# Anomalous Dimension and Infinitesimal Transformation

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?

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

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