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In Peskin & Schroeder, p.523, they give the diagram contributing to the gluon self-energy that arises from the 3-gluon vertex, and they claim that the $1/2$ factor is a symmetry factor:

Gluon loop contributing to 1-loop gluon self-energy

How can this symmetry factor be calculated? I get all sorts of things except the correct result. The relevant part of the Lagrangian is:

$$\mathcal{L} = -\frac{1}{4} F^a_{\mu\nu} F_a^{\mu\nu}\tag{1}$$

with

$$F^a_{\mu\nu} = \partial_\mu A^a_\nu - \partial_\nu A_\mu^a + gf^{abc} A_\mu^b A_\nu^c \tag{2}$$

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The matrix element follows from the Feynman rules \begin{align} \mathcal{M} &= \frac{1}{2} \int \frac{d^4p}{(2\pi)^4} \frac{-i\delta^{cd}g_{\rho\tau}}{p^2} \frac{-i\delta^{ef}g_{\sigma\kappa}}{(p+q)^2} \nonumber\\ &\qquad\times g f^{aec} \big[ g^{\mu\sigma} (q+p+q)^\rho + g^{\sigma\rho}(-p-q-p)^\mu + g^{\rho\mu}(p-q)^\sigma \big] \nonumber\\ &\qquad \times g f^{bdf} \big[ g^{\nu\tau} (-q+p)^\kappa + g^{\tau\kappa}(-p-p-q)^\nu + g^{\kappa\nu}(p+q+q)^\tau \big]\nonumber\\ &= \frac{1}{2} \int \frac{d^4p}{(2\pi)^4} \frac{-i}{p^2} \frac{-i}{(p+q)^2} g^2 f^{adc} f^{bcd} \big[ g^{\mu\kappa}(2q+p)^\tau + g^{\kappa\tau}(-2p-q)^\mu + g^{\tau\mu}(p-q)^\kappa \big] \nonumber\\ &\qquad \times \big[ \delta^\nu_\tau (-q+p)_\kappa + g_{\tau\kappa} (-2p-q)^\nu + \delta^\nu_\kappa (p+2q)_\tau \big]\nonumber\\ &= \frac{1}{2} \int \frac{d^4p}{(2\pi)^4} \frac{-i}{p^2} \frac{-i}{(p+q)^2} g^2 f^{acd} f^{bcd} \mathcal{N}^{\mu\nu} (p;q) \label{eq:ymkhgdd} \end{align} where \begin{align} \mathcal{N}^{\mu\nu}(p;q) & =\big[ g^{\mu\rho}(q-p)^\sigma + g^{\rho\sigma}(2p+q)^\mu + g^{\mu\sigma}(-p-2q)^\rho \big] \nonumber\\ & \times \big[ \delta^\nu_\rho (p-q)_\sigma + g_{\rho\sigma} (-2p-q)^\nu + \delta^\nu_\sigma (p+2q)_\rho \big] \end{align} I trust you can add the indices $(e,\sigma)$, $(f,\kappa)$, $(c,\rho)$ and $(d,\tau)$ in the appropriate places.

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  • $\begingroup$ I think that this answer is not addressing my question: where does the symmetry factor $1/2$ come from? $\endgroup$ – Jxx Sep 12 '19 at 12:06
  • $\begingroup$ Isn’t the symmetry factor just the same as in qed photon or $\phi^3$ self-energy? $\endgroup$ – Oбжорoв Sep 12 '19 at 21:01
  • $\begingroup$ Is it? I mean QED has no 3-photon vertex, and in $\phi^3$ there is a $1/3!$ that compensates the $3!$ arising from the diagrams. There must be an obvious subtlety that I am not seeing. $\endgroup$ – Jxx Sep 12 '19 at 21:05
  • $\begingroup$ You are right. But doesn’t it come from the fact that you can have a gluon or an anti-gluon in the loop? (And a gluon is identical to its antiparticle) $\endgroup$ – Oбжорoв Sep 13 '19 at 9:32
  • $\begingroup$ Well, that's the question. When I do the counting, I find that there are $(3!)^2=36$ different ways of coupling the 3-vertices to the external propagators, and this number should be divided by $2$ because of 2nd order in perturbation theory. But this gives me $18$, and not the $2$ of Peskin. $\endgroup$ – Jxx Sep 13 '19 at 9:39
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So I asked one of my professors about the issue, and he was able to explain me the symmetry factor. We have to rewrite the relevant part of the action in the following form:

$$S \propto -\int d^4p \int d^4q \int d^4k \frac{i}{3!} \tilde{A}_\mu^a(p) \tilde{A}_\nu^b(q) \tilde{A}_\rho^c(k) V^{abc}_{\mu\nu\rho}(p,q,k) \tag{3}$$

with $V^{abc}_{\mu\nu\rho}(p,q,k)$ the corresponding Feynman rule in momentum space. When we use this expression and expand, we get a $1/2$ coming from the expansion, a $(3!)^2$ coming from the different ways of coupling the two vertices and a $1/(3!)^2$ coming from expanding $(3)$ to second order. Thus we get the symmetry factor $1/2$ as claimed in Peskin.

If someone is interested in the details, take a look at chapter $8.1$ in the book by Ramond.

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  • 1
    $\begingroup$ IIRC, usually one proves that the symmetry factor is $1/|G|$, where $|G|$ is the order of the automorphism (symmetry) group $G$ of the diagram. In your diagram you have two equivalent propagators (the two in the loop, the external vertices are treated as distinct), so you have that $G$ is simply $\mathbb{Z}_2$ that exchanges them, and thus the symmetry factor is 1/2. $\endgroup$ – Peter Kravchuk Sep 19 '19 at 13:14
  • $\begingroup$ Say, if you had a self-energy in $\phi^4$, which is two loops and there is a configuration of propagators that looks like $\theta$, you have $G=S_3$ and $|G|=3!=6$. $\endgroup$ – Peter Kravchuk Sep 19 '19 at 13:15

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