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That is, $J_{i j}$ is a tensor. We can take this a step further, and let $R^{\prime}$ itself be an infinitesimal rotation, of the form $R^{\prime} \rightarrow 1+\omega^{\prime},$ with $\omega_{i j}^{\prime}=-\omega_{j i}^{\prime}$ infinitesimal. Then, to first order in $\omega^{\prime},$ Eq. $(4.1.9)$ gives $$ \frac{i}{2 \hbar}\left[J_{i j}, \sum_{k l} \omega_{k l}^{\prime} J_{k l}\right]=\sum_{k l}\left(\omega_{i k}^{\prime} \delta_{j l}+\omega_{j l}^{\prime} \delta_{i k}\right) J_{k l}=\sum_{k} \omega_{i k}^{\prime} J_{k j}+\sum_{l} \omega_{j l}^{\prime} J_{i l}. $$ Equating the coefficients of $\omega_{k l}^{\prime}$ on both sides of this equation gives the commutation rule of the $J$ s: $$ \frac{i}{\hbar}\left[J_{i j}, J_{k l}\right]=-\delta_{i l} J_{k j}+\delta_{i k} J_{l j}+\delta_{j k} J_{i l}-\delta_{j l} J_{i k}. \tag{4.1.10} $$

This is a paragraph from Steven Weinberg's lecture of quantum mechanics. Can anyone help me solve the equation (4.1.10)?

These are the commutation relation of generators of $SO(N)$

Qmechanic
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Ahsan Hayat
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1 Answers1

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I don't have your lecture, but, in the abstract, there is an infinity of solutions (representations) of (4.1.10) for the matrices $J_{ij}^{ab}$. The simplest one is the vector, N×N matrices, $$ J_{ij}^{ab}= -i\hbar (\delta_i^a \delta_j^b- \delta_i^b \delta_j^a ). $$ Can you check that?

Cosmas Zachos
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