# Beta-function for mass

I'm trying to find the renormalization group fixed points for a free, massive, scalar theory $$\begin{equation} S = \int_{\mathbb{R}^d}d^dx \left( \frac{(\partial_\mu\phi)^2}{2} + \frac{m^2\phi^2}{2} \right) \ . \end{equation}$$ That is, I'll treat $$m^2$$ as a quadratic interaction, and want to find its corresponding $$\beta$$-function, $$\beta_{m^2}$$. I know it's supposed to be $$\begin{equation} \beta_{m^2} = -2m^2 \ , \end{equation}$$ which gives me the fixed point $$\begin{equation} \beta_{m^2} = 0 \quad\Rightarrow\quad m^2 = 0 \ . \end{equation}$$ However, how do I find $$\beta_{m^2}$$? Should I do Feynman diagrams for the two-point function $$\langle\phi(x)\phi(y)\rangle$$? Can I find it from the path integral? Is it possible to find it from the Callan-Symanzik equation?

Since the free theory is Gaussian, I should not need to do perturbation theory, and be able to solve the theory (find $$\beta_{m^2}$$) exactly.

In this case, I would write your action with a bare mass $$\begin{equation} S = \int_{\mathbb{R}^d}d^dx \left( \frac{(\partial_\mu\phi)^2}{2} + \frac{m_0^2\phi^2}{2} \right) \end{equation}$$ and define a renormalized mass by $$\begin{equation} m_0^2 = \mu^2 m^2 \end{equation}$$ in order to get something dimensionless. We then arrange for $$m$$ to run in such a way that $$m_0$$ is independent of the renormalization scale. \begin{align} & \mu \frac{\partial}{\partial \mu} (\mu^2 m^2) = 0 \\ & 2\mu^2 m^2 + \mu^3 \frac{\partial m^2}{\partial \mu} = 0 \\ & \mu \frac{\partial m^2}{\partial \mu} = -2m^2 \end{align} As expected, this beta function has only a classical term. It's the one we would've guessed by seeing that $$\mu$$ appeared with an exponent of $$2$$ in the definition of the renormalized quantity. In an interacting theory, a shift in dimension by $$2$$ would only be part of the story because there would be further anomalous shifts in the dimensions from the loop corrections. This is what quantum terms in the beta function (which some sources misleadingly refer to as the whole beta function) represent.
• Thanks for the answer. I was thinking along the same lines at some point, but afterwards I got confused about $m_0^2 = \mu^2m^2$. Since the renormalization scale $\mu$ have mass dimension one, doesn't this mean that the renormalized mass $m^2$ is dimensionless? Dec 1, 2021 at 14:25
• This is the same reason why you set $g_0 = \mu^\epsilon Z_g g$ in $\phi^4$ theory. In four dimensions, $g$ and $g_0$ are both dimensionless but in $d = 4 - \epsilon$ only one of them is. Dec 1, 2021 at 14:40
• I see, thanks! I thought it was different for masses. Now, say I would like to generalize this a bit, e.g. by considering the theory near a boundary \begin{equation} S = \int_{\mathbb{R}_+^d}d^dx \left( \frac{(\partial_\mu\phi)^2}{2} + \frac{m_0^2\phi^2}{2} \right) + \int_{\mathbb{R}^{d - 1}}d^{d - 1}x_\parallel \frac{\mu_0^2\hat{\phi}^2}{2} \ , \end{equation} where $\mathbb{R}_+^d = \{ (x_\parallel, x_\perp): x_\parallel \in \mathbb{R}^{d - 1}, x_\perp > 0 \}$, and $\hat{\phi}$ denotes the boundary limit of $\phi$. Dec 1, 2021 at 14:55
• Then the $\beta$-function corresponding to the boundary mass $\mu_0$ should have fixed points at $\mu = 0$ (Neumann b.c.'s) and $\mu \rightarrow \pm\infty$ (Dirichlet b.c.'s). It seems like running the same arguments as you did for the bulk mass I only pick up the $\mu = 0$ fixed point. Dec 1, 2021 at 14:55