# How to interpret the extra indices of the generators of the Lorentz algebra in Peskin & Schroeder?

In Peskin & Schroeder p.39 they introduce the 4x4-matrices

$$\left(\mathcal{J}^{\mu\nu}\right)_{\alpha\beta} = i \left(\delta^{\mu}_{\;\alpha} \delta^{\nu}_{\;\beta} - \delta^{\mu}_{\;\beta}\delta^{\nu}_{\;\alpha}\right) \tag{3.18}$$

which they "pull out of a hat" and which represent the Lorentz algebra, that is they satisfy

$$\left[ \mathcal{J}^{\mu\nu} , \mathcal{J}^{\rho\sigma} \right] = i \left( \eta^{\nu\rho} \mathcal{J}^{\mu\sigma} - \eta^{\mu\rho}\mathcal{J}^{\nu\sigma} - \eta^{\nu\sigma}\mathcal{J}^{\mu\rho} + \eta^{\mu\sigma}\mathcal{J}^{\nu\rho} \right)\tag{3.17}$$

I am having difficulties interpreting the objects $\left( \mathcal{J}^{\mu\nu} \right)_{\alpha\beta}$. In his script about QFT, David Tong describes p.82 the objects $\left(\mathcal{J}^{\mu\nu}\right)$ as the six antisymmetric matrices describing the six transformations of the Lorentz group: 3 rotations $\mathbf{J}=\left(J_x,J_y,J_z\right)$ and 3 boosts $\mathbf{K}=\left(K_x,K_y,K_z\right)$, where as I understand it each $J_i$ and $K_i$ is a matrix. Then he says that the indices $\alpha\beta$ refer to the components of the matrices $\left(\mathcal{J}^{\mu\nu}\right)$, and gives as an example

$$\left(\mathcal{J}^{01}\right)^{\alpha}_{\;\beta} = \left[ \begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right]$$

which generates boosts in the $x^1$ direction. So far so good. I am interested into inserting (3.18) in (3.17) in order to check that the matrices satisfy the Lorentz algebra, however I am confused by the four indices. It seems to me that $\left(\mathcal{J}^{\mu\nu}\right)_{\alpha\beta}$ and $\left(\mathcal{J}^{\mu\nu}\right)$ cannot be the same objects really, so how do I retrieve the latter?

• – Qmechanic Sep 9 '18 at 19:45
• Tong and Peskin are using different definitions of $\mathcal{J}^{\mu\nu}$. You should multiply Tong's one by $i$ and lower the upper index to get Peskin's. Other than that, the indices stand for row and column number: $0$ stands for $x^{0}$'s row/column, hence the first row/column, $1$ stands for $x^{1}$'s row/column, hence the second row/column, and so on. – Giorgio Comitini Sep 9 '18 at 20:01
• @GiorgioComitini thanks for your answer. Yes when I used (3.18) I could see that. However it seems to me that the form (3.18) does not give me (at first glance at least) an expression for the elements of (3.17). – Jxx Sep 9 '18 at 20:09
• mike stone gave you the correct answer. You should interpret $J^{\mu\nu}$ as being a matrix labelled by $\mu$ and $\nu$ (which are not the matrix indices!), so that their commutator is given by matrix multiplication (together with the difference of course). Hence the sum over the lower indices (which are matrix indices!) comes from matrix multiplication. – Giorgio Comitini Sep 9 '18 at 20:18
• Ah I see now where the sum is coming from. Would it then be correct to rewrite the formula of mike stone as $\left( \left[ L_{\mu\nu},L_{\rho\sigma}\right]\right)_{\alpha}^{\;\beta} = \left(L_{\mu\nu}\right)_{\alpha\gamma}\left(L_{\rho\sigma}\right)^{\beta\gamma} - \left(L_{\rho\sigma}\right)_{\alpha\gamma}\left(L_{\mu\nu}\right)^{\beta\gamma}$? – Jxx Sep 9 '18 at 20:22

The indices are wired up as follows: $$([J^{\mu\nu},J^{\rho\sigma}])_{\alpha \gamma}= \sum_{\beta} (J^{\mu\nu}_{\alpha\beta}J^{\rho\sigma}_{\beta\gamma}- J^{\rho\sigma}_{\alpha\beta}J^{\mu\nu}_{\beta\gamma})$$ In other words, the upstairs indices tell you which matrix you are taking about, and the downstairs indices are the labels of the entries in that matrix. The sum on $\beta$ is the usual matric product sum: $$(AB)_{ac}= \sum_b A_{ab}B_{bc}$$
• Thanks for your answer. Could you elaborate a bit on the sum over $\beta$? I can see the commutator being taken on the upstairs indices, but I can’t see how you built the sum on the indices downstairs. – Jxx Sep 9 '18 at 20:11