I'm trying to understand how to decide if a minus sign is to be put in front of a vertex of interaction when dealing with Feynman diagrams at lowest order. At first I thought that you just take the interaction term in the lagrangian and multiply it by $-i$, but searching online I'm finding contradictory results. For example, from my professor's notes I've read that the interaction vertex between a fermion-antifermion pair and a $A$, $W$, $Z$ or $H$ boson are respectively $$\begin{array} -ie Q_f \gamma^\mu & \text{ for the $A$ boson}\\ \frac{-ig_w}{2\sqrt{2}}\gamma^\mu(1-\gamma^5) & \text{ for the $W$ boson}\\ \frac{-ig_w}{2\cos{\theta_w}}\gamma^\mu(V_f-A_f\gamma^5) & \text{ for the $Z$ boson}\\ \frac{-ig_wm_f}{2M_W} & \text{ for the $H$ (Higgs) boson} \end{array}$$ where $Q_f, A_f, V_f$ come from the charges of the fermion in the various symmetry groups and $m_f$ is the fermion's mass.

But this seems wrong. In the Lagrangian, the charged and neutral currents, in which the $A, W, Z$ bosons interactions with fermions appear, have an opposite sign with respect to the interaction term between the Higgs boson and the fermions. Overall, this is not a problem. But I'm trying to solve the scattering $\mu \mu \rightarrow HH$ at the lowest order, which involves the interaction of the Higgs boson with the muons and the triple self interaction $HHH$ of the boson with itself. The problem is that I don't know what is the relative sign between these two diagrams, and they must be summed in order to calculate the amplitude. Since both the $HHH$ term (proportional to $\lambda v$) and the $f\bar{f}H$ term (proportional to $\frac{m_f}{v}$) have a minus sign in the Lagrangian, I'd say their vertex interaction term should be $i\lambda v$ and $i\frac{m_f}{v}$, but this contraddicts the notes above.

  • $\begingroup$ Hi ultrapoci. Welcome to Phys.SE. Can you elaborate on your logic behind multiplying with $-i$? $\endgroup$
    – Qmechanic
    Nov 26 '21 at 11:58
  • $\begingroup$ @Qmechanic Interaction terms come from the perturbative expansion of the S-matrix, which is equal to the exponent of $-i$ times the time-ordered integral on the interacting Hamiltonian. Admittedly, this means that the Lagrangian's interacting terms at first order should be multiplied by $i$, not $-i$, because the interacting Lagrangian is minus the interacting Hamiltonian. But I've noticed that my professors multiplies by $-i$ and I stuck to it, since in the squared amplitude the overall sign doesn't matter. $\endgroup$
    – ultrapoci
    Nov 26 '21 at 12:51

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