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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!

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  • $\begingroup$ "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". They don't. They use the $\lambda^2$ contribution as well. Note that to linear (tree) order in $\lambda$ the counter-term is finite, and thus $\beta=0$ to this order. The first non-zero contribution comes from the $\mathcal O(\lambda^2)$ term in $G$. $\endgroup$ – AccidentalFourierTransform Jan 6 '18 at 14:58
  • $\begingroup$ @AccidentalFourierTransform: I'm sorry, still a bit confused... If the M(d/dM) G term in the Callan-Symnazik equation is quadratic in lambda then that means the \beta (d/dlambda) G term is also quadratic in lambda. And if \beta alone is quadratic in lambda then (d/dlambda) G must be independent of lambda, and thus G must be linear in lambda. So they didn't use the quadratic in lambda term, just the linear one, right? $\endgroup$ – JohnnyHotCakes Jan 6 '18 at 15:01
  • $\begingroup$ Anybody have any other ideas? Thanks very much ahead of time! $\endgroup$ – JohnnyHotCakes Jan 6 '18 at 15:27
  • $\begingroup$ I want to add that I think the top comment by AccidentalFourierTransform is not correct, just for other readers to be aware. The description below by Oбжорoв does seem correct. That said, still want to thank AccidentalFourierTransform for participating and note that its possible I misunderstood his/her answer. $\endgroup$ – JohnnyHotCakes Jan 7 '18 at 9:53
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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.

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  • $\begingroup$ Thanks very much for taking the time to add this, I appreciate the help. $\endgroup$ – JohnnyHotCakes Jan 7 '18 at 9:46

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