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Unformatted text preview: Phys 253a 1 Problem Set 12 Solutions December 4, 2010 1. (a) The kinetic term for is ( D ) * D = (  ieA ) * ( + ieA ) If we switch and * , then clearly we should take A  A to leave the Lagrangian invariant. (b) We have an expression for the correlation function in terms of a path integral h  T { A 1 ( q 1 ) ...A n ( q n ) } i = 1 Z Z D A D D * n Y j =1 A j ( q j ) e iS = 1 Z Z D A D D * n Y j =1 Z d 4 x j A j ( x j ) e ix j q j e iS where Z = R D A D D * e iS (and we could take the fourier transforms out of the path integral if we like). To prove Furrys theorem, consider the change of variables = * , *0 = , and A = A . The action is invariant S = S by construction. The measure is also invariant D [ A ] D * D = D A D D * . 1 Thus, we get Z D A D D * n Y j =1 A j ( q j ) e iS = Z D [ A ] D * D n Y j =1 A j ( q j ) e iS = Z D A D D * n Y j =1 A j ( q j ) e iS = ( 1) n Z D A D D * n Y j =1 A j ( q j ) e iS 1 This is true because path integrals are defined using a regulator, and a regulated path integral behaves like a finite dimensional integral (for instance, a latticeregularized path integral is finitedimensional). The measure is clearly invariant under the given transformation if the integral is finitedimensional (say if we integrate over only a finite number of modes of A , , and * ), so the path integral is invariant too. There are problems with this argument if the regulator itself somehow isnt invariant under the given transformation. Well learn about this phenomenon (called an anomaly) next semester. Phys 253a 2 So the correlation function vanishes if n is odd. (c) The correlation function is well defined if the q j are offshell, so Furrys theorem holds then too. 2 This is useful, since offshell correlators appear as internal parts of Feynman diagrams....
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This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.
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