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Why is an RG fixed point scale invariant?

Physics Asked by Jiahao Mao on November 28, 2020

I cannot understand why people say RG fixed point is scale invariant.

Scale invariant means the action $S[phi]$ of the theory is invariant under scale transformation like $phi(x)tolambda^{-Delta}phi(lambda x)$. Fixed point of RG is a theory with action invariant under RG transformation (i.e. integrating out fast mode followed by a scale transformation). Scale transformation $neq$ RG transformation. Then, why RG fixed point is scale invariant?

One Answer

Below I give a general sketch of how the arguments go. For the details, see the short book written by Zamolodchikov and Zamolodchikov "Conformal Field Theory and Critical Phenomena in Two-Dimensional Systems." They are only interested in the two-dimensional case because that's where the C-Theorem can be proved, but the part of the argument I describe here needs to assumptions to be made about the dimension of the theory.

Suppose that the operators $Phi_a$ form a basis of operators in the theory. Then it follows that the trace of the stress tensor may be represented by $$ T^mu_mu=beta^aPhi_a $$ for some coefficients $beta^a$ which have been suggestively named.

By playing some games with Ward-like (not quite symmetries, but good enough to obtain an identity) identities derived from translation invariance and applying the definition of scale transformations (we do not demand scale invariance, but can still discuss the operator), we can put our results together into the Callan-Symanzik equation: $$ beta^afrac{partial}{partial g^a}langle A_1cdots A_Nrangle=sum_{k=1}^Nlangle A_1cdots(x^mu_kpartial_mu+gamma_k(g))A_kcdots A_Nrangle $$ where the $A_k(x_k)$ are some local operators and $gamma_k(g)$ is some coupling-dependent scaling dimension (the renormalized scaling dimension of the field). The $g$'s are the couplings to the basis operators $Phi_a$.

From this equation, the coefficients $beta^a$, miraculously, may be identified as the beta functions associated with the RG flow. At a fixed point, by definition, the beta functions vanish, so $beta^a=0$. This now immediately says some interesting things about how the scale operator, $(x^mupartial_mu+gamma)$ which generates scale transformations, acts on fields in the theory. But more than that, we originally defined these coefficients $beta^a$ such that $$ T^mu_mu=beta^aPhi_a. $$ So, the vanishing of the beta functions implies the vanishing of the trace of the stress tensor.

Vanishing of the trace of the stress tensor may be shown to always imply conformal invariance (though the converse does not always hold). Scale transformations are one type of conformal transformation, and hence a traceless stress tensor also implies scale invariance.

Correct answer by Richard Myers on November 28, 2020

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