$\beta$-function in free massive scalar field

Question

It is well-known that interacting QFTs with conformal symmetry preserved at the quantum level have a vanishing $\beta$-function. Another common statement is that mass terms break conformal invariance.

The free massless scalar field is conformal, and adding a mass term breaks the conformal invariance. Although it is still commonly called "free", I think you could view the mass term $m^2 \phi \phi$ as an interaction term between two scalar fields. After all, what is the difference between $m^2 \phi^2$ and $\lambda \phi^4$, apart from the number of fields being coupled in the vertex?

So is there some sort of a $\beta$-function related to the mass? If yes, how would it look like? Or does the statement above only make sense when there is a coupling constant for $n \geq 3$ fields for some reason?

Answer 1

The concrete answer is actually in this post: Beta-function non-zero at classical level?

In short, the $\beta$-function for the mass is just:

$$\beta(m^2) = -2 m^2. \tag{1}$$

I am sure there are many ways to see that, e.g. the one described in the post. Here is another (trivial) way to check it. The renormalization group equation reads:

$$\left\lbrace p \frac{\partial}{\partial p} - \beta(m^2) \frac{\partial}{\partial m^2} + 2 - 2 \gamma_m \right\rbrace G^{(2)} (p;m^2) = 0\,, \tag{2}$$

where $g^{(2)} (p;m^2)$ is the exact propagator, i.e. (in Euclidean space):

$$G^{(2)}(p;m^2) = \frac{1}{p^2 + m^2}\,. \tag{3}$$

In a free theory the mass does not have an anomalous dimension, so $\gamma_m = 0$. The RGE can now be solved for $\beta(m^2)$, and we obtain the result mentioned above.