114
4 Four-Dimensional Superfield Supersymmetry
where we carried out the Wick rotation s = it (with t = −˜ s) and x 0 = i x 0E for
convenience.
It is convenient to split the indices n into odd, n = 2l + 1 and even, n = 2l, ones.
As a result, we can write K
(1) as
(1)
K =
d
4 θd
4 x 1
dt
t
∞
l=0
1
(4l + 2)!
(t
2 ¯
)
2l+1
2l +
1
(4l + 4)!
(t
2 ¯
)
2l+2
2l+1
×
× e
−t 2 δ
4 (x 1 − x 2 )| x1=x2 .
(4.233)
Now, let us consider the structure
n e
−t
2 δ
4
(x 1 − x 2 )| x 1 =x 2 . It is clear that the function V (t; x 1 , x 2 ) = e
−t
2 δ
4
(x 1 − x 2 ) which we will call the free heat kernel, satisfies
the equation
2 V (t; x 1 , x 2 ) = −
d
dt
V (t; x 1 , x 2 ),
(4.234)
hence
2l V (t; x 1 , x 2 ) = (−
d
dt
) l V (t; x 1 , x 2 ); 2l+1 V (t; x 1 , x 2 ) = (−
d
dt
) l V (t; x 1 , x 2 ).
(4.235)
In this subsection, the above expressions will be considered only in the limit x 1 = x 2 .
One can find that
V (t; x 1 , x 2 )| x 1 =x 2 =
d
4 k
(2π) 4 e
−tk
4 =
1
32π 2 t
;
V (t; x 1 , x 2 )| x 1 =x 2 =
d
4 k
(2π) 4 (−k
2
)e
−tk
4 = −
1
32π 3/2 t 3/2 ,
(4.236)
therefore
2l V (t; x 1 , x 2 )| x 1 =x 2 = (−
d
dt
)
l
1
32π 2 t
=
(−1)
l l!
32π 2 t l+1 ;
2l+1 V (t; x 1 , x 2 )| x 1 =x 2 = (−
d
dt
)
l
(−
1
32π 3/2 t 3/2 ) = −
(−1)
l+1
(2l + 1)!!
32π 3/2 2 l t l+3/2 . (4.237)
Replacing all this into (4.233), we arrive at
(1)
K =
1
32π 2
d
8 z
dt
∞
l=0
(−1)
l
t
3l l!(( ¯
)
2l+1
(4l + 2)!
−
− t
3l+3/2 (( ¯
)
2l+2
(4l + 4)!
√ π(2l + 1)!!
2 l
.
(4.238)