Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm taking a particle physics course and we're using Perkins Introduction to High Energy Physics as the text. I am looking at problem 1.7. It asks whether $$\pi^0\rightarrow e^- + e^+$$ is allowed or forbidden by the Standard Model based off of conservation laws. I feel it doesn't violate any of them. Energy and momentum can clearly be conserved, charge is conserved, lepton number is conserved, etc. However, I know Dalitz decay is a very similar process, but that the pion decays into $$\pi^0 \rightarrow e^- + e^+ + \gamma$$ so feel as though I must have overlooked something. I would be greatly appreciative if someone could point me in the right direction.

share|cite|improve this question
Consider parity and angular momentum, too. – dmckee Apr 7 '14 at 3:25
Thank you! I'll look at it a bit more and let you know if I come up with anything. – user33525 Apr 7 '14 at 4:00
up vote 3 down vote accepted

Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light quark). Therefore, this process is allowed in the SM, but highly suppressed.

On the other hand, if you have a photon in the final state, you can have an eletromagnetic decay at tree-level, as shown in the third diagram of the same figure.

The difference between these two processes can easily be seen from the experimental measurements, because the branching ratio is of order $10^{-8}$ for the first decay, while it is of order $10^{-2}$ for the Dalitz decay (c.f. PDG).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.