Feynman Path Integral Formulation

6.18 Lattice Invariance versus Continuum Invariance 221Furthermore it will be assumed that the functional measure [dA] is also locally gaugeinvariant, so that one has[dA Ω ]=[dA] S 0 (A Ω )=S 0 (A) . (6.207)Here and in the following a lattice regularization will be implicit, in order to makevarious functional manipulations well defined. In addition, the fact that the gaugetheory is compact will play a crucial role, as the proof generally does not go throughif the gauge variables are not compact. The compactness of the gauge group impliesfor the measure over the gauge parameters∫[dΩ] =finite , (6.208)and furthermore by invariance [dΩ]=[d(ΣΩ)] with Σ an arbitrary group element.Thus both [dA] and [dΩ] will be assumed to be the Haar measure over the group.Under a gauge transformation for the fields one hasS(A Ω −1) =S 0 (A)+δS(A Ω −1) , (6.209)since the first term is assumed to be invariant. As a consequence only the term δS(A)causes gauge breaking in Z.Using perturbation theory in δS(A) one can compute the vacuum expectationvalue of the gauge invariant quantity F(A)∫∫〈F(A)〉 = [dA] e −S(A) F(A) / [dA] e −S(A) . (6.210)For small δS(A) it is given by the expansion∫〈F(A)〉 = 〈F(A)〉 0 − dx 〈δL (x)F(A)〉 0+ 1 2∫dxdx ′ 〈δL (x)δL (x ′ )F(A)〉 c 0 + ... (6.211)Gauge invariance of the lowest order action S 0 (A) and measure [dA], usedintheaverage 〈...〉 0 , implies that the first correction on the r.h.s. vanishes. The secondorder correction only gives a contribution when x = x ′ , which is given by∫12dx〈 [δL (x)] 2 F(A) 〉 c 0 = 1 2∫dx〈 δI (x)F(A) 〉 c 0 , (6.212)where δI (x) is some gauge invariant operator. One concludes that a small gaugebreaking term leads to a correction which can be described by the insertion of a suitablegauge-invariant operator. Such effects could then be re-absorbed into a suitableredefinition of the bare coupling constants of the theory.