## K3sA[r[h[sbkirtllbk2r2

- Sp.ak[Ss.ir[ (o\bp[3[bkinbl,r,bk,r7 ... 10) . (9.115) Jsing this identity, the second order S'-matrix element becomes

X |[u(pi,s'1)75rtu(pi,si)] [u{p2,S,2)'r5Tju(p2,S2)]

x ^gi(p'l-pi) xi+i(p2-p2) x2 gi(pi-pi) i2 + i(p2-p2) xi^

X ^(Pl-Pllxi+'iPl-Pllxi + e'(Pl-P2)-X2+l(P2-Pl)xi ^ (9.116)

where it was assumed that (O|T0j(a;i)0j(x2)|O) depends on xi — x2 only and is symmetric in i,j. But this is clearly the case, because and <pj commute if j ^ i, so that

(0|T (Ux, )<f>j (x3)) |0> = Sij <0|T (0, (x2)) |0>

d4k (2tt)4

where, in the first line, we used the fact that the time ordered product can be evaluated using any one of the fields (because they all give the same answer) and the last step follows from Eq. (9.37). Finally, we call attention to the sign difference between the first and second term in Eq. (9.116), which is due to the anticommutation relations.

We now can simplify the result (9.116) for S if we introduce

With this substitution we can integrate out the overall energy conserving 6-function, and separate the Al-matrix, defined in Eq. (9.12). This gives the following result:

. . .. 2 [u(/'i!si)75riU(p1,Si)] [u(p'2,S,2)~f5Tiu(p2,S2)}

-exchange term with {p'js'i} <->• {p^^} ■ (9.119)

This result leads to the two Feynman diagrams shown in Fig. 9.6, except that in this case the states are antisymmetric so the second diagram comes in with a minus sign. The following Feynman rules apply to this example:

Rule 1: the operator gr^5 at each tiNN vertex.

Rule 2: a factor of

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