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, 2019 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, 2019 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, 2019 at 20:19

1 Answer 1


I guess you have been comparing Hitoshi Murayama's solution vs. Zhong-Zhi Xianyu's. Both solutions have the right (and essentially the same) approach. Unfortunately, both contain some critical errors here and there.

Xianyu's solution first covers the two-point function before treating the four-point function. This is great. We first get $\gamma$ out of the way.

As for the Feynman rule for the vertex, those $\epsilon$'s should really be $\delta$'s. They are nothing but identity matrices used to track which propagators should be connected to which vertices.

If you see something like $\delta_{\alpha\beta} (\frac{i}{\not{k}})_{\beta\gamma} \delta_{\gamma \delta} (\frac{i}{\not{k}})_{\delta\alpha} $, it is equivalent to $-\mathrm{Tr}(\frac{i}{\not{k}} \frac{i}{\not{k}}) = -2 \frac{1}{k^2}$. (The negative sign is because now we have a Fermion loop, while the factor of 2 is from the trace in two dimensions.)

If you see something like $\delta_{\alpha\beta} (\frac{i}{\not{k}})_{\beta\gamma} \delta_{\gamma \delta} (\frac{i}{\not{k}})_{\delta\epsilon} $, it is equivalent to $\frac{i}{\not{k}} \frac{i}{\not{k}}$, which is then meant to be sandwiched between $\bar{u}(p_1)$ and $u(p_2)$. Compared to the previous situation, there are two important differences. First, the sign is the opposite. Second, we don't have the extra factor of 2.

With this in mind, you should be able to see how to interpret (12.8) in Xianyu's solution. The integrand $(A+B)\frac{i}{\not{k}}(C+D)\frac{i}{\not{k}}$ can be decomposed into four parts.

  • $A\frac{i}{\not{k}}C\frac{i}{\not{k}}$: This is basically (58) in itoshi Murayama's solution.
  • $A\frac{i}{\not{k}}D\frac{i}{\not{k}}$: This is basically (59) in itoshi Murayama's solution.
  • $B\frac{i}{\not{k}}C\frac{i}{\not{k}}$: This is basically (59) in itoshi Murayama's solution.
  • $B\frac{i}{\not{k}}D\frac{i}{\not{k}}$: This is basically (61) in itoshi Murayama's solution (t-channel).

N.B.: I am not claiming (58) is error-free. In fact, the factor 3/2 should be 2 instead in the numerator of the second term in the integrand. (See (10.33) in Peskin and Schroeder for reference.) This error propagates to (62) and a bunch of other equations. Once this is fixed, you will get the desired $(N-1)$ instead of $(N-2)$ in the final result.

The vertex Feynmann rule given in Xianyu's solution is correct. As for how to derive it, I would refer to Section 4.7 in Peskin and Schroeder. You can write something like $\langle 1(p') 2(k') | \bar{\psi}_1(x)\psi_1(x)\bar{\psi}_1(x)\psi_1(x) \bar{\psi}_1(y)\psi_1(y) \bar{\psi}_2(y)\psi_2(y) | 1(p) 2(k) \rangle$ and draw lines to connect each of them. (1 and 2 are particle types, while $p$, $k$, etc. are momenta.) Depending on how you draw those lines, this particular one here can either contain a loop or not.

When you expand $(\bar{\psi}_i\psi_i)^2$ in the Langrangian, you get $N + N(N-1)/2$ terms for each vertex. Although a term like $2\bar{\psi}_1\psi_1\bar{\psi}_2\psi_2$ has an extra term of 2 compared to $\bar{\psi}_1\psi_1\bar{\psi}_1\psi_1$. The latter is in fact twice as large because it can be connected to propagators in four different ways.

A few extra notes about Xianyu's solution.

  • If you wonder what those $\gamma^\mu \gamma_\mu$ terms are about in e.g. (12.8), it has nothing to do with summation over $\mu$. It represents in fact $(\not{k}+\not{q})(\not{k}+\not{p})$ that needs to be sandwiched between $\bar{u}(q)$ and $u(p)$. Murayama's solution is a lot clearer in this regard.
  • The factor $1/2$ in (12.8), (12.9), and (12.10) really should not be there. In fact, a nice feature about this theory, just as Yukawa theory, is that you don't need to worry about symmetry factors.
  • $\begingroup$ A few questions about your answer -- (1) why don't terms like $\bar{\psi}_1\psi_1\bar{\psi}_1\psi_1$ vanish by fermi statistics? (2) in the Feynman rules from Xianyu (with all $\delta$'s rather than $\epsilon$'s), I see the two groups of delta functions come from matching in arrows with different out arrows. Why is there no relative sign between these terms from re-ordering the $\psi$'s? (3) how do you see that the 1/2 in the Lagrangian doesn't end up in the Feynman rule for the vertex? $\endgroup$
    – mkn
    Feb 13 at 20:05
  • $\begingroup$ @mkn It was a while ago that I answered the question. So the honest answer is that I don't remember. But for (1), keep in mind that each $\psi$ is a vector. You can expand $\bar{\psi}\psi$ as a sum of individual components and you will get your answer. $\endgroup$ Feb 13 at 23:35
  • $\begingroup$ Thanks, expanding the vector led me to realize what the issues are. I have just sorted out the rest of the issues in this answer. $\endgroup$
    – mkn
    Feb 17 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.