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

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Phys 253a1Problem Set 12 SolutionsDecember 4, 20101.(a) The kinetic term forφis(Dμφ)*Dμφ=(μ-ieAμ)φ*(μ+ieAμ)φIf we switchφandφ*, then clearly we should takeAμ→ -Aμto leave theLagrangian invariant.(b) We have an expression for the correlation function in terms of a path integralh0|T{Aμ1(q1). . . Aμn(qn)}|0i=1ZZDADφDφ*nYj=1Aμj(qj)eiS=1ZZDADφDφ*nYj=1Zd4xjAμj(xj)e-ixj·qjeiSwhereZ=RDADφDφ*eiS(and we could take the fourier transforms out ofthe path integral if we like). To prove Furry’s theorem, consider the change ofvariablesφ0=φ*, φ*0=φ, andA0=-A.The action is invariantS0=Sbyconstruction. The measure is also invariantD[-A]Dφ*Dφ=DADφDφ*.1Thus, we getZDADφDφ*nYj=1Aμj(qj)eiS=ZD[-A]Dφ*DφnYj=1-Aμj(qj)eiS0=ZDADφDφ*nYj=1-Aμj(qj)eiS=(-1)nZDADφDφ*nYj=1Aμj(qj)eiS1This is true because path integrals are defined using a regulator, and a regulated path integral behaveslike a finite dimensional integral (for instance, a lattice-regularized path integralisfinite-dimensional). Themeasure is clearly invariant under the given transformation if the integral is finite-dimensional (say if weintegrate over only a finite number of modes ofA,φ, andφ*), so the path integral is invariant too. There areproblems 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.
Phys 253a2So the correlation function vanishes ifnis odd.(c) The correlation function is well defined if theqjare off-shell, so Furry’s theoremholds then too.2This is useful, since off-shell correlators appear as internal partsof 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 transformationψ0ψ*,Aμ→ -Aμ, whereCis defined byC-1γμC=-γμT.
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