# Two-loop RG calculations using momentum shell

I am curious about how to calculate a two-loop integral in a momentum shell, as done in Wilsonian RG calculations. The simplest case would be a "sunset diagram" in (Euclidean) $$\varphi^4$$-theory, which would give a contribution of the form $$\int_{\Lambda(1 - \delta \ell)}^{\Lambda} \frac{\mathrm{d}^d q_1}{(2\pi)^d} \frac{\mathrm{d}^d q_2}{(2\pi)^d} \frac{\mathrm{d}^d q_3}{(2\pi)^d} \frac{\delta(q_1 + q_2 + q_3)}{(m^2 + q_1^2)(m^2 + q_2^2)(m^2 + q_2^2)} \sim \frac{\Lambda^{2d}}{(m^2 - \Lambda^2)^3} \int_{1-\delta\ell}^{1} \mathrm{d}^d x_1 \mathrm{d}^d x_2 \mathrm{d}^d x_3 \delta(x_1 + x_2 + x_3)$$ This (mostly) agrees with the only resource I have found on this, these notes. (Notes from Advanced Field theory, chapter 15, Eduardo Fradkin at University of Illinois.) However, the notes then claim to get a contribution proportional to $$\delta \ell$$. I do not see how this is possible. The delta function and the integration limits mean that the vectors $$x_i$$ form an (almost) equilateral triangle. Consider the 2D case, corresponding to the illustration below. One integral, corresponding to rotating the whole traingle around the circle, will give a sphere factor $$2\pi$$. This will generalize to $$S_d$$, the spherical factor, for higher dimensions. However, we have still left 3 degrees of freedom (3 2D integrals, minus 3 restrictions from the Dirac delta.) These correspond to integrating over the thickness of the shell. One way to do this while respecting the delta function is illustrated, where the two lower vectors are moved perpendicular to the third vector. A similar move can be done keeping any of the two lower vectors fixed, while moving the two that are left perpendicular.

This is handwavy, but to my understanding, this shows that the integral should be proportional to $$\delta \ell^3$$, in contradiction with the result in the notes. (they use $$\delta s$$, eq. (15.61)) I am not sure of my logic, but I do not see how you would possibly get a contribution proportional to $$\delta \ell$$ Am I doing something wrong? Are the results in the notes correct? How can this calculation be done properly?

You are getting a higher order infinitesimal contribution $$\delta\ell^3$$ from the diagram. By definition, the correct way is to evaluate the full three-variable integral from 0 to $$\Lambda$$ and differentiate the result against $$\log \Lambda$$. That by construction gives you a $$\propto \delta\ell$$ contribution. Equivalently, two integrals should be performed for all momenta, and only one of them is done over a momentum shell between $$\Lambda$$ and $$\Lambda(1-\delta\ell)$$.

• But this is not momentum shell, but rather cut-off regularization. In momentum shell, you split the field into high and low momentum degrees. The integral is over the high momentum modes, as you are integrating these modes out, and the integrand therefore only has support on a thin shell. Oct 25, 2023 at 20:53
• Indeed, you cannot read off the contribution to the $\beta$-function from putting all internal momenta in a shell in 2-loop RG. You need to evaluate the full integral, differentiate, and use Callan-Symanzik equation. Oct 26, 2023 at 20:35
• Ok, so you are saying momentum-shell is only valid to one loop? Have you seen this discussed in the literature? Thank you for you answers. Oct 27, 2023 at 12:42
• By momentum-shell RG if you mean reading off the beta function from the integrand, yes, it is only valid to one loop. Unfortunately I don't think textbooks spells out this caveat explicitly, and indeed I found your question by trying to understand this very issue myself. Nov 4, 2023 at 23:31