Computing the beta-function using the Wilsonian Effective action

Question

In these lecture notes, the beta-function for the $\phi^4$-theory is computed by identifying the change in the action when integrating modes with momenta $\mu'<|p|<\mu$, where $\mu$ and $\mu'$ are cut-offs as being the Wilsonian effective action.

In particular this turn out to be equal to $$S[\varphi; \mu,g_{2n}(\mu)]-S[\varphi; \mu + \delta \mu,g_{2n}(\mu+\delta\mu)]=a \mu^{d-1}\int \mathrm d^d x \, \log (\mu^2+V''(\varphi))\delta \mu \,.$$

The author argues that the beta-functions for the couplings $g_{2n}$ can be computed by 'expanding the right-hand side in powers of $\varphi$, leading to

$$\mu\frac{\mathrm d g_{2n}}{\mathrm d \mu}= (n(d-2)-d)g_{2n} - a \mu^{n(d-2)}\left.\frac{\mathrm d^{2n}}{\mathrm d \varphi^{2n}}\log (\mu^2+V''(\varphi))\right|_{\varphi=0}\,. $$

Could someone give a lead to how to begin the derivation of this result?

Answer 1

According to the paper the potential is assumed to be of the form

$$V(\varphi)=\sum_{n\ge 0}\frac{g_{2n} \varphi^{2n}}{(2n)!}\mu^{d-n(d-2)}.$$

In this setting we assume a constant field value $\varphi $, for which the LHS take the form $$\sum_{n\ge 0}\frac{\varphi^{2n}\mu^{d-n(d-2)}}{(2n)!}\left((d-n(d-2))g_{2n}\mu^{-1}+\frac{\partial g_{2n}}{\partial \mu}\right) \delta \mu \times V_d.$$ Expanding the RHS in powers of $\varphi $ gives $$\mu^{d-1}\sum_{n\ge 0}\frac{\varphi^{2n}}{(2n)!}\frac{d^{2n}}{d \varphi^{2n}}\log\left( \mu^2+V''\right)|_{\varphi=0} \delta \mu \times V_d .$$ The result follows after setting LHS=RHS and collecting factors of $\varphi $.