- #1
Manu_
- 12
- 1
- Homework Statement
- 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).
- Relevant Equations
- $$\mathcal{L} = \bar{\Psi}_{i}(i\not{\partial} )\Psi_{i}-g\sigma \bar{\Psi}_i\Psi_i-\frac{1}{2}\sigma^2$$
$$\delta_{Z_{\Psi}}=(ig)^2\int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i\not{k}}{k^2}\frac{i}{(k+p)^2}]=0 $$
$$\delta_{Z_{\sigma}}=-N(ig)^2\int \frac{d^{d}k}{(2\pi)^d}\frac{i\not{k}}{k^2} \frac{i(\not{k}+\not{p})}{(k+p)^2} =-Ng^2 \int \frac{d^{d}k}{(2\pi)^d}\frac{1}{(k+p)^2} =-Ng^2\frac{i}{4\pi\epsilon}+finite$$
$$\delta_{g}=(ig)^3 \int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i(\not{k}+\not{p_1})}{(k+p_1)^2} \frac{i(\not{k}+\not{p_2})}{(k+p_2)^2}] = ig^3 \int \frac{d^{d}k}{(2\pi)^d}\frac{-2k^2}{(k+p_1)^2 (k+p_2)^2}+...=-2ig^3\frac{i}{4\pi\epsilon} $$
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: ## $$\delta_{Z_{\Psi}}=(ig)^2\int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i\not{k}}{k^2}\frac{i}{(k+p)^2}]=0 $$
For ##\sigma:## $$\delta_{Z_{\sigma}}=-N(ig)^2\int \frac{d^{d}k}{(2\pi)^d}\frac{i\not{k}}{k^2} \frac{i(\not{k}+\not{p})}{(k+p)^2} =-Ng^2 \int \frac{d^{d}k}{(2\pi)^d}\frac{1}{(k+p)^2} =-Ng^2\frac{i}{4\pi\epsilon}+finite$$
- 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!
- 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: ## $$\delta_{Z_{\Psi}}=(ig)^2\int \frac{d^{d}k}{(2\pi)^d} Tr[\frac{i\not{k}}{k^2}\frac{i}{(k+p)^2}]=0 $$
For ##\sigma:## $$\delta_{Z_{\sigma}}=-N(ig)^2\int \frac{d^{d}k}{(2\pi)^d}\frac{i\not{k}}{k^2} \frac{i(\not{k}+\not{p})}{(k+p)^2} =-Ng^2 \int \frac{d^{d}k}{(2\pi)^d}\frac{1}{(k+p)^2} =-Ng^2\frac{i}{4\pi\epsilon}+finite$$
- 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!
