The Nucleon Mass in Chiral Perturbation Theory
Beyond one Loop
September 12, 2017
Institute of Theoretical Physics II
Hadron and Particle Physics
Contents
1
Introduction
2
The Lagrangian and Feynman Rules
3
Basics in QFT
4
Diagrams
5
Calculate the Diagrams
Contact Interaction
Preface
apply Feynman Rules
add Zeros
scalar integrals
Master Integrals
6
Extended-on-mass-shell Renormalization
7
Conclusion
Introduction
Quantum Chromodynamics (QCD)
ˆ
= strong interaction
→ nucleon mass
Lattice Calculation ˆ
= high energies
Chiral Pertubation Theory
ˆ
= low energies
calculations in baryon ChPT
Lattice calculation, Abdhel-Rehim et. al, 2015
•
BChPT power counting
7
•
HBChPT power counting
X
, Lorentz covariance
7
•
IR power counting
X
, Lorentz covariance
X
→ problems due to analytic properties of the loop integrals
Introduction
nucleon mass up to Order O(q
3
) (Scherer and Schindler 2012)
in terms of one loop integrals:
mN=m − 4c1M2+ 3g2 A 8F2(p · p)
n
/p m2− (p · p)Tπ+ −/p (p · p) + m2 − 2m (p · p)TN + /p − 2m2+ M2(p · p) + (p · p)2+ m4− m2M2− 2mM2(p · p)TπNo
renormalization of the integrals leads to:
m
N
= m − 4c
1
rM
2
+
3g
2
A
rM
2
32π
2
F
2
r
m −
3g
2
A
rM
3
32πF
2
r
+ O(q
4
)
nucleon mass up to order O(q
6
) with IR by Schindler 2007
goal: calculate nucleon mass up to order O(q
6
) with EOMS
The Lagrangian and Feynman Rules
The Lagragian
•
chiral Lagrangian up to chiral order O(q
4
) in SU(2) case
L = L
(2)
π
+ L
(4)
π
+ L
(1)
πN
+ L
(2)
πN
+ L
(3)
πN
+ L
(4)
πN
L
(1)
πN
= ¯
Ψ
i /
D − m +
g
A
2
/
uγ
5
Ψ
L
(2)
π
=
F
2
4
Tr
∂
µ
U (∂
µ
U)
†
M = O(q
1
)
∂
µ
U = O(q
1
)
∂
µ
Ψ = O(q
1
)
•
with nucleon and pion field:
Ψ = (p, n)
t
U =
1 +
i
F
~
τ ~
π −
1
2F
2
~
π
2
− iα
1
F
3
~
π
2
~
τ ~
π + (8α − 1)
1
8F
4
~
π
4
u =
12
+
i
2F
~
τ ~
π −
8
F
2
(~
τ ~
π)
2
+
i (8α − 1)
16F
3
(~
τ ~
π)
3
+
32α − 5
128F
4
(~
τ · ~
π)
4
The Lagrangian and Feynman Rules
The Lagrangian
expand the Lagrangian in pion fields (using for example trace rules)
L
(1)
πN
= − ¯
ΨmΨ + i ¯
Ψ /
∂Ψ
+
g
A
2F
¯
Ψγ
5
~
τ · /
∂~
πΨ +
1
4F
2
Ψ~
¯
π · ~
τ × /
∂~
π
Ψ
−
g
A
4F
3
Ψγ
¯
5
2α(~
π · ~
π)(~
τ · /
∂~
π) + (4α − 1)(~
π · ~
τ )(~
π · /
∂~
π)
Ψ
−
1
16F
4
Ψ(8α − 1)(~
¯
π · ~
π)~
π · ~
τ × /
∂~
π
Ψ + O(π
5
)
The Lagrangian and Feynman Rules
Feynman Rules
L
(2)
π
:
2
L
(4)
π
:
4
L
(1)
πN
:
1
1
1
1
L
(2)
πN
:
2
2
2
L
(3)
πN
:
3
3
3
L
(4)
πN
:
4
4
L
(6)
πN
:
6
The Lagrangian and Feynman Rules
Feynman Rules
q
→
=
ˆ
i
q
2
− M
2
+ i ε
p
→
=
ˆ
i
/
p − m + i ε
=
i /
p + m
p · p − m
2
+ i ε
p
1
→
p
2
→
q
1
→
, a
q
2
←
, b
1
ˆ
=(2π)
4
δ
4
(p
1
− p
2
+ q
1
+ q
2
)
1
4F
2
(/
q
1
− /q
2
)ε
abc
τ
c
Basics in QFT
Nucleon Mass
p
h0|T {Ψ
H
(x ) ¯
Ψ
H
(y )}|0i
p
=
any
inter-action
=
+
1PI
+
1PI
1PI
+...
=
i
/
p − m + i ε
+
i
/
p − m + i ε
(−i Σ)
i
/
p − m + i ε
+ ...
=
i
/
p − m + i ε
1 +
Σ
/
p − m + i ε
+
Σ
/
p − m + i ε
!2
+ ...
=
i
/
p − m − Σ + i ε
=
i
/
p + m + Σ
p · p − (m + Σ)
2
+ i ε
where the self energy Σ is defined
−iΣ :=
Basics in QFT
Nucleon Mass
Definition
The nucleon mass is a pole in the two-point
nucleon function.
h
p · p − (m + Σ)
2
+ i ε
i
p·p=m
2 N= 0
⇒ m
N
= m + Σ
Basics in QFT
Power Counting
q
m
< 1
and
q
m
N
< 1
M = O(q)
|~p| = O(q)
For integrals over pion and nucleon propagators one can show
•
integrals in d dimensions count as O(q
d
)
•
pion propagators count as O(q
−2
)
•
nucleon propagators count as O(q
−1
)
for example
m
Z
d
d
l
(2π)
d
1
(l · l − M
2
+ i ε) ((l − p) · (l − p) − m
2
+ i ε)
= O(q
d −3
)
Basics in QFT
Power Counting
for diagrams one can show that vertices of chiral order k count like
δ
d
(q)q
k
→ t
k−d
δ
d
(q)q
k
So a diagram with N
I
π
internal pion lines N
I
N
internal nucleon lines and
N
V
kvertices of chiral order k count in d -dimensions as
FD → t
D
FD
D = d + (d − 2)N
I
π
+ (d − 1)N
I
N
+
X
k
N
V
k(k − d )
with
N
L
− 1 = N
I
− N
V
k⇒ D = dN
L
− 2N
π
I
− N
I
N
+
X
k
kN
V
kBasics in QFT
Power Counting
For example (in four dimensions):
1
1
2
is of order
Diagrams
contact terms
2
O(q
2
)
4
O(q
4
)
6
O(q
6
)
Diagrams
one loop
1
1
(a) O(q
3
)
1
3
(b)O(q
5
)
3
1
(c) O(q
5
)
1
1
4
(d) O(q
5
)
2
(e) O(q
4
)
4
(f) O(q
6
)
2
4
(g) O(q
6
)
Diagrams
two loop
1
1
1
1
(a) O(q
5
)
1
1
1
1
(b) O(q
5
)
1
1
(c) O(q
5
)
1
2
1
(j) O(q
6
)
Diagrams
two loop
1
1
1
(d) O(q
5
)
1
1
1
(e) O(q
5
)
1
1
2
(f) O(q
6
)
2
1
1
(g) O(q
6
)
1
1
1
(h) O(q
5
)
1
2
1
(i) O(q
6
)
Diagrams
two loop
1
1
(k) O(q
5
)
1
1
(l) O(q
5
)
2
(m) O(q
6
)
2
2
(n) O(q
6
)
1
1
2
(o) O(q
5
)
Diagrams
two loop further
1
2
[f] O(q
6
)
2
1
Calculate the Diagrams
Contact Interaction
Σ
c
= −4c
1
M
2
− 2 e
115
+ e
116
+ 8e
38
M
4
+ ˆ
g
1
M
6
.
For the simplification of the calculations make a shift
m → m + Σ
c
⇒ m
N
− m = Σ
l
= O(q
3
).
This implies
lim
p→m
Np · p = m
2
+ m
2
N
− m
2
= m
2
+ O(mq
3
)
lim
p→m
N¯
u(~
p)/
pu(~
p) = ¯
u(~
p)
/
p − m
N
+ m + m
N
− m
u(~
p) = ¯
u(~
p) m u(~
p) + O(q
3
)
For diagrams, which are of minimal order O(q
4
) terms the replacments
/
p → m and p · p → m
2
Calculate the Diagrams
Preface
1.
The Feynman Rules are applied to get the matematical expression.
2.
Suitable zeros are added to reduce the tensor rank.
3.
All integrals are reduced to scalar integrals in d + ... dimensions
(Davydychev 1991, Tarasov 1996).
4.
The integrals are expressed with a short set of “Master Integrals”
(using the program TARCER based on Tarasov’s algorithm).
example:
→
l
1
, a
→
l
2
, b
−iΣ
(2)
c
=
→
p
p − l
→
1
− l
2
→
p
1
1
Calculate the Diagrams
apply Feynman Rules
•
apply FR
•
use {γ
µ
, γ
5
} = 0, γ
5
γ
5
=
1
d
, τ
a
τ
b
= δ
ab
12
+ i ε
abc
τ
c
to simplify the
expressions
•
simplify and order the expression using
{γ
µ
, γ
ν
} = 2g
µν
1
d
Calculate the Diagrams
apply Feynman Rules – Example
−2iΣ(2)c =
/l1⊗ (τcεabc) 4F2 − /l2⊗ (τcεabc) 4F2 i1 d⊗12 (l1· l1) + i ε − M2 i1d⊗12 (l2· l2) + i ε − M2 i (m1d⊗12+ (−/l1− /l2+ /p) ⊗12) ((−l1− l2+ p) · (−l1− l2+ p)) + i ε − m2 −/l1⊗ (τdεabd) 4F2 − −/l2⊗ (τdεabd) 4F2 −2iΣ(2)c = 3i 8F4 l1· l1− M2+ i ε l2· l2− M2+ i ε (l1+ l2− p) · (l1+ l2− p) − m2+ i ε 1dm(−2 (l1· l2) + (l1· l1) + (l2· l2)) − /p(−2 (l1· l2) + (l1· l1) + (l2· l2)) −/l1(2 (l1· l2) − 2 (l1· p) + (l1· l1) + 2 (l2· p) − 3 (l2· l2)) +/l2(−2 (l1· l2) − 2 (l1· p) + 3 (l1· l1) + 2 (l2· p) − (l2· l2)) ⊗12Calculate the Diagrams
add Zeros
•
idea: reduce tensor rank by adding suitable zeros
l · l
l · l − M
2
+ i ε
=
l · l − M
2
+ M
2
l · l − M
2
+ i ε
= 1 +
M
2
l · l − M
2
+ i ε
•
(in the case of one loop diagrams) rewrite in the denominator
q · q − m
2
+ i 0 → P[q, m]
and in the numerator
l · p → −
1
2
[(l − p) · (l − p) − (l · l + p · p)]
(l − p) · (l − p) → P[l − p, M] + m
2
l · l → P[l , M] + M
2
•
individual procedure – for example:
Calculate the Diagrams
add Zeros
further simplification:
•
scale-less integrals vanish in dimensional renormalization
Z
d
d
l
(2π)
d
l
ν1
l
ν2
→ 0
•
use T -notation
•
cancel odd integrals
Z
d
d
l
(2π)
d
l
ν
(p − l ) · (p − l ) − m
2
+ i ε
= 0
Calculate the Diagrams
add Zeros – Notation
T
(1),ν
1...ν
np,M,m
(d ; α
1
, α
2
) =
=
Z
d
d
l
(2π)
d
l
ν
1...l
ν
n(l · l − M
2
+ i ε)
α
1((l − p) · (l − p) − m
2
+ i ε)
α
2T
(2),ν
1 1...ν
n11ν
21...ν
2n2p,m
1,m
2,m
3,m
4,m
5(d ; β
1
, β
2
, β
3
, β
4
, β
5
) =
=
Z
d
d
l
1
(2π)
d
Z
d
d
l
2
(2π)
d
l
ν
1 11
...l
ν
1 n12
l
ν
212
...l
ν
2 n22
(l
1
· l
1
− m
2
1
+ i ε)
β
1(l
2
· l
2
− m
2
2
+ i ε)
β
21
((l
1
− p) · (l
1
− p) − m
2
3
+ i ε)
β
3((l
2
− p) · (l
2
− p) − m
2
4
+ i ε)
β
41
((l
1
+ l
2
− p) · (l
1
+ l
2
− p) − m
2
5
+ i ε)
β
5Calculate the Diagrams
add Zeros – Example
−2iΣ(2)c = − 3igµ1ν1 pµ1(1dm − /p) − 2M2γµ1 T(2),ν11 p,M,M,m,m,m(1, 1, 0, 0, 1) 4F4 − 3igµ1ν1 pµ1(1dm − /p) − 2M2γµ1 T(2),ν21 p,M,M,m,m,m(1, 1, 0, 0, 1) 4F4 + 3iM2(1 dm − /p)Tp,M,M,m,m,m(2) (1, 1, 0, 0, 1) 2F4 + 3i (1dm − /p)Tp,M,M,m,m,m(2) (0, 1, 0, 0, 1) 4F4 + 3i (1dm − /p)Tp,M,M,m,m,m(2) (1, 0, 0, 0, 1) 4F4 − 3i (1dm − /p)Tp,M,M,m,m,m(2) (1, 1, 0, 0, 0) 8F4 − 3i γµ1pµ2 gµ1ν2gµ2ν1+ gµ1ν1gµ2ν2T(2),ν11ν22 p,M,M,m,m,m(1, 1, 0, 0, 1) 2F4 + 3i γµ1gµ1ν1T(2),ν21 p,M,M,m,m,m(0, 1, 0, 0, 1) 2F4 + 3i γµ1gµ1ν1T(2),ν11 p,M,M,m,m,m(1, 0, 0, 0, 1) 2F4 − 3i γµ1gµ1ν1T(2),ν11 p,M,M,m,m,m(1, 1, 0, 0, 0) 8F4 − 3i γµ1gµ1ν1T(2),ν21 p,M,M,m,m,m(1, 1, 0, 0, 0) 8F4Calculate the Diagrams
scalar integrals
•
factorize into one loop integrals for example
T
ν
1 1...ν
1 n1ν
2 1...ν
2 n2p,{m
k}
5k=1(d ; α
1
, α
2
, 0, 0, 0) = T
ν
1...ν
n1p,m
1,0
(d ; α
1
, 0) T
ν
1...ν
n2p,m
2,0
(d ; α
2
, 0)
T
ν
1 1...ν
1n1ν
12...ν
n22p,{m
k}
5k=1(d ; 0, α
2
, 0, 0, α
5
) =
Z
d
d
l
1
(2π)
d
Z
d
d
l
2
(2π)
d
1
(l
1
· l
1
− m
2
5
+ i ε)
α
5(l
2
· l
2
− m
2
2
+ i ε)
α
2l
ν
111
...l
ν
n111
l
ν
2 12
...l
ν
2n22
l
1→l
1−l
2+p
•
reduce tensor integrals to integrals in d + ...-dimensions
1.
Rewrite the tensor structure by using derivatives.
2.
Rewrite the propagators with the Schwinger trick/ in λ-representation
3.
Evaluate the integral(s) over the loop momentum/ momenta per Gaussian
integration.
4.
Construct a operator and reverse the Gaussian integration and the
λ-parametrization for the scalar integral.
Calculate the Diagrams
scalar integrals
Tp,m1,m2(1),ν1...νn(d ; α1, α2) =Z
ddl (2π)d lν1...lνn (l · l ) − m2 1+ i ε α1 (l − p) · (l − p) − m2 2+ i ε α2•
first two steps yield
(−i )
n
n
Y
k=1
∂
∂a
ν
ka=0
(−i )
α
1+α
2Γ (α
1
) Γ (α
2
)
Z
∞
0
Z
∞
0
dλ
1
dλ
2
λ
α
1
1−1
λ
α
2−1
2
G
(1)
•
Gauss integration
G
(1)
=
1
(2π)
d
i
π
i
d21
(D(λ))
d2exp
i
Q(λ, a)
D(λ)
exp
i λ
1
−m
2
1
+ i ε + iλ
2
p
2
− m
2
2
+ i ε − i
λ
2
2
p
2
λ
1
+ λ
2
D(λ) = λ
1
+ λ
2
Q(λ, a, b) = (p · a) Q
1
+ a
2
Q
11
Q
1
= λ
2
Q
11
= −
1
4
Calculate the Diagrams
scalar integrals
G
(1)
=
1
(2π)
d
i
π
i
d21
(D(λ))
d2exp
i
Q(λ, a)
D(λ)
exp
i λ
1
−m
2
1
+ i ε + iλ
2
p
2
− m
2
2
+ i ε − i
λ
2
2
p
2
λ
1
+ λ
2
1
D(λ)
1
(2π)
d
π
i
d21
(D(λ))
d2= (2π)
2
i
π
1
(2π)
d +2
π
i
d +221
(D(λ))
d +22.
1
D(λ)
=i 4π
ˆ
2
d
+
λ
i
=i
ˆ
∂
∂m
2
i
(numerator)
Calculate the Diagrams
scalar integrals
one can build a operator
T
ν
1...ν
np,m
1,m
2(d ; α
1
, α
2
) =T
ν
1...ν
np, {∂/∂m
2
}, d
+
T
p,m
1,m
2(d ; α
1
, α
2
)
T
ν
1...ν
np, {∂/∂m
2
}, d
+
= (−i)
n
n
Y
k=1
∂
∂a
ν
kexp (iQ (λ, a) ρ)
a=0
λ
j=i
∂m2∂ jρ=i 4πd
+similar in the two loop case
T
ν
1 1...ν
n11ν
12...ν
n22p,{m
k}
5k=1(d ; α
1
, α
2
, α
3
, α
4
, α
5
)
=T
ν
11...ν
n11ν
12...ν
n22p, {∂/∂m
2
}, d
+
T
p,{m
k}
5k=1(d ; α
1
, α
2
, α
3
, α
4
, α
5
)
T
ν
11...ν
1 n1ν
2 1...ν
2 n2p, {∂/∂m
2
}, d
+
= (−i )
n
1+n
2n
1Y
k
1=1
n
2Y
k
2=1
∂
∂a
ν
k1∂
∂b
ν
k2exp (iQ (λ, a, b) ρ)
a=b=0
λ
j=i
∂m2∂ jρ=−16π
2d
+Calculate the Diagrams
scalar integrals – Example
−2iΣ(2)c = 12i π2 /p (p · p) + 2M2
−1dm (p · p) Tp,m1,m2,m3,m4,m5(2) (d + 2; 1, 2, 0, 0, 2) F4 + 12i π2 / p (p · p) + 2M2 −1dm (p · p) Tp,m1,m2,m3,m4,m5(2) (d + 2; 2, 1, 0, 0, 2) F4 + 3iM2(1dm − /p)Tp,m1,m2,m3,m4,m5(2) (d ; 1, 1, 0, 0, 1) 2F4 − 3iT (m5, 0)(d ; 1, 0)(/p(Tm1,0(1) (d ; 1, 0) + T (m2, 0)(d ; 1, 0)) −1dmT (1) m2,0(d ; 1, 0)) 4F4 − 3iT(1) m1,0(d ; 1, 0)((1dm − /p)Tm2,0(1) (d ; 1, 0) − 21dmTm5,0(1) (d ; 1, 0)) 8F4 −1536i π 4 / p (p · p) Tp,m1,m2,m3,m4,m5(2) (d + 4; 2, 2, 0, 0, 3) F4 + 24i π2/pTp,m1,m2,m3,m4,m5(2) (d + 2; 1, 1, 0, 0, 2) F4Calculate the Diagrams
Master Integrals – Integration by Parts
0 =
Z
d
d
l
1
(2π)
d
Z
d
d
l
2
(2π)
d
∂
∂l
1
µ
l
µ
1
P
α
1l
1,m
1P
α
2l
2,m
2P
α
5l
1−l
2,m
5=
Z
d
d
l
1
(2π)
d
Z
d
d
l
2
(2π)
d
dP
α
1l
1,m
1P
α
2l
2,m
2P
α
5l
1+l
2−p,m
5− 2α
1
(l
1
· l
1
) P
α
1+1
l
1,m
1P
α
2l
2,m
2P
α
5l
1−l
2,m
5−2α
2
(l
1
· l
2
) P
l
α
1,m
1 1P
α
2+1
l
2,m
2P
α
5l
1−l
2,m
5− 2α
5
((l
1
· l
1
) − (l
1
· l
2
)) P
α
1l
1,m
1P
α
2l
2,m
2P
α
5+1
l
1−l
2,m
5one can cancel the scalar products to obtain
d − 2α
1
1 + m
2
1
1
+
− 2α
2
1 + m
2
2
2
+
−2α
5
5
+
1
−
+ m
2
1
+
1
2
5
−
+ m
2
5
− 1
−
− m
2
1
− 2
−
− m
2
2
¯
T
p,{m
(2)
k}
5k=1(d ; α
1
, α
2
, 0, 0, α
5
) = 0
Calculate the Diagrams
Master Integrals – Dimension Reduction
T
p,m1,m2
(1)
(d ; α1
, α
2)
=
(−i )
α1
+α
2Γ (α1
) Γ (α2)
2
Y
j=1
Z
∞
0
dλ
j
λ
α
j−1
j
1
(2π)
d
i
π
i
d21
(D(λ))
d2exp
i λ
j
ˆl
j
− m
2
j
+ i ε
− i
λ
2
2
p
2
D(λ)
!
apply on both sides D(λ) = λ1
+ λ2
in operator form
D(λ) = i ∂
m
21
+ i ∂
m
22gives
i 4πT
p,m1,m2
(1)
(d − 2; α
1
, α
2
)
Calculate the Diagrams
Master Integrals
•
relations between integrals in same dimension
integration by parts
•
relations between integrals in different dimensions
Schwinger representation and Gauss integration
⇒ recurrence relations to reduce integrals to a set of
Calculate the Diagrams
Master Integrals – Example
−iΣ
(2)
c
=
−
1
2
i
2(d − 2)(3d − 4)F
4
m
n
2
h
2 2d
2
− 7d + 5 m
4
+
−7d
2
+ 23d − 16 m
2
M
2
+ d
2
− 5d + 6 M
4
T
(2)
M,m,0,0,M
(d ; 1, 1, 0, 0, 1)
−4M
2
M
2
− m
2
(d − 2)M
2
− 2(d − 1)m
2
T
(2)
M,m,0,0,M
(d ; 2, 1, 0, 0, 1)
i
+(d − 2) (d − 1)m
2
+ (d − 2)M
2
(T
M,0
(1)
(d ; 1, 0))
2
+(2 − d ) 4(d − 1)m
2
+ (d − 2)M
2
T
(1)
m,0
(d ; 1, 0)T
(1)
M,0
(d ; 1, 0)
o
Calculate the Diagrams
Master Integrals – one loop
−iΣ(1)a = 3g2 A 8F2(p · p)
n
/ p m2− (p · p)T (M, 0)({d }, 1, 0) + −/p (p · p) + m2− 2m (p · p)T (m, 0)({d }, 1, 0) + /p − 2m2+ M2(p · p) + (p · p)2+ m4− m2M2− 2mM2(p · p)T (M, m)({d }, 1, 1)o
−iΣ(1) b = − i Σ (1) c = 3gAmM2 F2n
M2(d18 − 2d16)T (M, m)({d }, 1, 1) + (d18 − 2d16)T (m, 0)({d }, 1, 0)o
−iΣ(1) d = 3g2 A 2F4 4m2− M2n
− 2mM4h
l3 2(d − 1)m2− (d − 2)M2 + l4 M2− 4m2i
T (M, m)({d }, 1, 1) + 2m (d − 2)l3M4+ l4 4m2M2− M4 T (m, 0)({d }, 1, 0) − (d − 2)l3mM4T (M, 0)({d }, 1, 0)o
Calculate the Diagrams
Master Integrals – one loop
−iΣ(1)e =1 2 6M2T (M, 0)({d }, 1, 0) dm2(2c1 − c3) − c2m2
dF2m2 −iΣ(1) f = − 1 2 24M4T (M, 0)({d }, 1, 0) d (d + 2)F2n
d2(2e14+ 2e19− e36− 4e38) + 2d (2e14+ e15+ 2e19+ e20+ e35− e36− 4e38) + 4e15+ 6e16+ 4e20+ 4e35o
−iΣ(1)g = −1 2 6M4(2c1d (2l4− (d − 2)l3) + (c2 + c3d )(dl3− 2l4))T (M, 0)({d }, 1, 0) dF4Calculate the Diagrams
Master Integrals – two loop
−iΣ(2) k = − i Σ (2) l = 1 2 6i (10α − 1)gA2mT (M, 0)({d }, 1, 0) M2TM,m(1) (1, 1) + Tm,0(1)(1, 0)
4F4 −iΣ(2)m = −1 8 12iM2(TM,0(1)(1, 0))2 (2 − 20α)c2m2+ dm2(5(8α − 1)c1− 20αc3+ 2c3) dF4m2 −iΣ(2)n =1 4 3iM2(TM,0(1)(1, 0))2 dm2(2c1(40α + d − 6) − c3(40α + d − 4)) − c2(40α + d − 4)m2 dF4m2 −iΣ(2)o = −1 2 6ig2 A 8F4 4m2− M2 mT(1) M,0(1, 0)