Consider QED with an additional charged (complex) scalar field, $\phi$:$$\require{cancel} \mathcal{L} = -{1\over4} F^{\mu\nu} F_{\mu\nu} + (D^\mu \phi)^*(D_\mu \phi) - \mu^2 \phi^* \phi - \overline{\psi}(i\cancel{D} - m) \psi.$$I know what the Feynman rules are for this theory and can derive them. My question is, though, are there any scalar-fermion vertices in this theory?

  • 6
    $\begingroup$ In your Lagrangian there is no term that couples $\psi$ and $\phi$ directly, so no, there are no scalar-fermion vertex in the Feynman rule. You can effectively generate coupling between $\psi$ and $\phi$ by an intermediate photon, but that is not a fundamental Feynman rule, and the coupling is different from something like $\bar{\psi}\psi\phi$. $\endgroup$ – Meng Cheng Dec 1 '15 at 3:10

Your Answer

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

Browse other questions tagged or ask your own question.