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Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

The Dirac field has the global symmetry $\psi \mapsto e^{i\alpha }\psi$, $\bar{\psi }\mapsto e^{-i\alpha }\bar{\psi }$, while the real scalar is unchanged. Both the free Dirac action and the pseudoscalar Yukawa interaction preserve this symmetry. Therefore Dirac fermion number conservation holds: its charge counts particles minus antiparticles.

The initial state has charge $+1$, whereas a final antiparticle and a neutral scalar have charge $-1$. Since the S-matrix commutes with that charge,This holds at every order, not just tree level. In the Feynman rules, a continuous fermion arrow cannot connect these specified external states. Replacing only the outgoing $\bar{u}$ by a $v$ in the previous expression would not give a physical amplitude. Changing both external fermions to antiparticles would instead produce an allowed process with reversed fermion flow and the appropriate $v$ spinors; moving a leg between initial and final states is a different operation governed by crossing symmetry.