 Solution

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.