Mass renormalization counterterm in Wilsonian effective action diagram

Question

this is my first post on stack exchange so please pardon me any mistakes or bad format. My question is about counterterms and the Wilson effective action. We started in Euclidean spacetime from $$\int \mathcal{D}\phi \exp \left(- \int d^dx~ \frac{1}{2} (\partial_\mu \phi)^2 + \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{4!} \phi^4 \right) $$

and then separated our scalar field $\phi$ in a "fast" and "slow" field $\phi = \Phi + \varphi $, where $\Phi$ is slow and $\varphi$ is the fast field, with momenta in the scales $$ \Phi: 0<k<\mu $$ $$ \varphi: \mu<k<M.$$

The Wilson effective action was defined as $$ \exp(-S_{eff}[\Phi]) = \int [\mathcal{D}\varphi]_\mu^M \exp(-S[\Phi + \varphi]) $$

and after some calculation $$ \exp(-S_{eff}[\Phi]) = \exp(S[\phi]) \int [\mathcal{D}\varphi]_\mu^M \exp \left( - \int d^dx ~\frac{1}{2}(\partial_\mu \varphi)^2 + \frac{1}{2} m^2 \varphi^2 + \frac{\lambda}{4!} \varphi^4 + \frac{\lambda}{4} \Phi^2 \varphi^2 + \frac{\lambda}{6} \Phi \varphi^3\right) $$

The ansatz was to do perturbation theory in both $m$ and $\lambda$:

$$ S_{int}[\Phi,\varphi] = \int d^dx \left( \frac{1}{2} m^2 \varphi^2 + \frac{\lambda}{4!} \varphi^4 + \frac{\lambda}{4} \Phi^2 \varphi^2 + \frac{\lambda}{6} \Phi \varphi^3\right) $$

To first order in $\lambda$ one gets the following Feynman diagram which is chosen to give the mass renormalisation counterterm $\delta m^2$

via $ \langle \Omega| T \left( -\int d^dx ~\frac{\lambda}{4} \Phi^2(x) \varphi^2(x) \right) |\Omega \rangle = -\frac{1}{2} \int d^dx ~\Phi^2(x)\delta m^2 $.

I do not understand why this Feynman diagram was chosen to give the mass renormalization, since as I get it, we want to get the mass ren. for the $\varphi$ field. So why do we take the diagram resulting from the $\Phi^2\varphi^2$ interaction, why does the slow field play any role?

EDIT: Sorry, I forgot to clarify which one is the fast and slow field in the diagram. The dashed line represents the fast field $\varphi$, the full line the slow one $\Phi$.

Answer 1

Let me give you some empirical rules about Feynman diagrams, and then a proper reference. First, the renormalization of the mass is happening for the mass terms, which is you case, is only the $\sim m^2\phi^2$ term in your action. Second, in the Feynman diagram context, a $\sim\phi^n$ term is renormalized via diagrams with $n$ external legs. In your case, for the mass term it is then two legs, the first diagram of that kind (in presence of quartic interactions) is the « tadpole », a common name for the diagram you drawn. More generally, for a term like $\sim\alpha\phi_1\phi_2…\phi_n$, $\alpha$ will be renormalized via diagrams with $n$ external legs of a kind $1,2,…,n$. Third, this diagram is not « chosen », it is a direct result of perturbation theory applied to your problem, interpreted in a graphical form. I recommend you the Kleinert monograph about phi4 theory for a detailed derivation of the diagrammatic in quartic theories Kleinert.

Note that, for the action you wrote, you have to consider two different tadpoles at the first order. One with a $\phi$ loop, from the $\phi^4$ interaction term, and one with the $\Phi$ loop (the one you drawn), from the $\phi^2\Phi^2$ interaction term. The interaction term $\Phi\phi^3$ do not give a tadpole since you cannot « close the loop » if that make sense.