Consider the $\phi^4$ theory. The two divergent Feynman diagrams, namely the two point function and the 4 point function have been isolated and by putting a cut off on their momentum integrations, their divergent behavior has been isolated at one loop. Consider the two point function. At one loop, its diverges as: \begin{equation} \frac{-i\lambda}{32{\pi}^2}[{\Lambda}^2-M^2\log{\frac{{\Lambda}^2+M^2}{M^2}}+O({\Lambda}^{-2})] \end{equation} Here $M$ is the mass of the boson and $\Lambda$ is the momentum cutoff. This function is clearly divergent as $\Lambda$ goes to infinity. To remedy this a counter term is added to the mass term in the Lagrangian which has the exact singularity structure of the two point function at one loop.

Now, I have two sources of confusion. Firstly, let's assume that the entire calculation for the two point function is carried out again at one loop. All the $M$'s in the divergent two point function will be replaced by the new mass, $M_{renormalized}$. Hence, at the end of the calculation, the only difference will be that the $M$ in the above expression will be changed to $M_{renormalized}$. How will you show that it's finite as Λ goes to infinity?

Secondly if you assume that the theory now has another free propagator with mass $M-M_{renormalized}$, how will you prove that the contribution of the additional free propagator will cancel out the divergence in the two point function?

TL;DR: Is it possible to make the scalar self energy of phi 4 theory finite at one loop by adding a constant (possibly infinite) to the mass term of the Lagrangian?


1 Answer 1


You say that "All the $M$'s in the two-point function will be replaced by $M_\text{ren}$" but that's too fast. By definition, the renormalized mass is something like $$ M_\text{ren}^2 \equiv \lim_{p^2 \to 0} G(p)^{-1} $$ where $G(p)$ is the two-point function in momentum space, and as such it's a function of the bare couplings $M$ and $\lambda$. Next, you're supposed to invert this relationship order by order in $\lambda$, i.e. you write the bare couplings as a function of the renormalized mass and the renormalized coupling $\lambda_\text{ren}$. This is not the same as simply relabeling $M \mapsto M_\text{ren}$! Almost by construction, your physical observables will now have a finite limit as $\Lambda \to \infty$.


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