# On renormalization of $\phi^4$

I am reading Schwartz's chapter on renormalizing the $$\phi^4$$ theory and I have two questions. We define the renormalized coupling to be the matrix element of all contributing diagrams at a given energy scale

$$\lambda_R=\lambda+\frac{\lambda^2}{32 \pi^2} \ln\frac{s_0}{\Lambda^2}+...\tag{15.65}$$

Now we want to get an expression $$\lambda=\lambda(\lambda_R)$$ to substitute this in the matrix element for general energies. To this end, Schwartz now writes a series expansion

$$\lambda=\lambda_R+a \lambda^2_R+...\tag{15.66}$$

Why can we do this?

Second, the resulting expression for $$\mathcal{M}(s)$$ reads

$$\mathcal{M}(s)=-\lambda_R-\frac{\lambda^2_R}{32 \pi^2}\ln\frac{s}{s_0}+...\tag{15.69}$$

He then writes

This equation gives us an expression for $$\mathcal{M}(s)$$ for any $$s$$ that is finite order-by-order in perturbation theory.

I dont really understand this claim. He showed that its finite at one-loop-order. Why is it clear from the preceding calculation that there won't be divergences at some higher order?

Edit: It can be found on page 298 in Schwartz "QFT and the SM"

• Feynman diagrams can be expanded as a Taylor series in the external momentum. The polynomial representing the diagram will have divergent coefficients, reflecting the divergence of the integral, and we set external momentum to p = 0. QFT for the Gifted Amateur (p 289) covers the same ground as in your post. – StudyStudy Jun 21 '20 at 15:06

The expression for $$\lambda$$ in terms of $$\lambda_R$$ is just perturbation theory. Given an expression for $$\lambda_R$$ in terms of $$\lambda$$ $$\lambda_R = \lambda + \frac{\lambda^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+...$$ we solve perturbatively, assuming that $$\lambda$$ can be written as a power expansion in $$\lambda_R$$. That is, we guess a solution of the form $$\lambda = \sum_{i} a_i\lambda^i$$ plug it into our expression for $$\lambda_R$$ $$\lambda_R = \left(\sum_{i} a_i\lambda^i\right) + \frac{\left(\sum_{i} a_i\lambda^i\right)^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+...$$ then solve for the $$a_i$$ at each order in $$\lambda_R$$. At order 1, we get $$\lambda_R = \lambda_R$$ indicating that $$a_1 = 1$$. Plugging this in, we go to order 2, getting $$0 = a_2\lambda_R^2+ \frac{\lambda_R^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}\longrightarrow a_2 = -\frac{1}{32\pi^2}\ln\frac{s_0}{\Lambda^2}$$ and so forth for the other orders.