Now we need to sum over all possible contractions However we must remind

# Now we need to sum over all possible contractions

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even. Now, we need to sum over all possible contractions. However, we must remind ourselves to keep the non-trivial sign of every contraction correctly. Since, every flip of the fermionic fields gives an extra minus sign, it should not be hard to convince yourselves that the net contraction is proportional to X (all full contractions) det V where V ij = ¯ u ( p i ) · u ( p j ) (47) Explicitly V = ¯ u ( p 1 ) · u ( p 2 ) ¯ u ( p 1 ) · u ( p 4 ) · · · · · · ¯ u ( p 1 ) · u ( p 2 n ) ¯ u ( p 3 ) · u ( p 2 ) ¯ u ( p 3 ) · u ( p 4 ) · · · · · · ¯ u ( p 3 ) · u ( p 2 n ) . . . . . . . . . . . . . . . . . . . . . . . . ¯ u ( p 2 n - 1 ) · u ( p 2 ) ¯ u ( p 2 n - 1 ) · u ( p 4 ) · · · · · · ¯ u ( p 2 n - 1 ) · u ( p 2 n ) (48) 7
Now, since u ( p ) is a 4-component object, 4 of these are enough to span the entire space of u ’s. In particular, we can write u ( p 2 n ) = α n, 2 u ( p 2 ) + α n, 4 u ( p 4 ) + α n, 6 u ( p 6 ) + α n, 8 u ( p 8 ) , n > 4 (49) where the α 2 , 4 , 6 , 8 could depend on any of the momenta. Given this, we can see that if n > 4, the determinant of V would be zero, since we can easily perform some row operations on the first row to make it equal to any of the n th rows ( n > 4). 1 So far we have shown that as long as we don’t consider self contractions of ( ¯ ψψ ) n , these terms to do not contribute to the Feynman diagrams. It is not hard (but tedious) to show in a similar way that it does not contribute even when we do consider self-contractions. 3(b) We are interested in computing Z O = Z D A μ D φe i R d 4 x L [ A μ ] O ( x ) (52) Consider the function f ( ξ ) = Z D πe i R d 4 x 1 ξ ( 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 would indeed produce gauge invariant operators. This would however not be a useful way of gauge-fixing, but it is technically possible. In order to look for gauge invariance of the term 1 ξ A 2 μ , we must consider the function. f ( ξ ) = Z D πe i R d 4 x 1 ξ ( μ π ) 2 (54) However, for the Faddeev Popov procedure to work, we need to be able to invert the equation A μ = μ α (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 not possible to add a term of the form 1 ξ A 2 μ and yet maintain gauge invariance. Another way to see that such a term cannot be added is as follows. Consider the adding such a term. This is basically a mass term in the Lagrangian with m 2 A = 2 ξ . The propagator of the photon is then modified to - i k 2 - 2 η μν - k μ k ν 2 . (56) Our physical calculations should not depend on the gauge parameter ξ . The Ward Identity can get rid of the k μ k ν 2 term in the propagator, but there is no way to get rid of the ξ dependence in the denominator. Thus, physical processes will depend on the value of ξ . This is clearly not allowed, and therefore it should not be possible to add a term of the form 1 ξ A 2 μ in the Lagrangian and yet maintain gauge invariance.

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