Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chiral anomaly, pion to photon decay

  1. Jan 28, 2008 #1
    Hello, I understand that the non-zero (or non-small) rate for [tex]\pi^0 \rightarrow \gamma\gamma[/tex] 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 [tex]\partial_\mu J_5^\mu = 0[\tex] and [tex]m_\pi = 0[/tex]. 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.

    I understand heuristically what he's asking: show that the rate for [tex]\pi^0 \rightarrow \gamma\gamma[/tex] 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 [tex]m_\pi = 0[/tex], 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 [tex]m_\pi[/tex] in the kinematics. But Peskin doesn't assume that the axial current is conserved and fixes terms based on the existence of the anomaly.

    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 [tex]\partial_\mu J^\mu_5 = 0[/tex] since the amplitude is proportional to: (by Lorentz invariance)

    [tex]\langle 0| J^\mu_5 | \pi(k) \rangle = fk^\mu[/tex]

    (which defines the constant [tex]f[/tex]), and hence

    [tex]\langle 0| \partial_\mu J_5^\mu | \pi(k) \rangle = f m^2_\pi[/tex].

    Thus a conserved current ([tex]\partial_\mu J^\mu_5 = 0[/tex]) means the pion has to be massless.

    I'm just not really sure what series of steps Zee wants us to take.

    Any tips would be greatly appreciated!
  2. jcsd
  3. Jan 29, 2008 #2
    I can say that although kinematically a massless pion decaying to two massless photons is impossible, the amplitude for the process doesn't necessarily forbid it.

    Recall, that the transition operator is factored into a 4-momentum conserving delta function and the amplitude:

    [tex]iT=(2\pi)^4\delta^{(4)}(p_1+p_2-k_1-k_2)\,i\mathcal{M}(p_1\,p_2\rightarrow k_1\,k_2) [/tex]​

    So, [itex]\mathcal{M}(p_1\,p_2\rightarrow k_1\,k_2)[/itex] may not be zero, but the kinematics is partly taken care of by the momentum conserving delta function.
    Last edited: Jan 29, 2008
  4. Jan 30, 2008 #3


    User Avatar
    Science Advisor

    The result you want is known as the Sutherland-Veltman theorem. I believe the approach is to work with a massive pion, and show that the amplitude M has a factor of m_pi^2, so that M vanishes in the massless limit.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Chiral anomaly, pion to photon decay
  1. Pion Decay (Replies: 4)

  2. Pion Decay (Replies: 1)

  3. Decay of pion (Replies: 8)

  4. Pion decays (Replies: 2)

  5. Pion decay (Replies: 5)