# Derivation of the Renormalization Group from Renormalized Coupling?

At page 303 in the book Quantum Field Theory for the Gifted Amateur by Blundell and Lancaster, they argue that the renormalization group equation for the coupling $$\lambda$$ in $$\phi^4$$ theory can be derived as follows. Firstly, we introduce the renormalized coupling $$\lambda_R(s_0) = \lambda +\lambda^2 C \ln \left( \frac{s_0}{\Lambda^2}\right) + \ldots ,$$ and rewrite the matrix element in terms of $$\lambda_R$$: $$iM = - \lambda_R(s_0) -C \ln \left( \frac{s }{s_0}\right) \lambda_R^2(s_0) + \ldots \tag{1}$$ For a different scale $$s_1$$, we find $$\lambda_R(s_1) = \lambda +\lambda^2 C \ln \left( \frac{s_1}{\Lambda^2}\right) + \ldots ,$$ $$iM = - \lambda_R(s_1) -C \ln \left( \frac{s }{s_1}\right) \lambda_R^2(s_1) + \ldots \tag{2}$$ They then argue that by "subtracting" it follows that $$\lambda_R(s_1) = \lambda_R(s_0) +C \ln \left( \frac{s_1 }{s_0}\right) \lambda_R^2(s_0) + \ldots \tag{3}$$ I'm failing to see why this should be true. If we subtract Eq. 2 from Eq. 1 we find a term of the form $$-C \ln \left( \frac{s }{s_1}\right) \lambda_R^2(s_1)+C \ln \left( \frac{s_0}{\Lambda^2}\right)\lambda_R^2(s_0)$$ which is not equal to the one in Eq. 3 unless $$\lambda_R^2(s_0) = \lambda_R^2(s_1)$$.

Am I missing something or is the derivation shown here wrong?

• In eq.(2), why should the first term on the RHS depend on $s_0$ instead of $s_1$? – Jon Nov 18 '19 at 18:39
• @Jon that was a typo, thanks! – jak Nov 18 '19 at 20:33
• You subtracted Eq. 1 from Eq. 2 incorrectly. The answer starts with the answer they give you and a term proportional to $\lambda_R^2 (s_0) -\lambda_R^2 (s_1)$, which, by above, you must appreciate is of higher order in $\lambda$, cubic. can you work it out? – Cosmas Zachos Nov 18 '19 at 21:07
• @CosmasZachos unfortunately no. I still can’t see it. Any further hint would be much appreciated! – jak Nov 19 '19 at 7:47

After subtraction you get $$\lambda_R(s_1)=\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)-C\ln\left(\frac{s}{s_1}\right)\lambda_R^2(s_1)+O\left(\lambda_R^3\right).$$ This is an implicit equation for $$\lambda_R(s_1)$$ that can be solved iteratively. Then, $$\lambda_R(s_1)=\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)-C\ln\left(\frac{s}{s_1}\right)\left[\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)+\dots\right]^2+\ldots.$$ From this, neglecting powers higher than 2 of the coupling, you are able to recover the book's result.