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?

  • $\begingroup$ 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

1 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.