3
$\begingroup$

I'm trying to learn the spinor-helicity formalism from Schwartz's QFT book.

His equation 27.44 is describes the annihilation of an electron(1)-positron(2) pair to a muon(3)-antimuon(4) pair. He writes,

$$iM(1^-2^+3^-4^+)=(-ie)^2 \langle 2\gamma^{\mu}1]\frac{-i g_{\mu\nu}}{s}\langle 3 \gamma^{\nu} 4] = 2 \frac{ie^2}{s} [41] \langle 23 \rangle \tag{27.44}.$$

On the RHS, we have $1]$, $4]$ and $2\rangle$, $3\rangle$ which correspond to right-handed particles 1 and 4, left-handed particles 2 and 3.

However on the LHS, the labels 1 and 2 appear to label them oppositely. Furthermore, considering the symbol on the LHS is defined for all incoming momenta, then we seem to have the opposite helicity label for 3 and 4 also.

If anyone could explain, I'd appreciate that thanks.

$\endgroup$
  • $\begingroup$ Why do you feel like the RHS labels the particles inversely? $\endgroup$ – Prahar Mar 20 '16 at 22:27
  • $\begingroup$ Well a $1]$ for instance means a right-handed electron. But on the left it is labelled $1^-$ which means negative helicity. I thought right-handed meant the same as positive helicity. $\endgroup$ – Kris Mar 20 '16 at 22:39
0
$\begingroup$

There is no problem here. You may refer to (27.42) in Schwartz's book, which is usually called Fierz identity. Or by direct matrix calculation you can get the same result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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