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)?


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.

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  • $\begingroup$ I do not understand why you can expand in Taylor series the denominator if that quantity includes a divergence. We are making $(1 + x)^{-1} = 1 - x$ but $x$ diverges while $(1 + x)^{-1} \rightarrow 0$. Is all this method just a trick that we know gives the same result as Callan-Symanzik eqs. or is there something that allows us to make it? I think mathematician wouldn't be happy with this. Of course, thank you for your answer and do not take this as a complaint, I'm just trying to understand it. Thanks! $\endgroup$ – Vicky Feb 25 '19 at 3:03
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    $\begingroup$ This is at the heart of perturbation theory. You have to think the regulator as a finite number throughout all manipulations, and then you are allowed to send it to infinity only in the final expressions where the dependence on the regulator necessarily disappears. $\endgroup$ – MannyC Feb 25 '19 at 3:32
  • $\begingroup$ As for $\lambda$, in perturbation theory any quantity $f(\lambda)$ is though of as a formal power series in $\lambda$:$\sum_k \alpha_k\lambda^k$, which doesn't have to converge (normally one has asymptotic series). If you compute diagrams up to $\ell$ loops you'll obtain an approximate answer $f^{(\ell)}(\lambda)$ which can also be expressed as a formal power series $\sum_k \alpha_k^{(\ell)}\lambda^k$. Perturbation theory tells you that $\alpha_k = \alpha_k^{(\ell)}$ for all $k\leq \ell$ and tells you nothing for the remaining ones. [continues...] $\endgroup$ – MannyC Feb 25 '19 at 3:36
  • $\begingroup$ So when we throw away $\alpha_k^{(\ell)}$ for $k>\ell$ we are just saying that those coefficients are not useful for determining the true coefficients in the power series at this particular order. We are not claiming that they are "small" or "negligible." Hope this makes sense for you. $\endgroup$ – MannyC Feb 25 '19 at 3:37
  • $\begingroup$ I agree with your explanation about the power series on λ but I think that reasoning is only valid if we had developed the denominator of $\beta(\lambda)$ without using the already known Taylor expansion $(1 + x)^r = \sum_n {{r\choose{n}}}x^n$ which is only right for $|x| \ll 1$, i.e., $\lambda_p/\epsilon \ll 1$. If we had done it from pure Taylor series formula (without assuming anything about $x$) I would agree to that expansion in powers of $\lambda$, but the expansion we used in denominator of $\beta$ is restricted to $|x| \ll 1$ $\endgroup$ – Vicky Feb 25 '19 at 4:02

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