How to get the relation for dependence of anomalous dimension on regularization?

Question

Here is the anomalous dimension: $$ \gamma_{\Gamma}(t, g) = \left[\frac{\partial }{\partial t}\ln \left(Z_{\Gamma}(t , g) \right)\right]_{t = 1}, $$ where $Z_{\Gamma}$ is renormalization factor which arises in n-point functions $\Gamma $, $t$ denotes change of renormalization parameter $t = \frac{\mu{'}}{\mu}$. $Z_{\Gamma}$ arises explicitly after making shift of renormalization parameter $\mu$ (for fixed type of renormalization):

$$ \Gamma (xt , g) = Z_{\Gamma}^{-1}(t , g) \Gamma (x, \bar{g}(t , g)), \quad x = \frac{k}{\mu}, \quad t = \frac{\mu}{\mu{'}}. $$

Let's change type of regularization (coupling constant will change to $g \to \tilde {g}(g)$. Then n-point function will change as $$ \Gamma \left(\frac{k}{\mu} , g \right) = q(g) \tilde {\Gamma}\left( \frac{k}{\mu} , \tilde {g}(g) \right). $$ How to get from these equations that $\gamma_{\Gamma}$ will change to $$ \tilde{\gamma}_{\Gamma}(\tilde {g}(g)) = \gamma_{\Gamma}(g) - \beta (g)\frac{d\ln (q(g))}{dg} $$ (the definition for $\beta$-function see here)?

Answer 1

I'm not quite sure where the details of the last equations come from, but I think that the step that you are missing is to identify,

$$q(g) = \frac{1}{Z_\Gamma(g)} \, .$$

Answer 2

Two functions, $\Gamma \left(\frac{k}{\mu}, g\right)$ and $\Gamma \left( \frac{k}{\mu}, \tilde{g}(g)\right)$, satisfy so called Callan–Symanzik equations: $$ \tag 1 \left(t\partial_{t} - \beta (g) \partial_{\beta} + \gamma_{\Gamma}(g)\right)\Gamma (t , g) = 0, \quad \left(t\partial_{t} - \tilde{\beta} (\tilde{g}) \partial_{\tilde{\beta}} + \tilde{\gamma}_{\Gamma}(\tilde{g})\right)\tilde{\Gamma} (t , \tilde{g}) = 0. $$ Let's use identities (the second one is the result of the definition of $\beta$-function) $$ \Gamma(t, g) = q(g)\tilde{\Gamma}(t, \tilde{g}), \quad \tilde{\beta}(\tilde{g}) = \frac{d \tilde {g}}{dg}\beta (g) $$ and let's insert them into the second equation of $(1)$ with using the first one: $$ \left(t\partial_{t} - \beta (g) \partial_{\beta} + \tilde{\gamma}_{\Gamma}(\tilde{g})\right)\frac{1}{q(g)}\Gamma (t , g) = \frac{1}{q(g)}\left( t\partial_{t} - \beta (g) \partial_{\beta} + \gamma_{\Gamma}(g) \right) + $$ $$ + \Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right) \right] = $$ $$ =\Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right) \right]= 0 \Rightarrow $$ $$ \tilde{\gamma}_{\Gamma}(\tilde{g}) = \gamma_{\Gamma}(g) - q(g) \beta (g)\frac{q'(g)}{q^{2}(g)} = \gamma_{\Gamma}(g) - \beta(g)\partial_{g}ln (q(g)). $$