How to get Peskin and Schroeder's Equation 12.46?

Question

I'm trying to use the one-loop expression for the 4 point Greens function to calculate the beta function of massless $\phi^4$ theory. In fact, Peskin and Schroeder give the result in Eq. 12.46, but it is not clear to me how they did it... can someone tell me where my error is? Specifically, I don't understand why P&S only use the linear-in-lambda term of the 4-point Green's function to derive the beta function? The one-loop 4-point Green's function is given just below Eq. 12.43 and reproduced here:

$$G^{\left(4\right)}\left(p_1,p_2,p_3,p_4\right) = \left[-i \lambda + (-i \lambda)^2\left\{iV(s) + iV(t) + iV(u)\right\} - i \delta_{\lambda} \right] \cdot \prod_{i=1}^4 \frac{i}{p_i^2}$$

where \begin{equation} \begin{aligned} s&=(p_1 + p_2)^2\\ t&=(p_3 - p_1)^2\\ u&=(p_4 - p_1)^2 \end{aligned} \end{equation} are the Mandelstam variables, and where

$$V(p^2) = -\frac{1}{2}\int_0^1 dx \frac{\Gamma(2-d/2)}{(4 \pi)^{d/2}}\frac{1}{\left[ -x(1-x) p^2 \right]^{2-d/2}}$$

using dimensional regularization. Also, the vertex counterterm (determined using Peskin and Schroeder's renormalization scheme where the 4-point scattering amplitude is forced to equal $i \lambda$ at the spacelike momentum interval $p^2 = -M^2$) is

$$\delta_{\lambda} = (-i \lambda)^2 3 V(-M^2).$$

Now back to my questions... If you substituted both the linear and quadratic-in-lambda term in the 4-point Green's function into the Callan-Symnazik equation for a massless theory:

$$\left[ M \frac{\partial}{\partial M} + \beta \frac{\partial}{\partial \lambda} + 4 \gamma \right] G^{\left(4\right)}\left(p_1,p_2,p_3,p_4\right) = 0 $$

then that would lead to a contribution to beta that was linear in lambda, right? And wouldn't this be the leading order contribution to beta, instead of what P&S write in Equation 12.46? P&S's Equation 12.46 is

$$\beta = \frac{3 \lambda^2}{16 \pi^2} $$ and another important intermediate step is P&S's Equation between 12.45 and 12.46:

$$ M \frac{\partial}{\partial M} G^{\left(4\right)}\left(p_1,p_2,p_3,p_4\right) = \frac{3 i \lambda^2}{16 \pi^2}\prod_{i=1}^4 \frac{i}{p_i^2}$$

Many thanks everyone!

Answer 1

You seem to agree that $M\partial_M G^{(4)}= 3 i\lambda^2/16\pi^2 + o(\lambda^3)$ and that $G^{(4)} = -i\lambda + o(\lambda^2)$. You don't question the assumption that $\gamma$ has no linear term in $\lambda$ so we take it to be $o(\lambda^2)$. Putting it all in the CS equation gives \begin{equation} \Big[ \frac{3 i\lambda^2}{16\pi^2} + o(\lambda^3) \Big] +\Big[ \beta(\lambda) \frac{\partial}{\partial \lambda} \left (-i \lambda + o(\lambda^2) \right) \Big] + \Big[ o(\lambda^2) o(\lambda) \Big] =0 \end{equation} The lowest order is in $\lambda^2$ and tells us that \begin{equation} \beta(\lambda) = \frac{3 \lambda^2}{16 \pi^2} + o(\lambda^3) \end{equation} If you assume that $\beta$ has a term linear in $\lambda$ then it has to be zero, as there is no other term in the CS equation to cancel it.