# $\beta$ function for the Gross-Neveu model

In the Peskin & Schroeder textbook, the $$\beta$$-function for the Gross-Neveu model is discussed in problem 12.2. After computing it, I have tried checking my results with some solutions found online. My problem is that they all disagree among each other (something quite recurrent for this book actually). I have basically followed the original paper from Gross and Neveu:

• We start from the Lagrangian $$\mathcal{L} = \bar{\Psi}_{i}(i\not{\partial} )\Psi_{i}-g\sigma \bar{\Psi}_i\Psi_i-\frac{1}{2}\sigma^2$$.
• We compute the field strength renormalisation for $$\Psi$$.

For $$\Psi: \quad$$ $$\delta_{Z_{\Psi}}=(-ig)^2\int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i\not{k}}{k^2}\frac{i}{(k+p)^2}]=0$$

For $$\sigma: \quad$$ $$\delta_{Z_{\sigma}}=-N(-ig)^2\int \frac{d^{d}k}{(2\pi)^d}\frac{i\not{k}}{k^2} \frac{i(\not{k}+\not{p})}{(k+p)^2} =-Ng^2 \int \frac{d^{d}k}{(2\pi)^d}\frac{1}{(k+p)^2} =-Ng^2\frac{i}{4\pi\epsilon}+finite$$

• We compute the coupling renormalisation (though in the Gross-Neveu original paper, the authors argue the vertex need not be renormalised. That is a point I don't get)

$$\delta_{g}=(-ig)^3 \int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i(\not{k}+\not{p_1})}{(k+p_1)^2} \frac{i(\not{k}+\not{p_2})}{(k+p_2)^2}] = -ig^3 \int \frac{d^{d}k}{(2\pi)^d}\frac{-2k^2}{(k+p_1)^2 (k+p_2)^2}+...=2ig^3\frac{i}{4\pi\epsilon}$$

• We use relation (12.53)

$$\beta (g) = M\frac{\partial}{\partial M}\left(-\delta_g +\frac{1}{2}g\sum_i \delta_{Z_i} \right)=\frac{2g^3}{4\pi}+g\frac{-Ng^2}{4\pi}=-\frac{g^3}{4\pi}(N-2)$$

It is asymptotically free, but I do fully trust my result. Any comment welcome!

Also, as a second question, the $$\beta$$ function derived by Gross & Neveu is $$\beta(g)=-\frac{g^2N}{2\pi}$$. The dependence on g is only $$g^2$$ and not $$g^3$$. Do they use a different definition of the $$\beta$$ function? If so, what is the advantage of this other definition?

• When the $\beta$-function was first introduced by Gell-Mann and Low, they actually used a related quantity $\Psi(g)=\beta(g)/g$, and there was, early on, some disagreement about whether the renormalization group function should include that extra power of $g$ or not. Personally, I actually think $\Psi(g)$ is a little more useful than $\beta(g)$, since the absolute sign of $\Psi(g)$ determines whether the is asymptotically free. – Buzz Jun 1 at 0:27
• Thanks for your clarifying this definition, Buzz. However, I am not sure I see the difference between the two definitions. The sign of $\beta$ also determines asymptotic freedom, right? – Free_ion Jun 5 at 11:21
• The sign of $\beta(g)$ also includes the sign of $g$ itself. So if you define the charge $e=-|e|$ to be negative (as Peskin & Schroeder do), then $\beta(e)$ for QED is actually negative, even though the theory becomes more strongly coupled at higher momenta. – Buzz Jun 5 at 20:19