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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?

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  • $\begingroup$ Yes you can treat the mass term as an interaction. I haven't done the calculation, but I am pretty sure that the $\beta$ function would be proportional to the mass (squared). $\endgroup$ Commented Jan 16, 2021 at 12:58
  • $\begingroup$ @Oбжорoв Thanks! I would do the computation, but I am not sure where to start. The free massive two-point function is known exactly and is a Bessel function in position space. Can I read the exact $\beta$-function off that expression somehow? $\endgroup$
    – Pxx
    Commented Jan 16, 2021 at 14:18
  • $\begingroup$ I mean, it is clear that the 1st order correction to a free massless scalar field is $-m^2/p^4$, which can easily be obtained either by expanding the exact propagator $1/(p^2+m^2)$ (I'm in Euclidean space) or by writing down the (rather trivial) diagram. But how is there a dependence on the scale? $\endgroup$
    – Pxx
    Commented Jan 16, 2021 at 16:20
  • $\begingroup$ @ChiralAnomaly Which part exactly? (the link brings me to the actual question by the way, but I found the post you meant) $\endgroup$
    – Pxx
    Commented Jan 17, 2021 at 0:48
  • $\begingroup$ @Jxx Wow, sorry about the incorrect link in my first comment. I guess you figured it out, but here's the correct link for convenience: physics.stackexchange.com/q/303611. I was referring to this sentence in AccidentalFourierTransform's answer: "we are not considering on-shell parameters (e.g., pole masses), as these are defined at a particular (fixed) energy scale." It may not completely answer what you're asking, but I thought it might be a helpful perspective. $\endgroup$ Commented Jan 17, 2021 at 2:04

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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.

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  • $\begingroup$ Usually in the Callan-Symanzik equation we have a term $\mu\frac{\partial}{\partial\mu}$. Why is this term replaced with $p\frac{\partial}{\partial p}$? Also, where does the $2$ comes from? $\endgroup$
    – A.Dunder
    Commented Nov 30, 2021 at 10:50

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