I am reading the 1986 paper [1] where Zamolodchikov proves the c-theorem and I would like to understand how equations (7a), (7b) and (8) are derived from the Callan-Symanzik equation.

For self-containedness, let me describe the beginning of the proof.

We start from a unitary renormalizable 2d (Euclidean) QFT invariant under translations and rotations, described by a local action $S = \int L(g,a,x) d^2 x$ which depends on coupling constants $g^i$ and on a UV length cutoff $a$. The goal is to construct a quantity $c(g)$ which decreases under RG flow.

Consider the scalar fields $\Phi_i = \partial L / \partial g^i$ and the beta function $\beta^i(g) = a \partial g^i / \partial a$ (sign warning: $a$ is a length cutoff). Let $T=T_{zz}$ and $\Theta=T_{z\bar{z}}$ be components of the stress energy tensor in complex coordinates $(z,\bar{z})=(x^1+ix^2,x^1-ix^2)$.

Then $\beta^i \Phi_i = \frac{1}{4} T_\mu^\mu = \Theta$ because $a(\partial g^i/\partial a) (\partial / \partial g^i)$ generates a scale transformation. The conservation law $\partial^\mu T_{\mu\nu} = 0$ reads $\partial_{\bar{z}} T + \partial_z \Theta = 0$.

Zamolodchikov then introduces the correlation functions $$ C = 2 z^4 \langle T(z,\bar{z}) T(0) \rangle \\ H_i = z^3 \bar{z} \langle T(z,\bar{z}) \Phi_i(0) \rangle \\ G_{ij} = z^2 \bar{z}^2 \langle \Phi_i(z,\bar{z}) \Phi_j(0) \rangle . $$

Combining $\Theta = \beta^i \Phi_i$ with $\partial_{\bar{z}} T + \partial_z \Theta = 0$ and the Callan–Symanzik equation, which I am not sure how to write in this case (I expect the scaling dimensions of $T$ and $\Phi_i$ to be unknown), Zamolodchikov then obtains $$ \frac{1}{2} \beta^i \frac{\partial}{\partial g^i} C = - 3 \beta^i H_i + \left(\beta^k \frac{\partial}{\partial g^k}\right) (\beta^i H_i) $$ and a similar equation relating $H_i$ and $G_{ij}$. I would like an explanation of how to derive this equation.

[1] A. B. Zamolodchikov (1986), "Irreversibility" of the Flux of the Renormalization Group in a 2D Field Theory, JETP Lett. 43: 730–732. (See pdf or nLab (identical).)


1 Answer 1


First, notice that dimensionless and rotation invariant correlation functions $f(g,a,x)$ only depend on $g$ and $x^2/a^2=z\bar{z}/a^2$, such that $-a\partial_af/2=z\partial_zf=\bar{z}\partial_{\bar{z}}f$.

Notice also that the Callan-Symanzik equation implies that any correlation function $f(g, a, x)$ in the regime $a\ll{x}$ is invariant under the renormalization group flow $0=df/dt=\beta^i\partial_if+a\partial_af/2$.

Hence, using $\partial_{\bar{z}}T+\partial_z\Theta=0$ and $\Theta=\beta^i\Phi_i$, we find that

$$ \frac12\beta^i\partial_iC(g,x)= \frac12\bar{z}\partial_{\bar{z}}\left(2z^4\left<T(x)T(0)\right>\right) = \left(z\partial_z - 3\right)\left(z^3\bar{z}\left<T(x)\Theta(0)\right>\right)$$ $$= 3\beta^iH_i-\beta^i\beta^k\partial_kH_i-\beta^k(\partial_k\beta^i)H_i, $$

and analogously

$$ \left(\bar{z}\bar{\partial} - 1\right)\left(z^3\bar{z}\left<T(x)\Theta(0)\right>\right) + \left(z\partial - 2\right)\left(z^2\bar{z}^2\left<\Theta(x)\Theta(0)\right>\right) = 0, $$

as desired.

  • $\begingroup$ How did you find the Callan-Symanzik equation? That factor 1/2 is a bit strange to me... Moreover, there is a global sign in the derivative of C with respect to the original paper. $\endgroup$ Jul 13 at 18:04
  • $\begingroup$ @MarcoVenuti I indeed made some deviations from the definitions/notations in the paper, because I didn't see another way to reconcile the equations. In particular I used $(g,a)\mapsto(R_tg,e^{t/2}a)$ and $c=C-4H-6G$, instead of $(g,a)\mapsto(R_tg,e^ta)$ and $c=C+4H-6G$, which leads to a factor $1/2$ and a minus sign. I believe this also has been done in the book String Theory (Polchinski, 1998, vol. II, p. 262) $\endgroup$
    – Johannes
    Jul 21 at 18:01

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