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"

  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

For the loop divergences: Schwartz doesn't actually prove that higher orders don't diverge in this section. For that, check out 21.1.3 in Schwartz, where he extends the argument from one loop to multiple loops.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.