For a real scalar field $\phi$ after performing all the 1-loop renormalization
for dimensional regulator $d = 4 – \epsilon,\ \epsilon \rightarrow 0^+$, I have found that the renormalized coupling $\lambda$ can be related to the bare one by
$$\lambda\Bigg(1 + \frac{3}{(4\pi)^2\epsilon}\lambda_p\Bigg) = \lambda_0 \tag1$$
I'm stuck trying to get the beta function from that equation. We call beta function to
$$\beta(\lambda) = \mu\frac{\partial\lambda}{\partial\mu}$$
Taking into account that
$$\lambda = \lambda_p\mu^\epsilon,\qquad [\lambda_p] = 0, \qquad [\mu] = 1$$
My problem is that any way I use to get $\beta(\lambda)$ from Eq. (1) gives a dependence on $\epsilon$, but in Peskin it is used a different way to solve this and the solution is
$$\beta(\lambda) = \frac{3\lambda^2}{(4\pi)^2}$$
How can I get the beta function via Eq. (1)?
Best Answer
The idea is that the bare quantities explicitly do not depend on $\mu$, thus one has the equation $$ 0 = \mu\frac{d \lambda_0}{d\mu} = \left(\frac{\partial}{\partial\mu}+ \beta(\lambda_p)\frac{\partial}{\partial \lambda_p}\right)\left(\mu^\epsilon\lambda_p \,Z_\lambda\right)= \epsilon \mu^\epsilon\lambda_p Z_\lambda+\mu^\epsilon \beta(\lambda_p)\frac{d(\lambda_p Z_\lambda)}{d\lambda_p}\,. $$ Where $$Z_\lambda = 1 + \frac{3\lambda_p}{(4\pi)^2\epsilon}\,.$$ This determines the $\beta$ function as $$ \beta(\lambda_p) = - \frac{\epsilon \lambda_p Z_\lambda}{Z_\lambda + \lambda_p\frac{dZ_\lambda}{d\lambda_p}}\,. $$ Obviously you need to do this perturbatively in $\lambda$, which means that even if $\lambda$ multiplies a pole in $\epsilon$ (which is divergent), you still series expand it as if it was small. If $Z_\lambda \equiv 1 + \lambda_p^2\,z/\epsilon$ one has $$ \beta(\lambda_p) = - \frac{\epsilon \lambda_p + z\lambda_p^2}{1+2z\lambda_p/\epsilon} =z \lambda_p^2 +O(\epsilon)= \frac{3\lambda_p^2}{(4\pi)^2}\,. $$ A consistency check is that the result should never be divergent as $\epsilon \to 0$, this can be proved with the Callan-Symanzik equations.