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In the theory of Renormalization Group (RG) transformations I ended up with the following equation

$$\left( \frac{Z(\mu)}{Z(\mu / s)} s^{d-2}\right)^{1/2} = 1+\left(\frac{1}{2}(d-2)-\gamma_{\phi}\right)\delta s$$

where $Z(\mu)$ is the wave function renormalization, $s$ is a parameter , $d$ is the dimension of the theory and

$$\gamma_{\phi} = -\frac{\mu}{2Z}\frac{dZ}{d\mu}$$

is the anomalous dimension. How do I get this equation? I know I am supposed to consider an infinitesimal transformation $s = 1+\delta s$, but I do not see how to expand the LHS. Any ideas?

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    $\begingroup$ Yes. Write $Z(\mu / s)$ as $Z(\mu - \mu \delta s)$ first since you are only working to first order. $\endgroup$ Jun 14, 2021 at 17:09

1 Answer 1

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Expanding the LHS=$\left(\displaystyle \frac{Z(\mu)}{Z(\mu/s)s^{d-2}} \right)^{1/2}$ in $s=1+\delta s$ to first order in $\delta s$ goes as follows:

  1. $\displaystyle \frac{\mu}{s}=\frac{\mu}{1+\delta s} \sim \mu (1-\delta s)$

  2. $Z(\mu/s) \sim Z(\mu (1-\delta s))\sim Z(\mu) - \displaystyle \mu \frac{dZ(\mu)}{d\mu} \delta s\quad$ Taylor expanded around $\delta s$ hence the factor $\mu$

  3. $\displaystyle \frac{Z(\mu)}{Z(\mu/s)} \sim 1 + \frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \delta s$

  4. $\left[\displaystyle \frac{Z(\mu)}{Z(\mu/s)} s^{(d-2)}\right]^{1/2} \sim \left( 1 + \displaystyle \frac{1}{2}\frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \delta s\right) \left(1+\displaystyle \frac{(d-2)}{2} \delta s \right)\qquad$ because $(1+\delta s)^{(d-2)/2}=e^{(d-2)/2\ln (1+\delta s)} \sim 1 + \displaystyle \frac{(d-2)}{2} \delta s$

Expanding the last product again to $\cal O (\delta s^2)$ the fourth equation yields the RHS:

RHS $= 1 + \left( \displaystyle \frac{(d-2)}{2} +\frac{1}{2}\frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \right) \delta s = 1 + \left( \displaystyle \frac{1}{2}(d-2) -\gamma_\phi \right) \delta s$

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