3.4 Quantum Description for the Superfield Models
31
i S
(1)
1a ( p) =
1
2
d
2
θ 1 d
2
θ 2
d
3 k
(2π) 3 I (k, p)
×
δ 12 (k
2
+ m
2
)(k αβ − C αβ D
2
)δ 12 A
α
(− p, θ 1 )A
β
( p, θ 2 )
+ δ 12 (−D
2
+ m)(k αβ − C αβ D
2
)δ 12 (D
2 A
α
(− p, θ 1 ))A
β
( p, θ 2 )
. (3.105)
The only terms giving non-zero contributions are those containing just one D
2
since δ 12 D
2
δ 12 = δ 12 , see (3.19). Indeed, by employing this identity and after integrating over θ 2 with the help of the delta function, we obtain
i S
(1)
1a ( p) = −
1
2
d
2
θ
d
3 k
(2π) 3 I (k, p)
(3.106)
×
(k
2
+ m
2
)C αβ A
α
(− p, θ)A
β
( p, θ)
+ (k αβ + mC αβ )(D
2 A
α
(− p, θ))A
β
( p, θ)
.
The second term of (3.101) is
i S
(2)
1a ( p) =
1
4
d
2
θ 1 d
2
θ 2
d
3 k
(2π) 3 I (k, p)
×
(D
2
1 + m)δ 12 (D
2
1 + m)D β1 δ 12 (D
α A α )(− p, θ 1 )A
β
( p, θ 2 )
. (3.107)
In this expression we must keep only the term proportional to D
2
1 δ 12 (D
2
1 + m)D β1 δ 12
(the remaining part is a trace of an odd number of derivatives which clearly vanishes).
Thus, after manipulations similar to those performed for S
(1)
1a , we find
i S
(2)
1a ( p) = −
1
4
d 2 θ
d 3 k
(2π) 3 I (k, p)
D γ D α A α (− p, θ)(k γβ + mC γβ )A β ( p, θ)
.
(3.108)
By adding (3.106) and (3.108) we can write the total contribution from Fig. 3.1a as
i S 1a ( p) = −
1
2
d
2 θ
d 3 k
(2π) 3 I (k, p)
×
(k
2 + m
2 )C αβ A
α (− p, θ)A
β ( p, θ) + (k αβ + mC αβ )(D
2 A
α (− p, θ))A
β ( p, θ)
+
1
2
D
γ D
α A α (− p, θ)(k γβ + mC γβ )A
β ( p, θ)
.
(3.109)
The algebraic manipulations for the graph Fig. 3.1b are much more simpler and
yield
i S 1b ( p) =
1
2
d
3 k
(2π) 3
1
(k + p) 2 + m 2 C αβ A
α
(− p, θ)A
β
( p, θ). (3.110)