Feynman rules after a Hubbard-Stratanovich transformation, i.e. for a field with no kinetic term

Question

I am trying to calculate the beta function of the 2D Gross-Neveu model after performing a Hubbard-Stratanovich transformation. Of course, you can calculate it without this transformation, but I am doing this as a warm-up for something else.

My initial Lagrangian is

\begin{equation} \mathcal{L}=\bar{\psi}_i(i\partial\kern-0.3em\raise0.1ex\hbox{/}-m)\psi_i+\frac{1}{2}g^2(\bar{\psi}_i\psi_i)^2 \end{equation}

and I integrate in a new field $\phi$ to get

\begin{equation} \mathcal{L}_{\phi}=\bar{\psi}_i(i\partial\kern-0.3em\raise0.1ex\hbox{/}-m)\psi_i+\frac{1}{2}\phi^2+ig\phi\bar{\psi}_i\psi_i \end{equation}

I am confused about the Feynman rules for $\phi$ since there is no kinetic term. I see in a comment here that perhaps there are no internal lines at all for $\phi$. But then there are no loop diagrams to calculate and I don't know how to do the usual procedure to find counterterms.

Even if there actually are internal lines for $\phi$, I don't see how they could have momentum dependence. My naive guess is that the propagator would just be a constant like $i$ in Minkowski signature.

So, how do the Feynman diagrams and rules work out for this scalar with no kinetic term?

Answer 1

The quickest route to the Feynman rules for weird situations like this is almost always via the path integral. If $\langle \cdot \rangle_0$ stands for expectation values with respect to just the quadratic part of the Lagrangian, then the propagator for $\phi$ is found immediately from the two-point function $\langle \phi(x) \phi(y) \rangle_0$, which is ultra-local: $$ \langle \phi(x) \phi(y) \rangle_0 = \frac{1}{Z_0} \int D\phi \ \phi(x) \phi(y) \ e^{\frac{i}{2} \int \phi^2} = i \delta^d(x-y) = \int \frac{d^d p}{(2\pi)^d} i e^{i p (x-y)} $$ In the last part equality I wrote the two-point funciton in momentum space, to emphasize that you can still derive momentum-space Feynman rules with this propagator as well. Represent the propagator with a dotted line carrying momentum $k$, and assign the value $G(k) = i$ to each such propagator. It's a useful exercise to check that this propagator and the $\phi \bar{\psi}_i \psi_i$ vertex gives the same perturbative expansion for the fermionic correlation functions as the original Lagrangian; as a hint, note that you will need to sum $s$, $t$, and $u$-channel diagrams (with appropriate symmetry factors) in the $\phi \bar{\psi}_i \psi_i$ theory to recover the bare vertex of the $(\bar{\psi}_i \psi_i)^2$ theory.

Answer 2

A partial answer to my own concerns, now that Zack clarified that the propagator is indeed momentum independent:

The $\phi$ line still carries a momentum when you draw the diagrams even though the propagator ends up being momentum independent. So there still is an integral to be done over the loop momentum even though it does not show up in that particular term of the integral. Thus the loop calculations to find the beta function can proceed as usual.