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PS12-solutions - Phys 253a 1 Problem Set 12 Solutions...

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Phys 253a 1 Problem Set 12 Solutions December 4, 2010 1. (a) The kinetic term for φ is ( D μ φ ) * D μ φ = ( μ - ieA μ ) φ * ( μ + ieA μ ) φ If we switch φ and φ * , then clearly we should take A μ → - A μ to leave the Lagrangian invariant. (b) We have an expression for the correlation function in terms of a path integral h 0 | T { A μ 1 ( q 1 ) . . . A μ n ( q n ) }| 0 i = 1 Z Z D A D φ D φ * n Y j =1 A μ j ( q j ) e iS = 1 Z Z D A D φ D φ * n Y j =1 Z d 4 x j A μ j ( x j ) e - ix j · q j e iS where Z = R D A D φ D φ * e iS (and we could take the fourier transforms out of the path integral if we like). To prove Furry’s theorem, consider the change of variables φ 0 = φ * , φ *0 = φ , and A 0 = - A . The action is invariant S 0 = S by construction. The measure is also invariant D [ - A ] D φ * D φ = D A D φ D φ * . 1 Thus, we get Z D A D φ D φ * n Y j =1 A μ j ( q j ) e iS = Z D [ - A ] D φ * D φ n Y j =1 - A μ j ( q j ) e iS 0 = Z D A D φ D φ * n Y j =1 - A μ j ( q j ) e iS = ( - 1) n Z D A D φ D φ * n Y j =1 A μ 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 lattice-regularized path integral is finite-dimensional). The measure is clearly invariant under the given transformation if the integral is finite-dimensional (say if we integrate over only a finite number of modes of A , φ , and φ * ), so the path integral is invariant too. There are problems with this argument if the regulator itself somehow isn’t invariant under the given transformation. We’ll learn about this phenomenon (called an anomaly) next semester.
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Phys 253a 2 So the correlation function vanishes if n is odd. (c) The correlation function is well defined if the q j are off-shell, so Furry’s theorem holds then too. 2 This is useful, since off-shell correlators appear as internal parts of Feynman diagrams. (d) The proof works essentially the same as before; we use charge conjugation in- variance of the action and measure. The only difference now is that charge con- jugation is slightly more complicated. We use the transformation ψ 0 ψ * , A μ → - A μ , where C is defined by C - 1 γ μ C = - γ μT .
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