# Gamma Matrices in Dimensional Regularization

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Prove that $tr\left(\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)=0$ when the spacetime dimension is not 4.

What I have tried:

We know that $\gamma_\alpha\gamma^\alpha=d\mathbb{1}$, so we can write:

$tr\left(\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)=\frac{1}{d}tr\left(\gamma_\alpha\gamma^\alpha\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)$

Then I thought I could commute $\gamma^\alpha$ past two gammas because if $\alpha\notin\left\{\mu,\,\nu,\,\rho,\,\sigma\right\}$, then $\left\{\gamma_\alpha,\,\gamma_\mu\right\}=0$ and somehow show that I get minus of what I started with, using the cyclicality of the trace and that $\left\{\gamma_5,\,\gamma_\mu\right\}=0$.

However, what I am not sure about is why can we always find such $\alpha$ so that $\alpha\notin\left\{\mu,\,\nu,\,\rho,\,\sigma\right\}$. I understand this is generally possible when $d\in\mathbb{R}\wedge d>4$, however, when $d\in\mathbb{C}$, this claim doesn't make any sense for me.

Can anyone provide a rigorous proof of this claim which avoids the hurdle I mentioned above?

This post imported from StackExchange Physics at 2014-09-14 21:38 (UCT), posted by SE-user PPR
edited Sep 15, 2014

This question is ambiguous, as the definition of $\gamma_5$ in dimensions other than 4 is not specified. Do you mean that $\gamma_5$ is the same product of gamma matrices as in 4d?

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The "proof" that is being looked for is, I suppose, along the following lines: the appearance of $\gamma_5$ means that you need something which changes sign under parity, and the 4 indices means you need a 4-index tensor. To get antisymmetry, you need the $\epsilon$ tensor, all of whose contractions with itself or with the metric vanish, so that in all dimensions greater than 4 you can't make anything. In 3d you can't make anything either, and in 2d, you can only make $\epsilon_{\mu\nu} \delta{\rho\sigma}$ and other index combinations on the bottom, but it doesn't work because the result has the same symmetry structure in $\mu\nu\rho\sigma$, and the $\epsilon$ and $\delta$ parts have mixed symmetry, so they will vanish in the appropriate combination with the right symmetry between all 4 indices. You can show this in a straightforward way.

But this question is really nonsense as posed, although it is probably derived from grad-school homework, because there is no unambiguous continuation of $\gamma_5$ to other dimensions. One tradition is to consider $\gamma_5$ as the same product of $i\gamma_0\gamma_1\gamma_2\gamma_3$ as in 4d, so that it is not a higher dimensional rotationally invariant construction. In this case, the statement is false, you get a non-rotationally invariant answer for the contraction in dimensions 5 and above.

But a better answer is that all this is really about voodoo. The process of dimensional regularization can only be understood and internalized after you understand that it is derived from an earlier method of propagator modification called analytic regularization. Analytic regularization simply modifies the $k^2$ at the bottom of Feynman propagators to $k^{2-\epsilon}$ in each field, and takes the limit as $\epsilon$ goes to zero. It makes sense non-perturbatively, as it is converting the Schwinger representation of random walks to Schwinger representations of Levy flights, and in Stochastic quantization it just replaces the Laplacians with fractional Laplacians appropriate to describing the propagation of Levy flights. It is very intuitive, and rather straightforward, but it gives a mess when you sit down to do actual calculations.

The reason it gives a mess is because the propagator modification modifies the perturbation integrals in a nonuniform way, with a different regularizing power of k at large k depending on the particle type in the loops, and the number of particle in the loop, and so on. This doesn't make any difference, ultimately the only thing that matters to get a finite analytically continuable answer is that you replace $d^4k$ with $d^4 k /k^\epsilon$ for some $\epsilon$. This is what dimensional regularization is all about. Once you make the loop integrals, you just introduce a little bit of extra falloff at large k, and make it smooth, and if you introduce it in the exponent, it's an analytic function of $\epsilon$, and you can throw away the pole parts with appropriate subtraction.

This process of decorating integrations with denominators is reinterpreted as continuing the dimension as a useful intuition trick for Veltman and 'tHooft. This way, they can make an integral table for various rotationally invariant k integrals integrated over a measure with an extra completely unimportant nondiverging constant factor C(a) multiplying the integration measure:

$$C(a) d^4 k \over k^a$$

where $C(a) = S(4-a)/S(4)$, where

$$S(d) = {\pi^{d/2}\over\Gamma(d/2+1)}$$

is the volume of a d dimensional sphere.

With this convention, you use Feynman parameters to make a diagram rotationally invariant, and then the integrations get a small denominator, and any rotationally integral gets a value which is a smooth interpolation of the values it takes in any dimension less or greater than 4.

But you don't need to continue any of the numerator decorations. You can do the numerator operations as if in 4d. The point is that you don't need to make sensible interpolations between the dimensions for any of the other quantities in the numerator, and focusing on this is an endless source of confusion for students. Best to just leave the numerator in 4d.

answered Sep 15, 2014 by (7,545 points)

With your answer as a prelude, what is the correct analytic continuation of $\gamma_5$ that makes it work in chiral theories, and what does it imply for the trace the OP was asking about?

Honestly, I don't remember anymore! I just remember that I had trouble with this as a student, then I sorted it out when I realized the above, and stopped thinking about consistent continuation between different dimensions altogether. I could work it out again, but it's in books, and OP should do it himself anyway.

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As shown in chapter 47 of the book by Srednicki, using the definition of $\gamma_5$ and the relations $(\gamma^i)^2=-1$ and $(\gamma^0)^2=1$, we can show that

$\text{Tr}(\gamma_5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\epsilon^{\mu\nu\rho\sigma}.$

We can now use a fundamental property of the Levi-Civita symbol, namely the fact that the number of its indices has to agree with the number of spacetime dimensions. If this is not the case, it vanishes. Hence, this proves the initial assertion.

This post imported from StackExchange Physics at 2014-09-15 08:35 (UCT), posted by SE-user Frederic Brünner
answered May 4, 2014 by (1,120 points)
So, should we neglect this kind of term when we do dimensional regularization with $d=4-\varepsilon$?

This post imported from StackExchange Physics at 2014-09-15 08:35 (UCT), posted by SE-user Melquíades
Even though it is formally zero, it might make sense to keep it in some cases. I am, however, not experienced enough in that area to make a general statement.

This post imported from StackExchange Physics at 2014-09-15 08:35 (UCT), posted by SE-user Frederic Brünner
@FredericBrünner, Thanks, but I am not so happy with this solution. To show that $Tr(\gamma_5\gamma^\mu\gamma^\nu \gamma^\rho\gamma^\sigma)=-4i\varepsilon^{\mu\nu\rho\sigma}$ I use the fact that if two indices in ($\mu,\nu,\rho,\sigma$) are the same, then we pick some new index, $\lambda\notin\{\mu,\nu,\rho,\sigma\}$, and then anti-commute it to the end to get the trace is minus itself and thus zero. However, this sort of trick is exactly what I was trying to avoid when I first posed my question because when $d\neq4$ I don't feel comfortable to pick such an index.

This post imported from StackExchange Physics at 2014-09-15 08:35 (UCT), posted by SE-user PPR
@PPR : There is a interesting discussion in this paper, chapter $7.4$ p.$213$, the precise definition of $\gamma^5$ p.$214$, and the formula $(7.22)$ p. $215$ might interest you.

This post imported from StackExchange Physics at 2014-09-15 08:35 (UCT), posted by SE-user Trimok

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