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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Buzz Jun 1 at 0:27
  • $\begingroup$ 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? $\endgroup$ – Free_ion Jun 5 at 11:21
  • $\begingroup$ 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. $\endgroup$ – Buzz Jun 5 at 20:19

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