The relation between anomalous dimensions and renormalization constants

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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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moha
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Updated on June 09, 2020

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  • moha
    moha over 3 years

    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.

    • Abhinav
      Abhinav over 7 years
      Could the downvoter explain? I cannot understand the question because I don't know RG, but there is clearly some effort put into it.