2
$\begingroup$

The main Feynman diagram I've seen corresponding to this decay is the following:

enter image description here

What I don't understand is why is it not possible for the up-antiup quarks that form the pion to annihilate into a single virtual photon, and then have that virtual photon decay to a positron-electron pair, similar to the annihilation of electron-positron pairs into a muon-antimuon pair in the following fashion?

enter image description here

EDIT: My question was closed and flagged as a duplicate of a question that did not answer what I asked so I repost it. I repeat, it is NOT a duplicate, I am not asking why the loop contribution is suppressed, I am asking why a different Feynman diagram is not acceptable.

$\endgroup$
2

1 Answer 1

3
$\begingroup$

Charge conjugation, C, contrasts even and odd numbers of photons in states and amplitudes.

The $𝜋^0$ has C=+, but one photon has C=—, and hence two photons +.

By contrast, the $\rho^0$, with C=—, can and does couple to one photon, the heart of the Vector Dominance Model.

Note the Z has no well-defined C, as the weak interactions break C. Such a coupling to one Z exists, as the corresponding axial current of PCAC couples to the Z in its inimitable cockeyed way...

All charged fermions couple to an indefinite number of photons, in principle, of course. But the overall amplitude must preserve the C of the incoming and outgoing states, in QED which preserves C (unlike the weak interactions). The daughter $e^+e^-$ of the triangle diagram above are in an even C state, +; unlike the $e^+e^-$ that annihilate to one virtual photon, and then resolve to $\mu^+\mu^-$ in your latter diagram.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.