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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?

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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.).

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  • $\begingroup$ Thank you. This makes more sense. This would only be applicable for theories where the first loop corrections are $\mathcal{O}(\lambda^2)$, right? $\endgroup$
    – saad
    Nov 30, 2023 at 8:46
  • $\begingroup$ Because since the theory is massless, the snail diagram is 0. $\endgroup$
    – saad
    Nov 30, 2023 at 8:50
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    $\begingroup$ @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. $\endgroup$ Nov 30, 2023 at 19:35
  • $\begingroup$ Sorry I had meant to say at least $\mathcal{O}(\lambda^{2})$. $\endgroup$
    – saad
    Dec 4, 2023 at 18:51
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    $\begingroup$ @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. $\endgroup$ Dec 7, 2023 at 0:21

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