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I know the commutator of angular momentum operators are $$ [J_i,J_j]=\mathrm i\hbar \varepsilon_{ijk}J_k. $$ For spin-1/2 particles, $J_i=\frac\hbar2\sigma_i$ where $\sigma_i$ are Pauli matrices, and I can compute $\{J_i,J_j\}$ from the algebra of $\sigma_i$'s, $$ [\sigma_i,\sigma_j]= 2\mathrm i\varepsilon_{ijk}\sigma_{k}, \quad \{\sigma_i,\sigma_j\}=2\delta_{ij}1 \!\!1_2. $$ When it comes to higher dimensons, these relations are broken.

I want to know what are the relations in higher dimensions.

This answer shows the case for a fundamental representation of $SU(N)$, $$ \{t^{A},t^{B}\} = \frac{2N}{d}\delta^{AB}\cdot 1_{d} + d_{ABC}t^{C},\tag{1} $$ where $$ \mathrm{Tr}[t^{A}t^{B}] = N\delta_{AB}\quad d_{ABC} = \frac{1}{N}\mathrm{Tr}[\{t^{A},t^{B}\}t^{C}].\tag{2} $$ If I substitude Eq.(1) into Eq.(2), I get $$ d_{ABC}=\frac1N \frac{2N}{d}\frac1N\mathrm{Tr}[t^At^B] \cdot \mathrm{Tr}[t^C] +\frac1N d_{ABD}\mathrm{Tr}[t^Dt^C] $$ where $\mathrm{Tr}[t^C]=0$. Above equation dosen't tell me how to compute $d_{ABC}$, since $\mathrm {Tr}[t^D t^C]=\delta_{DC}\mathrm{Tr}[1_d]=N\delta_{DC}$.

Cosmas Zachos
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Luessiaw
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  • What I'm dealing with is actually $SU(2)$, not for arbitrary $N$. – Luessiaw Jan 21 '23 at 05:02
  • The d coefficient of su(2) always vanishes. Your right hand side is *not in the Lie algebra*, but, instead, in the universal enveloping algebra, addressed in answers to the linked questions. There is no good formal theory for it. – Cosmas Zachos Jan 22 '23 at 18:31

1 Answers1

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For the adjoint (3d irrep) of su(2), it is straightforward to compute all anticommutators explicitly, $$ \{J^a,J^b \}_{mk}= -\epsilon_{amn} \epsilon_{bnk} -\epsilon_{bmn}\epsilon_{ank}= 2\delta_{ab}\delta_{mk} -(\delta_{am}\delta_{bk}+\delta_{bm}\delta_{ak}), $$ in variant normalization, $(J^a)_{mn}=i\epsilon_{amn}$, which is the least of your problems.

You see by inspection that, for $a\neq b$, the right-hand side is a symmetric traceless matrix, so not a linear combination of the three generators! Your equation (1) fails.

Keep reading other answers in that question. su(2) irreps are real/pseudoreal, so anomaly coefficients vanish, as your QFT text must have emphasized.

But I am unaware of a generic shortcut expression for M.

Experiment with higher spin representations, all explicit. Can the r.h.s. of $\{S_x,S_z\}$, traceless, be represented by a linear combination of the three generators? (No! You may readily compute it for arbitrary spin s, given item 4 of this link, and confirm it is symmetric with vanishing diagonal elements, but is not $\propto S_x$, the only generator with these properties!)

Cosmas Zachos
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