# Beta function calculation in massless minimal subtraction $\phi^4$ theory

I'm trying to understand how to calculate the beta function in massless phi^4 theory using dimensional regularisation and minimal subtraction. I'm struggling to understand:

• Is it possible to renormalise the massless $\phi^4$ theory using the minimal subtraction scheme? Or do we need some other scheme.
• When do we introduce a new mass parameter $\mu$? Is it to make the coupling constants dimensionless or is it to fix some coupling constant renormalisation scheme (e.g. renormalising such that $\Gamma(p_i = \mu) = -\lambda$). Is there actually a distinction here?

I've tried to illustrate these points with the following calculation:

I understand we start with a (Euclidean) Lagrangian with counter terms:

$$\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi + \frac{1}{2}m^2 \phi^2 + \frac{\lambda}{4!}\phi^4 \\ + \frac{1}{2}\delta_z\partial_{\mu}\phi\partial^{\mu}\phi + \frac{1}{2}\delta_m \phi^2 + \frac{\delta_{\lambda}}{4!}\phi^4$$

Where the counter terms in terms of the bare parameters are: $\delta_z = Z -1$, $\delta_m = m_0^2 - m^2$, $\delta_{\lambda} = \lambda_0 Z^2 - \lambda$ with $Z$ being the field renormalisation $\phi_0 = Z \phi$.

In minimal subtraction and dimensional regularisation ($d = 4-\epsilon$) from the one loop diagram of $\Gamma^{(4)}$ in the minimal subtraction scheme I find that $\delta_{\lambda} = -\frac{3\lambda^2}{16\pi^2\epsilon}$.

Now the definition of the beta function (in terms of the dimensionless renormalised coupling constant $g = \lambda \mu^{-\epsilon}$:

$$\beta(g) = \mu \frac{\partial g}{\partial \mu}$$

Where $\lambda_0$ is held constant.

So in this case since $Z = 1$ to one loop order we have that $\lambda = \lambda_0 + \frac{3\lambda^2}{16\pi^2\epsilon}$. Inverting this (using the fact $\lambda = \lambda_0$ to quadratic order):

$$g = \lambda_0 \mu^{-\epsilon} + \frac{3\lambda_0^2}{16\pi^2\epsilon} \mu^{-\epsilon}$$

I'm struggling to see how to recover the usual beta function from this. In particular how to substitute back in terms of $g$. I'm expecting $\beta(g) = -\epsilon g + \frac{3}{16 \pi^2}g^2$.

• So is the correct answer for the one loop beta function of $\phi^4$ in dimensional regularisation not $\beta(g) = -\epsilon g + \frac{3}{16 \pi^2}g^2$? – Wooster Jan 3 '16 at 13:17