I learned recently how to compute the basic generating functionals. My next goal is to understand how I can use them to compute effective interactions. I started with the Yukawa interaction because it is quadratic and, a priori, exactly solvable. On the other hand, the examples in books and lecture notes are in the great majority based on non-exactly solvable theories, like $\phi ^4$ theory ( e.g. https://arxiv.org/abs/2006.16285).

As Yukawa is quadratic I am performing the functional integration over the bosonic field and then obtaining something I imagine to be the effective potential. On the other hand, it is quite common to see books dealing with this kind of problem with the steepest descent approximation (I did not study this topic yet). I am afraid to be forgetting something important by just doing the integration.

Here is what I did. The Lagrangian of the full theory is given by

$$\mathcal{L} = \mathcal{L}_{\psi} + \frac{1}{2} (\partial_\mu \phi)(\partial^\mu \phi) - \frac{1}{2} m^2 \phi^2 + g \bar{\psi} \phi \psi , $$

where $\mathcal{L}_{\psi}$ is the Dirac Lagrangian and $\phi$ is the scalar field. Then, the generating functional $Z$ of the full theory is given by

$Z = \int \mathcal{D} \bar{\psi}\mathcal{D}\psi \mathcal{D}\phi \exp \Big[i \Big(\int d^4x \mathcal{L}_{\psi} - \int d^4 x \frac{1}{2} \phi (\square + m^2) \phi + g \bar{\psi} \phi \psi \Big) \Big] = \\ = \int \mathcal{D} \bar{\psi}\mathcal{D}\psi \exp \Big[i \int d^4x \mathcal{L}_{\psi} \Big] \int \mathcal{D}\phi \exp[\frac{i}{2} \int d^4x \phi (\square + m^2) \phi + 2g \bar{\psi} \psi \phi] $.

In the book of Lancaster & Blundell (Quantum Field Theory for the Gifted Amateur) we find the solution for a free scalar, which is

$Z_{scalar} = C \exp \Big[ - \frac{1}{2} \int d^4x d^4y J(x) [i A^{-1}(x,y)J(y)] \Big]$,

where $J$ are "sources", $A^{-1}(x,y)$ is the "inverse" of $- (\square + m^2)$ and $C$ is a (divergent) constant. If I consider $J(x) \equiv \bar{\psi}(x)\psi(x)$ in the Yukawa potential, I obtain

$Z = C \int \mathcal{D} \bar{\psi}\mathcal{D}\psi \exp \Big[i \Big(\int d^4x \mathcal{L}_{\psi} (x) + \int d^4x d^4 y \bar{\psi}(x)\psi(x) A^{-1}(x,y) \bar{\psi}(y)\psi(y) \big]$

and I am interpreting $\int d^4y \bar{\psi}(x)\psi(x) A^{-1}(x,y) \bar{\psi}(y)\psi(y)$ as an effective interaction between two fermions. Is it right? Why? If it is right, can I always do it for any quadratic Lagrangian? For instance, in the electromagnetic coupling?

  1. Yes, in the path integral one can in principle integrate out a field that only appears quadratically in the action via a Gaussian integration. However, the reduced theory is non-local, and there might be an issue with zero-modes.

  2. As an example of integrating out the EM field, see e.g this Phys.SE post.


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