Physics Asked by Ahsan Hayat on April 13, 2021
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 $SO(N)$
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, $$ J_{ij}^{ab}= -ihbar (delta_i^a delta_j^b- delta_i^b delta_j^a ). $$ Can you check that?
Correct answer by Cosmas Zachos on April 13, 2021
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