# Commutation relation of gamma matrices

While going through the third chapter of Peskin's QFT book, I am stuck at the following proof:

$$[\gamma^\mu, S^{\rho\sigma}] = (\mathcal{J}^{\rho\sigma})^\mu_{~\nu} \gamma^\nu,$$

where,

$$S^{\mu\nu} = \frac{i}{4} [\gamma^\mu, \gamma^\nu]\qquad;\qquad\qquad(\mathcal{J}^{\mu\nu})_{\alpha\beta} = i (\delta^\mu_\alpha \delta^\nu_\beta - \delta^\mu_\beta \delta^\nu_\alpha).$$

Using these above definitions, I found from R.H.S. of the first equation,

$$(\mathcal{J}^{\rho\sigma})^\mu_{~\nu} \gamma^\nu = i (\delta^{\rho\mu} \gamma^\sigma - \delta^{\sigma\mu} \gamma^\rho)$$

and from L.H.S. of the first equation,

$$[\gamma^\mu, S^{\rho\sigma}] = i (g^{\rho\mu} \gamma^\sigma - g^{\sigma\mu} \gamma^\rho).$$

Clearly there is a big difference between the two sides. One side has the metric tensors, whereas other side has delta functions. I think this will be fine if in the definition of $$(\mathcal{J}^{\mu\nu})_{\alpha\beta}$$ the delta functions are replaced by the metric tensors. It would be great if anyone could shade some light on this.

You forgot to lower the $$\mu$$ index in $${(\mathcal{J}^{\rho\sigma})^\mu}_\nu$$ before using its definition. Doing it makes $$g$$ appear as you need.
• Okay. Thanks a lot. I have a doubt though. In this case, what will be the difference between $g^{\mu\nu}$ and $\delta^{\mu\nu}$? Don't they both act as raising and lowering operators? I understand, there is difference in the sign. In Minkowski space, $g^{\mu\nu} = diag(1,-1,-1,-1)$ whereas, $\delta^{\mu\nu} = (1,1,1,1). But doesn't this difference create any observable difference? Oct 14, 2020 at 19:32 •$\delta^{\mu\nu}$never appears in Lorentz-covariant expressions. Oct 14, 2020 at 21:26 • But the above expression of$(\mathcal{J}^{\mu\nu})_{\alpha\beta}$is true and it contains$\delta$-function. So, I don't understand your point/objection. Oct 14, 2020 at 21:40 • What I meant is that only$g_{\mu\nu}$,$g^{\mu\nu}$, and$\delta^\mu_\nu$are allowed. When both indices are upper or lower, it’s the metric or inverse metric; when the indices are mixed, it’s the Kronecker delta. Oct 15, 2020 at 3:51 • And they are all linked by$g_{\mu\rho}g^{\rho\nu} = \delta^\nu_\mu$, which is how you define$g^{\mu\nu}\$ Oct 15, 2020 at 19:21