# Contracting gamma matrices with explicit indices

So I was calulating the matrix element of an interaction and arrived at the following contraction

$$\gamma^\mu_{ab}\gamma_{\mu\,cd}$$

With $$a,b,c,d$$ spinor indices that are never contracted with each other through other matrices. I know that in $$D$$ dimensions we have the representation independent result

$$\gamma^\mu\gamma_\mu=D$$

But that's just when the matrices are contracted with each other, which in my case they aren't. Is there a representation independent result for this contraction I need?

In Schwartz there is the relation

$$\bar u_s(p)\gamma^\mu u_{s'}(p)=2\delta_{ss'}p^\mu$$

So there must be something similar to what I want.

• You’re not doing any averaging over spins? Commented Dec 9, 2019 at 0:49

To derive an expression for $$\sum_\mu\gamma^\mu_{ab}\gamma_{\mu\ cd}$$, note that this is equivalent to requesting an expression for $$\sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu)$$ for aribtrary matrices $$X$$ and $$Y$$. To derive such an expression, express $$X$$ and $$Y$$ as sums of products of Dirac matrices (with arbitrary coefficients), and use the anticommutation relations to get an expression for $$Y' := \sum_\mu\gamma^\mu Y\gamma_\mu.$$ The keywords Fierz identities should lead you to more information about manipulations like this. The book Supergravity by Freedman and van Proeyen (2012) includes a relatively thorough treatment of such identities.

• That does seem to work out fine, thanks! However, the Fierz decomposition exploits the fact that there are 16 4x4 matrices, so it should only work on spacetimes for which the Clifford algebra has a 4-dimensional representation. Commented Dec 9, 2019 at 18:22
• Is there a representation-free equivalent of this? Commented Dec 9, 2019 at 18:22
• @GabrielGolfetti This approach generalizes to $N$-dimensional spacetime for arbitrary $N$ and arbitrary signature. For $N=2n$, there are $2^N$ linearly independent products of $\gamma$-matrices (including the identity), and the matrices have size $2^n\times 2^n$. For $N=2n+1$, we have to be careful because the matrices still have size $2^n\times 2^n$, and the product $\prod_\mu \gamma^\mu$ is $\pm 1$ times the identity matrix, so only half of the products of $\gamma$-matrices are linearly independent. Commented Dec 10, 2019 at 3:13
• @GabrielGolfetti I don't think a completely representation-free approach is possible, because the requested quantity $\sum_\mu \gamma^\mu\otimes\gamma_\mu$ is not invariant under similarity transforms of the $\gamma$s. However, it is possible to organize the derivation in such a way that representation-dependent details enter only at the very end, as in the answer by MadMax. Commented Dec 10, 2019 at 3:26

Let's expand on @Chiral Anomaly's answer via taking a look at a concrete example.

In 2D, one would have the traceless property: $$\sum_\mu\gamma^\mu\gamma^\sigma\gamma_\mu = 0$$ which can be verified easily, e.g. $$\sum_\mu\gamma^\mu\gamma^0\gamma_\mu=\gamma^0\gamma^0\gamma_0 + \gamma^1\gamma^0\gamma_1 = (\gamma^0\gamma_0)\gamma^0 - (\gamma^1\gamma_1)\gamma^0= \gamma^0- \gamma^0 = 0$$ This identity is representation-independent, since it only invokes the anti-commuting properties of the Gamma matrices. The identity can be extended to any vector $$l = \sum_\sigma l_\sigma \gamma^\sigma$$: $$\sum_\mu\gamma^\mu l\gamma_\mu = 0$$

The identity implies that the QED vacuum polarization is traceless: $$\sum_{\mu\nu}\Pi^{\mu\nu}(p)g_{{\mu\nu}} \sim \sum_{\mu\nu}\int_l Tr\left(\gamma^\mu\frac{1}{\not l}\gamma^\nu\frac{1}{\not l+\not p}\right)g_{{\mu\nu}} = \int_l \frac{Tr\left(\sum_{\mu}\gamma^\mu{\not l}\gamma_\mu({\not l+\not p})\right)}{l^2(l+p)^2} = 0$$

But for 2D massless QED (Schwinger model), it turns out that a Ward identity compatible vacuum polarization is NOT traceless $$\sum_{\mu\nu}\Pi^{\mu\nu}(p)g_{{\mu\nu}} \sim \sum_{\mu\nu}(g^{{\mu\nu}} - \frac{p^\mu p^\nu}{p^2})g_{{\mu\nu}} = 2 - \frac{p^2}{p^2} = 1 \neq 0$$ Confused? Welcome to the wonderful QFT club which is purportedly self-consistent!

Hint: $$0*\infty \neq 0$$ Infinities and dimensional regularization can do magic whereby Ward identity $$\sum_{\nu}p_\nu \Pi^{\mu\nu}(p) \sim p^{{\mu}} - \frac{p^\mu p^2}{p^2}=0$$ is preserved at the cost of tracelessness.

For $$D=4$$, let's say you want to calculate $$\sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu)$$, and Y can be expressed as: $$Y = AI + B^\mu\gamma_\mu + C^{\mu\nu}\gamma_\mu\gamma_\nu + D^\mu\gamma_\mu \gamma_5 + E \gamma_5,$$ then $$Y' := \sum_\mu\gamma^\mu Y\gamma_\mu \\ =4AI -2 B^\mu\gamma_\mu +2 D^\mu\gamma_\mu \gamma_5 -4E \gamma_5.$$ Note that the antisymmetric portion $$C^{\mu\nu}\gamma_\mu\gamma_\nu$$ drops out for $$D=4$$, as opposed to vector portion dropping out in 2D.

The last step is to calculate: $$\sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu)= \text{trace}(XY')$$