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Is it possible to dimensionally regularize an amplitude which contains the totally antisymmetric Levi-Civita tensor $\epsilon^{\mu\nu\alpha\beta}$?

I don't know if it's possible to define $\epsilon^{\mu\nu\alpha\beta}$ in e.g. $$d-\eta$$-dimensions where $\eta$ is considered small and which we set to zero in the end.

So what are your thoughts?

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The chiral anomaly falls into this category, and this is usually cited as an example where dim reg fails (because of the epsilon problem), but allegedly there is a workaround (but paywalled, so I haven't read it) –  twistor59 Apr 12 '13 at 6:44
    
Thanks for the link...I'll check it out. –  Faraday Apr 12 '13 at 11:23

2 Answers 2

up vote 2 down vote accepted

This problem was already mentioned in the original 't Hooft-Veltman article and solved by Breitenlohner and Maison. This solution is known by the name "HVBM scheme" (after 't Hooft, Veltman, Breitenlohner and Maison).

A clear description of this regularization procedure is given for example in the following dissertation by Barbara Jäger. It consists basically of splitting the metric into a $4$-dimnsional and $(d-4)-$ parts, assuming the Levi-Civita tensor to have non-vanishing components only in the 4-dimensional subspace. In addition $\gamma_5$ is assumed to anti-commute with the $\gamma$ matrices of the 4-dimnsional subspace and commutes with the others. This procedure leads to consistent Ward identities.

As mentioned in Jäger 's thesis, this procedure leads to a higher complexity in the computation of the Feynman diagrams, but there exist computer algebra programs implementing this scheme.

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Awesome...that's exactly what I needed. Thanks a bunch. –  Faraday Apr 15 '13 at 10:54

this fails since the tensor $ \epsilon ^{a,b,c,d} $ is diemnsion dependent

however in the case of zeta regularization of integrals $ \int_{a}^{\infty}dx x^{m-s} $ with 's' a regulator we can overcome this problem

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