I am trying to understand the general strategy and technical details of calculating $\beta$-function at higher orders. $\beta$-function is the anomalous dimension of the coupling constant and there is a complete set of anomalous dimensions corresponding to different fields, propagators (gauge field), and vertices.

The anomalous dimension corresponding to a renormalization constant could be defined as

$$\gamma=-\mu^2\frac{d \log Z}{d\mu^2}$$

And in the minimal subtraction scheme, one could expand the renormalization constants as

$$Z=1+\sum_{i=1}^\infty \frac{z_i (a_s,\xi)}{\epsilon^i}$$

where $\xi$ is the gauge parameter, i.e. we do not need to fix the gauge before computing the anomalous dimension, and the number of space-time dimensions $D=4-2\epsilon$. Now, let's consider the anomalous dimension of the coupling constant, and assume that scale dependence of the corresponding renormalization constant $Z_{a_s}$ happens through $a_s$ and $\xi$. How does the following relation hold?

$$-\beta(a_s)=\left(-\epsilon + \beta(a_s) \right) a_s \frac{\partial \log Z_{a_s}}{\partial a_s} $$

What about the anomalous dimension of $\xi$. What would the relation be?

note: Please refer to http://arxiv.org/abs/hep-ph/0405193v3 for the conventions; This article http://link.springer.com/article/10.1007%2FBF01079292 also contains valuable points such as (2.4) and (2.6) which I believe are related to the problem at hand.

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    $\begingroup$ Could the downvoter explain? I cannot understand the question because I don't know RG, but there is clearly some effort put into it. $\endgroup$
    – Abhinav
    Commented May 30, 2016 at 4:52

1 Answer 1


In $\overline{MS}$ the $\beta$ function does not depend on the gauge parameter. This means that the dependence on $\mu$ in $Z$ only comes from the coupling constant $a \propto \mu^{-2 \epsilon}$.

For general $Z(a_s,\xi)$, the relation is as follows (eq. 21 in the Chetyrkin paper):

$$-\gamma =\left(-\epsilon + \beta(a_s) \right) a_s \frac{\partial \log Z}{\partial a_s}+\gamma_3(a_s,\xi)\xi \frac{\partial \log Z}{\partial \xi}$$


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