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The contribution to the one loop beta function for the WZW model

+ 3 like - 0 dislike
26 views

When the Wess-Zumino-Witten model

$$S_{WZW}=\frac{k}{4\pi}\int d^2 z \, \, \mathrm{Tr}[\partial u \bar{\partial}u^{-1} ]+ \frac{k}{12\pi}\int d^3 \sigma \epsilon^{ijk}\, \mathrm{Tr}[(u^{-1}\partial_i u)(u^{-1}\partial_j u)(u^{-1}\partial_k u)]$$

is expanded around a solution of the equations of motions $u=u_0 e^{iT^a\pi^a}$ one gets

$$S_{WZW}=\frac{k}{4\pi}\int d^2 z \lbrace \, \mathrm{Tr}[\partial u_0 \bar{\partial}u_0^{-1} ]+ \frac{1}{2}\partial_\mu \pi^a\partial ^\mu \pi^a \\ +\frac{1}{2}(\eta^{\mu\nu}-\epsilon^{\mu\nu}) \, \mathrm{Tr}\lbrace (u_0^{-1}\partial_\mu u_0)[T^a\pi^a,T^b\partial_\nu \pi^b]\rbrace + \mathcal{O}(\pi^3)\rbrace$$

The one loop renormalization diagram is like

enter image description here

In "Non-Perturbative Field Theory" by Y.Frishman and J.Sonnenschein (chapter 4.2 page 65) I read that the non vanishing contributions come only from the diagrams with both vertices proportional to $\eta^{\mu\nu}$ or to $\epsilon^{\mu\nu}$. Could someone explain me how one comes to that conclusion?

This post imported from StackExchange Physics at 2014-12-09 15:13 (UTC), posted by SE-user Anne O'Nyme
asked Nov 28, 2014 in Theoretical Physics by Anne O'Nyme (170 points) [ no revision ]

1 Answer

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One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is antisymmetric.

This post imported from StackExchange Physics at 2014-12-09 15:13 (UTC), posted by SE-user Frederic Brünner
answered Nov 28, 2014 by Frederic Brünner (1,060 points) [ no revision ]

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