the delta function identity 1 Z D \u03b1 \u03b4 G A \u03b1 det \u03b4G A \u03b1 \u03b4\u03b1 whereA \u03b1 \u03bc A\u03bc 1 e \u03bc \u03b1

# The delta function identity 1 z d α δ g a α det

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the delta function identity 1 = Z [ D α ] δ ( G ( A α ))det δG ( A α ) δα , where A α μ = A μ + 1 e μ α is the potential gauge transformed by α . (This is just the infinite dimensional generalization of the delta function composition identity you’re used to.) Now let’s insert this factor of 1 into the path integral, Z [ D A ] e iS [ A ] O ( A ) = Z [ D α D A ] e iS [ A ] O ( A ) δ ( G ( A α ))det δG ( A α ) δα For the G that we chose above, δG ( A α ) δα = 1 e 2 , so the determinant is independent of A and α , and we can pull it out of the integral. Thus we have Z [ D A ] e iS [ A ] O ( A ) = det 1 e 2 Z [ D α D A ] e iS [ A ] O ( A ) δ ( G ( A α )) = det 1 e 2 Z [ D α D A α ] e iS [ A ] O ( A ) δ ( G ( A α )) = det 1 e 2 Z [ D α D A α ] e iS [ A α ] O ( A α ) δ ( G ( A α )) = det 1 e 2 Z [ D α D A ] e iS [ A ] O ( A ) δ ( G ( A )) In the second line we changed integration variables from A to A α , in the third we used the invariance of the action and of O under gauge transformations, and in the fourth we simply changed the name of the dummy variable A α to A . Thus we’ve shown that Z [ D A ] e iS [ A ] O ( A ) = det 1 e 2 Z [ D α ] Z [ D A ] e iS [ A ] O ( A ) δ ( μ A μ - ω )
Phys 253a 8 Since everything is independent ofα, the path integral overαjust produces aninfinite constant, which will drop out of the normalized correlation functions.The determinant will also drop out.Let’s pause for a second to consider what we’ve done. We’ve essentially changedvariables in the functional integral into ones that integrate over gauge inequiva-lent configurations, and one,α, which integrates over the gauge slices. We couldplug inω= 0, have the expression for Lorentz gauge, and call it a day. However,this isn’t a very useful form of the functional integral. In particular, it isn’t onewe know how to do perturbation theory with; we have a delta function sittingnext to our exponentiated action. Luckily, there’s a nice trick we can use to getthis into a more useful form. Since the above equality holds for any functionω,it will also hold for a weighted sum of differentω’s. Thus, we have in particularZ[DA]eiS[A]O(A)=N(ξ)Z[Dω]e-iRd4xω22ξdet1e2Z[Dα]Z[DA]eiS[A]O(A)δ(μAμ-ω)=N(ξ)det1e2Z[Dα]Z[DA]eiS[A]e-iRd4x(∂μAμ)22ξO(A)whereN(ξ) is a normalization factor and delta function was used to get rid oftheωintegral. The correlation functions are then given byh0|T{O(A)}|0i=R[DA]eiS[A]e-iRd4x(∂μAμ)22ξO(A)R[DA]eiS[A]e-iRd4x(∂μAμ)22ξThus gauge fixing the path integral was equivalent to inserting a term-(μAμ)22ξinto the Lagrangian. Formally, since we are always normalizing our path integrals,the gauge fixed and un-fixed path integrals are equal, so we can say that addingthe term-(μAμ)22ξto the Lagrangian didn’t change our correlation functions.Now we’re asked about the effect of adding either a term (μAμ)4or a termAμAμto the Lagrangian. We see from the above that the first would just amount tochoosing a different weighting function for the integral overω’s.Since we arefree to choose any weighting function we like, everything would follow throughthe same way if we had chosenω42ξinstead, and that would effectively add a term(μAμ)4to our Lagrangian instead of (μAμ)2.
Phys 253a 9

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