When you have higher order correlation functions you get contact terms for each

When you have higher order correlation functions you

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When you have higher order correlation functions, you get contact terms for each additional field. For m = 0 , square ( 0 | T { φ ( x ) φ ( y 1 ) φ ( y n ) }| 0 ) = ( 0 | T { square φ ( x ) φ ( y 1 ) φ ( y n ) }| 0 ) (13.138) i summationdisplay j δ (4) ( x y j ) ( 0 | T { φ ( y 1 ) φ ( y j 1 ) φ ( y j +1 ) φ ( y n ) }| 0 ) (13.139) Since the correlation functions define the theory, these Schwinger-Dyson equations allow us to show that two theories are equal by matching their correlation functions. For example, go back to the canonically quantized theory and define Z ˆ [ J ] = ( 0 | T { e i integraltext φJ }| 0 ) (13.140) This is not a functional integral, just a generating functional for correlation. Expanding out the exponen- tial, it is just a set of correlation functions of the second quantized φ and an external current J , which we already know hot to calculate. But it has the same properties as the functional integral ( 0 | T { φ ( x 1 ) φ ( x n ) }| 0 ) = ( i ) n d n Z ˆ dJ ( x 1 ) dJ ( x n ) | J =0 (13.141) 13.12 Schwinger-Dyson equations 159
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We would like to show that Z = Z ˆ , which will then provide another proof of the path integral. To do that, take the SD equations and multiply by J ’s: square ( 0 | T { φ ( x ) φ ( y 1 ) φ ( y n ) J ( y 1 ) J ( y n ) }) = ( 0 | T { square φ ( x ) φ ( y 1 ) φ ( y n ) J ( y 1 ) J ( y n ) }) (13.142) i summationdisplay j δ (4) ( x y j ) ( 0 | T { φ ( y 1 ) φ ( y j 1 ) φ ( y j +1 ) φ ( y n ) J ( y 1 ) J ( y n ) }) (13.143) The first term comes from the expansion of square ∂Z ˆ [ J ] ∂J ( x ) . For the second term, note that these are the full inter- acting φ s which evolve with H , or equivalently satisfy square φ = V ( φ ) by the Heisenberg equations of motion. So this term is part of the expansion of V parenleftBig ∂J ( x ) parenrightBig Z ˆ [ J ] . Integrating the last term over the δ -function gives ( 0 | T { φ ( y 1 ) φ ( y j 1 ) φ ( y j +1 ) φ ( y n ) J ( y 1 ) J ( y j 1 ) J ( x ) J ( y j +1 ) J ( y n ) }) (13.144) This comes from the expansion of J ( x ) Z [ J ] . So if we sum up the whole Taylor series, we get square ∂Z ˆ [ J ] ∂J ( x ) = bracketleftbigg V parenleftbigg ∂J ( x ) parenrightbigg + J ( x ) bracketrightbigg Z ˆ [ J ] (13.145) This is a concise form of the Schwinger-Dyson equations. More importantly, it is a differential equation for Z ˆ [ J ] which no longer makes reference to the Hilbert space. We showed this equation holds for Z ˆ [ J ] . You can show in the same way that it holds for the path integral Z [ J ] . Therefore, with the same physical boundary conditions, the two are equivalent. This is another way to derive the path integral. 13.13 BRST invariance There is a beautiful symmetry called BRST invariance, which is a residual exact symmetry of the Lagrangian even after gauge fixing. It is particularly useful for studying more complicated gauge theories, but it is a little easier to understand in the QED case. Notice that the gauge fixing term ( μ A μ ) 2 is invariant under gauge transformations A μ A μ + μ α for square α = 0 . So this is a residual symmetry of the entire Lagrangian. As we have already seen, it is awkward to deal with constrained theories, and constraints on symmetries are no different. So consider adding to the Lagrangian two new free fields c and d L = L [ A,φ i ] + 1 ξ ( μ A μ ) 2 d square c (13.146) So the path integral becomes Z O =
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