JosephButler
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Hello, I understand that the non-zero (or non-small) rate for \pi^0 \rightarrow \gamma\gamma was historically a big motivation for the non-conservation of the axial current. I've been trying to work on problem IV.7.2 (p. 252) in Zee which asks to show that this amplitude vanishes if \partial_\mu J_5^\mu = 0[\tex] and m_\pi = 0. He suggests following the argument he used in a previous section where he motivated the pion as a goldstone boson (sec IV.2), leading up to the Goldberger-Treiman relation. <br />
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I understand heuristically what he's asking: show that the rate for \pi^0 \rightarrow \gamma\gamma is much larger than what would be expected without the chiral anomaly. However, I don't quite understand the limiting case that he's asking us to confirm in the problem. In the case m_\pi = 0, the decay is impossible kinematically. Peskin (ch 19.3, p. 675-676) does a similar thing where he takes the limit of the pion mass to be zero and then fills in factors of m_\pi in the kinematics. But Peskin doesn't assume that the axial current is conserved and fixes terms based on the existence of the anomaly.<br />
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So what I'm confused about is how to approach the problem in the 1950's point of view, the way that Zee wants. I want to assume the axial current is conserved and that the pion is a goldstone boson (massless), and I want to show that the amplitude for pion decay into photons vanishes. Is it necessary to assume that the pion has a small mass and then go to the massless limit after deriving a result? At any rate, the pion having a mass explicitly violates \partial_\mu J^\mu_5 = 0 since the amplitude is proportional to: (by Lorentz invariance)<br />
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\langle 0| J^\mu_5 | \pi(k) \rangle = fk^\mu<br />
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(which defines the constant f), and hence<br />
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\langle 0| \partial_\mu J_5^\mu | \pi(k) \rangle = f m^2_\pi.<br />
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Thus a conserved current (\partial_\mu J^\mu_5 = 0) means the pion has to be massless. <br />
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I'm just not really sure what series of steps Zee wants us to take.<br />
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Any tips would be greatly appreciated!<br />
Cheers,<br />
JB