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I have seen this mentioned in various papers/lecture notes/Stack Exchange questions [1-4]: $$\Theta = \beta^i(g) \Phi_i.$$

Here $\Theta$ is the trace of the stress tensor, $g=\{g^i\}$ is the set of all (dimensionless) coupling constants of the theory, and the fields $\Phi_i$ are defined as (following [3]): \begin{equation} \Phi_i(g,x) = \frac{\partial}{\partial g^i}\sigma(g,a,x) \end{equation} where sigma is defined as the action density, i.e. $ S(g,a) = \int \sigma(g,a,x) d\mathbf{x}$, and $a$ is a UV length cutoff.

This answer proves this in the case of a conformal field theory. But it seems like all the other sources are stating this as a general property of RG flow. In fact Zamolodchikov [3] states: "The exact meaning of the assertion that a field theory is renormalizable is that for all g the field $\Theta$ can be expanded in (the above mentioned) basis: $\Theta = \beta^i(g)\Phi_i$"


So my questions are:

  1. If this identity is true for all $g$, as Zamolodchikov states, that means that at a fixed point $g=g^*$, satisfying $\beta(g^*)=0$, we must have $\Theta=0$. But this is equivalent to conformal invariance. This line of argument can't be correct because then this would be a trivial proof of scale invariance $\implies$ conformal invariance, without any of the assumptions that usually go into the proof (e.g. Polchinski showed that we need to assume a discrete spectrum of scaling dimensions in $d=2$ in order to conclude SI $\implies$ CI). How do we reconcile this with the identity in question?

  2. What is the correct way to interpret this identity and how would we go about writing a proof for the same?


References:

[1] (between eqs 5.5 and 5.6) https://arxiv.org/abs/1302.0884

[2] (eq 6.50) https://arxiv.org/abs/1511.04074

[3] (eq 5) Zamolodchikov, a B. 1986. “Irreversibility of the Flux of the Renormalization Group in a 2D Field Theory.” JETP Lett. 43 (1986) 730-732, Pisma Zh.Eksp.Teor.Fiz. 43 (1986) 565-567

[4] Stack exchange question: Zamolodchikov's c-theorem paper

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  • $\begingroup$ your assertion in 1 is correct, but the action appearing is not the classical action but the quantum effective action. this removes any discrepancies with the statements in Polchinski etc. $\endgroup$ Commented May 6, 2020 at 8:21
  • $\begingroup$ @Wakabaloola do you mean that the identity in question is not true as an operator equation but only as a classical equation? $\endgroup$
    – Arkya
    Commented May 6, 2020 at 17:54
  • $\begingroup$ all i’m saying is that the action appearing is the quantum effective action. so it’s the legendre transform of the generating function of renormalised connected greens functions. it formally contains all quantum corrections $\endgroup$ Commented May 6, 2020 at 17:58
  • $\begingroup$ Do you mean $\int dx \sigma(g,a,x)$ is the effective 1PI action? $\endgroup$
    – Arkya
    Commented May 6, 2020 at 19:10
  • $\begingroup$ yes, that's it. $\endgroup$ Commented May 6, 2020 at 19:14

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