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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?

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  • $\begingroup$ In eq.(2), why should the first term on the RHS depend on $s_0$ instead of $s_1$? $\endgroup$ – Jon Nov 18 '19 at 18:39
  • $\begingroup$ @Jon that was a typo, thanks! $\endgroup$ – jak Nov 18 '19 at 20:33
  • $\begingroup$ 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? $\endgroup$ – Cosmas Zachos Nov 18 '19 at 21:07
  • $\begingroup$ @CosmasZachos unfortunately no. I still can’t see it. Any further hint would be much appreciated! $\endgroup$ – jak Nov 19 '19 at 7:47
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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.

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