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

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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)?

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user PhysiXxx
asked Sep 14, 2014
retagged Sep 18, 2014
Try plugging into Callan–Symanzik equation and then read off the anomalous dimension from there.

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Orbifold

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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,

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

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Steven Mathey
answered Sep 15, 2014 by (350 points)
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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: $$\left(t\partial_{t} - \beta (g) \partial_{\beta} + \tilde{\gamma}_{\Gamma}(\tilde{g})\right)\frac{1}{q(g)}\Gamma (t , g) = \frac{1}{q(g)}\left( t\partial_{t} - \beta (g) \partial_{\beta} + \gamma_{\Gamma}(g) \right) +$$ $$+ \Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right) \right] =$$ $$=\Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right) \right]= 0 \Rightarrow$$ $$\tilde{\gamma}_{\Gamma}(\tilde{g}) = \gamma_{\Gamma}(g) - q(g) \beta (g)\frac{q'(g)}{q^{2}(g)} = \gamma_{\Gamma}(g) - \beta(g)\partial_{g}ln (q(g)).$$

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Andrew McAddams
answered Sep 15, 2014 by (340 points)
The CS equation for an n-point function $\Gamma$ contains a factor of $n$ along with the anomalous dimension. How did you neglect that? It would result in a $1/n$ factor in the final answer!

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Orbifold
Also could you explain the relative '$-$' sign in $\gamma_{\Gamma}(g)-\beta(g)\partial_gln(q(g))$

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Orbifold
@Orbifold : as for the second, the relative $-$ sign have arised from derivation of $\frac{1}{q(g)}$ function: $$q(g)\partial_{g}\frac{1}{q(g)} = -q(g)\frac{q(g)'}{q^{2}(q)} = -\partial_{g}ln(q(g)).$$ As for the first, the OP definition

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Andrew McAddams
@Orbifold : as for the first one, it's only the redefinition of the definition of $Z$-constant (as you can see from OP definition, there is $Z^{-1}$, not $Z^{-n}$ at n-point function.

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Andrew McAddams
regarding the minus sign: thats fine! but when you evaluate for $\tilde{\gamma}$ will is not become $+$ when going on the other side of the equation?

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Orbifold
@Orbifold : here is the set of actions: $$\tilde {\gamma}_{\Gamma}(\tilde{g}) - \gamma_{\Gamma} (g) - q(g)\beta (g)\partial_{g} \frac{1}{q(g)} = 0 \Rightarrow \tilde {\gamma}_{\Gamma}(\tilde{g}) = \gamma_{\Gamma}(g) + q(g)\beta (g)\partial_{g} \frac{1}{q(g)} =$$ $$= \gamma_{\Gamma}(g) - \beta(g)q(g)\frac{q'(g)}{q^{2}(q)} = \gamma_{\Gamma}(g) - \beta (g)\partial_{g}ln (q(g)).$$

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Andrew McAddams
@Orbifold : and please check my comment about $n$ factor.

This post imported from StackExchange Physics at 2014-09-18 08:02 (UCT), posted by SE-user Andrew McAddams

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