Contracting gamma matrices with explicit indices

Question

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.

Answer 1

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.

Answer 2

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: $$ \sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu)= \text{trace}(XY') $$