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

+ 2 like - 0 dislike
1446 views

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 in Theoretical Physics by PhysiXxx (45 points) [ no revision ]
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

2 Answers

+ 0 like - 0 dislike

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 Steven Mathey (350 points) [ no revision ]
+ 0 like - 0 dislike

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 Andrew McAddams (340 points) [ no revision ]
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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