TOPOLOGY OF GAUGE FIELDS AND RELATED PROBLEMS
101
natural cut-off, by the quenching factor
The
is needed here,
since £„ in (6.62) can be both positive and negative.
After these explanations let us compute the quantity:
which is just the regularized axial current. We have:
_^i5,(iA„y,r5W
— L
¡ 7
^
(6.65)
= - 2 X
= 2 tr,
(6.66)
= (iy.v,)^)
Here we have used an identity, following straight from (6.63);
i
= -2£„('A„y5'/'»)
The last term in (6.66) is easily calculable as e 0. We have:
®
+ -4^)^ -
= Ky„. y.]
(6.67)
The most singular term in e
will come from 8^ in (6.67). If we neglect
all fields, then:
< x |e - ‘®|x> =
1
(2n)*
I6n^e
2,2
(6.68)
Expansion in and
will give us less singular terms. These terms are
important since for A^ = 0:
Tr(75C-^) =
rTrys = 0
(6.69)
\6n^e^
We see that it will be useless to expand (d^ -h A^)^ in terms of A^, since
Tr 75 = 0 will remain in our formulas. Expansion up to the first term in
^Iso does not help since Tr 75
= 0. The first nonzero contribution arises from:
T r .y ^ e -® « < x|e‘^^|x> - TT(y,cr^,c,^) ■ T r (F ,,F ,,)
(6.70)
If we notice that:
Tr(y5(7„,(T,i.) = 4t, H vX p