y (y + z − 1))4−d
ˆ
1
dy
yε
ˆ
1
1
ε ≡
1−ε/2
(y + z − 1)
z
ˆ
dy dz z −1+ε/2
ε =
(y (y + z − 1))
z
1
0
1
ˆ
0
dy dz z −1+ε/2
ε.
(y (y + z − 1))
0
dz
z 1−ε/2
0
dz
f (z) ≡
0
.
Γ (4 − d) 1
\$
#
Γ 4 − d2 ∆4−d
1
ε.
(y + z − 1)
z=0
ˆ
1
f (z) =
1−ε/2
ˆ
1
=
ˆ
dz
z
(4π)
dy dz z 1−d/2
d/2
ε→0
ˆ
1
i (−1)
0
ˆ
i λ3 Γ (ε) 1
i M2 = −
2 (4π)4 sε
ˆ
Γ (4 − d) 1
\$
#
,
Γ 4 − d2 ∆4−d
1
i (−1)
&quot;4−d/2 =
d !
(2π) ⃗
(4π)d/2
ℓ2 − ∆
ˆ
0
dz
z 1−ε/2
f (z) ,
1
ε.
(y + z − 1)
1
dz
(f (z) − f (0))
1−ε/2
z
0
%
&amp;
ˆ 1
dz
1
dz
1
1
.
ε +
ε −
ε
1−ε/2
z 1−ε/2 (y − 1)
(y + z − 1)
(y − 1)
0 z
z
0
0
dz
1
f (0) +
1−ε/2
ˆ
ε→0
ˆ
0
1
dy
yε
ˆ
0
1
dz
1
ˆ
0
−−−→
ε→0
ˆ
1
dz
1
ε
z 1−ε/2 (y − 1)
ˆ 1
ˆ 1
dy
dz
=
ε (y − 1)ε
1−ε/2
y
z
(%0&amp;
'0
2
Γ (1 − ε)2
=
Γ (2 − 2ε)
ε
1
ε =
z 1−ε/2 (y + z − 1)
1
0
1
dz
=
,
n
z
1−n
dy
yε
ˆ
0
2
,
ε
Re n &lt; 1.
i λ3 Γ (ε) 1
i M2 = −
4
(4π) sε
% &amp;
1
.
ε
Γ (ε) =
# \$
1
s−ε ≡ e−ε log s = 1 − ε log s + ε2 log2 s + O ε3 ,
2
1
− γ + O (ε) ,
ε
%
&amp;% &amp;
&amp;%
# 3\$
1
1 2
1
2
i M2 = −
− γ + O (ε)
1 − ε log s + ε log s + O ε
4
ε
2
ε
(4π)
&amp;
&amp;%
%
3
# \$
1
1
iλ
1
− γ + O (ε)
− log s + ε log2 s + O ε2
=−
4
ε
ε
2
(4π)
%
&amp;
3
iλ
1
1
1
=−
− log s + log2 s ,
4
2
ε
ε
2
(4π)
i λ3
ε→0
s→∞
&amp;
&amp;
%
1
1
i λ3
1
1
1
1
2
2
−
log
s
+
log
s
−
2
&middot;
−
log
s
+
log
s
4
4
ε2
ε
4
ε2
ε
2
(4π)
(4π)
%
&amp;
3
iλ
3
3
5
=−
− log s + log2 s .
4
2
ε
ε
4
(4π)
i (M1 + M2 + M3 ) = −
%
i λ3
2
i M4 ≡
(− i λ) (− i δλ )
2
ˆ
dd⃗k
d
i
i
⃗k 2
(2π)
# \$
≡ (− i λ) (− i δλ ) &middot; i V p⃗2 .
−
m2
(⃗k + p⃗)2 − m2
% &amp;
2
,
δλ =
2
ε
2 (4π)
3λ2
% &amp;
# \$
1
V ⃗p2 .
i M4 = −
2
ε
(4π)
# 2\$
V p⃗
ε
i 3λ3
1/ε
m→0
dd⃗k 1
1
(2π)d ⃗k 2 (⃗k + p⃗)2
ˆ
ˆ
i 1
dd⃗k
1
=
dx
&quot;2
d !
2 0
(2π) (1 − x) ⃗k 2 + x(⃗k + p⃗)2
ˆ
ˆ
i 1
dd⃗k
1
=
dx
!
&quot;2
d
2 0
(2π) ⃗k 2 + 2x⃗k &middot; ⃗p + x⃗
p2
ˆ
ˆ
dd ℓ⃗
1
i 1
dx
=
!
&quot;2 ,
d
2 0
(2π) ⃗ℓ2 − ∆
# \$
i
V p⃗2 =
2
ˆ
⃗ℓ ≡ ⃗k + x⃗
p,
∆ ≡ −x (1 − x) p⃗2 .
dd ⃗
ℓ
ˆ
1
!
&quot;2 =
(2π) ⃗
ℓ2 − ∆
d
1
#
\$
i Γ 2 − d2
d/2
(4π)
1
∆2−d/2
,
#
\$
i Γ 2 − d2
1
2−d/2
∆
(4π)
0
#
\$
ˆ 1
Γ 2 − d2
1
1
=−
dx
d/2
2−d/2
2−d/2
2
0
2 (4π)
(⃗
p )
(x (1 − x))
#
\$
\$2
#
Γ −1 + d2
Γ 2 − d2
1
.
=−
2 (4π)d/2 (⃗
p2 )2−d/2 Γ (−2 + d)
# \$
i
V p⃗2 =
2
d→4−ε
ˆ
dx
d/2
ε→0
# \$
# 2\$
Γ 2ε
1
V ⃗
p =−
2
ε/2
2
2 (4π) (⃗
p )
&amp;
&amp;%
%
1
ε
ε2
2
2 2
2
− γ + O (ε)
1 − log p⃗ +
log p⃗
=−
2
8
2 (4π)2 ε
&amp;
%
1
ε
2
2 2
2
=−
,
−
log
p
⃗
+
log
p
⃗
2
ε
4
2 (4π)
p⃗2
ε
% &amp;'
%
&amp;(
1
2
1
ε
2 2
2
i M4 = −
−
− log p⃗ + log p⃗
2
2
ε
ε
4
(4π)
2 (4π)
%
&amp;
1
i 3λ3
1
1
=
−
log ⃗p2 + log2 ⃗p2
4
2
ε
2ε
8
(4π)
&amp;
%
3
1
1
i 3λ
1
2
−
log s + log s .
=
2ε
8
(4π)4 ε2
i 3λ3
i
5
+
i=1
&amp;
&amp;
%
3
5
i 3λ3
1
1
3
1
2
2
−
log
s
+
log
s
+
2
&middot;
−
log
s
+
log
s
ε
4
2ε
8
(4π)4 ε2
(4π)4 ε2
%
&amp;
i λ3
3
1
=
− log2 s .
4
ε2
2
(4π)
Mi = −
i λ3
s t
u = −s
%
s
u
s→∞
− i δλ
iM = −i
δλ
3λ3
4
2 (4π)
iM
t
s
1/ε2
log2 s,
!
```