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even. Now, we need to sum over all possible contractions. However, wemust remind ourselves to keep the non-trivial sign of every contraction correctly. Since, every flip of the fermionicfields gives an extra minus sign, it should not be hard to convince yourselves that the net contraction is proportionaltoX(all full contractions)∼detVwhereVij= ¯u(pi)·u(pj)(47)ExplicitlyV=¯u(p1)·u(p2)¯u(p1)·u(p4)· · ·· · ·¯u(p1)·u(p2n)¯u(p3)·u(p2)¯u(p3)·u(p4)· · ·· · ·¯u(p3)·u(p2n)........................¯u(p2n-1)·u(p2)¯u(p2n-1)·u(p4)· · ·· · ·¯u(p2n-1)·u(p2n)(48)7
Now, sinceu(p) is a 4-component object, 4 of these are enough to span the entire space ofu’s. In particular, we canwriteu(p2n) =αn,2u(p2) +αn,4u(p4) +αn,6u(p6) +αn,8u(p8),n >4(49)where theα2,4,6,8could depend on any of the momenta. Given this, we can see that ifn >4, the determinant ofVwould be zero, since we can easily perform some row operations on the first row to make it equal to any of thenthrows (n >4).1So far we have shown that as long as we don’t consider self contractions of(¯ψψ)n, these terms to do not contributeto the Feynman diagrams. It is not hard (but tedious) to show in a similar way that it does not contribute even whenwe do consider self-contractions.3(b) We are interested in computingZO=ZDAμDφeiRd4xL[Aμ,φ]O(x)(52)Consider the functionf(ξ) =ZDπeiRd4x1ξ(π)4(53)Using this function, one can now proceed in a way similar to that done in class and show that such a term wouldindeed produce gauge invariant operators. This would however not be a useful way of gauge-fixing, but it is technicallypossible.In order to look for gauge invariance of the term1ξA2μ, we must consider the function.f(ξ) =ZDπeiRd4x1ξ(∂μπ)2(54)However, for the Faddeev Popov procedure to work, we need to be able to invert the equationAμ=∂μα(55)This is clearly not possible since we have 4 degrees of freedom on the LHS and only 1 on the RHS. Thus, it is notpossible to add a term of the form1ξA2μand yet maintain gauge invariance. Another way to see that such a termcannot be added is as follows. Consider the adding such a term. This is basically a mass term in the Lagrangian withm2A=2ξ. The propagator of the photon is then modified to-ik2-2/ξημν-kμkν2/ξ.(56)Our physical calculations should not depend on the gauge parameterξ. The Ward Identity can get rid of thekμkν2/ξterm in the propagator, but there is no way to get rid of theξdependence in the denominator.Thus, physicalprocesses will depend on the value ofξ. This is clearly not allowed, and therefore it should not be possible to add aterm of the form1ξA2μin the Lagrangian and yet maintain gauge invariance.