# Calculation of $\gamma(\lambda)$ in massless renormalizable scalar field theory

In Peskin & Schroeder p.413 and 414, the Callan-Symanzik equation for a 2-point Green's function is used to calculate $$\gamma(\lambda)$$ for a massless renormalizable scalar field theory. The two-point Green's function for such a theory is given by

$$G^{(2)}(p)=\frac{i}{p^2}+\frac{i}{p^2}\left(A \log \frac{\Lambda^2}{-p^2}+\text { finite }\right)+\frac{i}{p^2}\left(i p^2 \delta_Z\right) \frac{i}{p^2}+\cdots \tag{12.49}$$

$$\delta_{z}$$ has the energy scale dependence. Peskin & Schroeder then goes on to make the following assumption: Neglect the $$\beta$$ term because it is always smaller by at least one power of the coupling. Why is this so? We just saw for massless renormalizable $$\phi^4$$ theory that $$\beta$$ is $$\mathcal{O}(\lambda^{2})$$ and $$\gamma$$ is also $$\mathcal{O}(\lambda^2)$$.

Furthermore, Peskin & Schroeder writes the Callan-Symanzik equation for the two-point function as:

$$-\frac{i}{p^2} M \frac{\partial}{\partial M} \delta_Z+2 \gamma \frac{i}{p^2}=0.\tag{p.414}$$

The $$\gamma$$ is multiplied by $$\frac{i}{p^{2}}$$ only. Why isn't it multiplied by the rest of the Green's function?

At first one has to realize that the calculation of (12.49-51) is a generalization of the $$\phi^4$$ case. A couple of theories could be concerned like $$\phi^3$$ in 6 dimensions described by Srednicki or the Yukawa-theory and of course $$\phi^4$$ included. For $$\phi^3$$ and Yukawa theory the lowest order loop diagrams contain two vertices, where the $$\phi^4$$ contains only one, but the one-loop diagram of the $$\phi^4$$ has no contribution to the field-strength renormalization, therefore the next interesting loop-diagram of $$\phi^4$$ (the so called sunset or Saturn diagram) has to be considered which yields a nonzero contribution to the field-strength renormalization based on two vertices.
This means that the coefficient $$A$$ of the term which comes from these loop-diagrams

$$A \log \frac{\Lambda^2}{-p^2}$$

$$A \sim \lambda^2$$(2 vertices) where $$\lambda$$ is the coupling constant. One could also imagine even higher-order loop-diagrams where $$A\sim \lambda^3$$ or $$A\sim \lambda^4$$ or even higher, but at the end the argumentation works well with for all these cases. We will just assume that $$A=a\lambda^2$$ where $$a$$ is just another coefficient.

In the next step we will already substitute the counterterm $$\delta_Z$$ by the requirement that it cancels the divergence, we will set

$$\delta_Z = A\log \frac{\Lambda^2}{M^2}+ \text{finite}$$

Then we get for $$G^{(2)}$$:

$$G^{(2)} =\frac{i}{p^2}\left[ 1 + a\lambda^2 \log\frac{M^2}{-p^2}\right]$$

Plugging this in the Callen-Symanzik equation we get:

$$0=\left[M\frac{\partial}{\partial M} + \beta + 2\gamma\right]G^{(2)} = \frac{i}{p^2}\left[2a\lambda^2 + 2a\beta\lambda\log\frac{M^2}{-p^2} + 2\gamma\left( 1 + a\lambda^2 \log\frac{M^2}{-p^2}\right)\right]$$

We already see that the term with the $$\beta$$-function can be neglected since $$\lambda\beta \sim \lambda^3$$ or even $$\lambda\beta \sim \lambda^4$$. We finally only need to solve for $$\gamma$$:

$$\gamma = -\frac{a\lambda^2}{1 + 2a\lambda^2\log\frac{M^2}{-p^2}}\approx -a\lambda^2(1 - 2a\lambda^2\log\frac{M^2}{-p^2}) \approx -a\lambda^2 \equiv -A$$

We have developed the denominator in a geometric series truncated after the linear term and finally observe that we can even neglect this (linear) term when it is multiplied with $$a\lambda^2$$. Because we only need to compute the coefficients $$\beta$$ and $$\gamma$$ for the lowest order which appears in the calculation. Of course if $$A\sim \lambda^3$$ or even $$A\sim \lambda^4$$ the argumentation would not change.

By the way this result is in line with the result of problem 13.2 mentioned by P&S whose result is $$\gamma \sim \lambda^2$$ for $$\phi^4$$-theory. It also agrees with P&S statement that equation (12.51) is also valid for the $$\phi^4$$-theory although on first sight (in the book) it does not seem to be the case.

The key for the correct computation of $$\gamma$$ is the observation that $$A \sim \lambda^2$$ (or for higher loop contribution $$A\sim \lambda^4$$ etc.).

• Thank you. This makes more sense. This would only be applicable for theories where the first loop corrections are $\mathcal{O}(\lambda^2)$, right?
• @saad Not necessarily, the argument works also for $A\sim\lambda^3$ or $\sim\lambda^4$ which correspond to higher loop correction. Or even a in case of a series $A\sim a \lambda^2 + b\lambda^4 + c\lambda^6 +\ldots$ corresponds to whole series of loop diagrams. Nov 30, 2023 at 19:35
• Sorry I had meant to say at least $\mathcal{O}(\lambda^{2})$.
• @saad yes, the snail diagram has no contribution in $\phi^4$ theory (no renormalization at all at this level, the theory is massless). Only diagrams with at least 2 vertices contribute. Dec 7, 2023 at 0:21