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We know that in the spin-1/2 representation the anticommutation relation of the Pauli matrices is $\{\sigma_{a},\sigma_{b}\}=2\delta_{ab}I$. Does a similar relation hold for the spin-1 representation?

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Of course not, in general, as the anticommutator is in the universal enveloping algebra: it is not even in the Lie algebra augmented by the identity, as evident in the specific example below.

For the spin 1 representation of the algebra, $J^a_{~~bc}=-i\epsilon_{abc}$, consisting of hermitean, imaginary, antisymmetric 3×3 matrices, i.e. the adjoint representation, it is straightforward to compute all anticommutators explicitly, $$ \{J^a,J^b \}_{mk}= -\epsilon_{amn} \epsilon_{bnk} -\epsilon_{bmn}\epsilon_{ank}= 2\delta_{ab}\delta_{mk} -(\delta_{am}\delta_{bk}+\delta_{bm}\delta_{ak}). $$

You then see the r.h.s. are symmetric real matrices.

  • For $a\neq b$, they are traceless off-diagonal ones;

  • For $a=b$, they are diagonal, traceful ones, but with a 0 in the ab entry.

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What would the higher-spin analogue of the Pauli matrices be? Thanks to a coincidence in $3$-dimensional space, two different generalizations are possible. This answer consideres both definitions, and the answer in both cases is no.

In $D$-dimensional space, the number of generators of rotations (in canonical planes) is $D(D-1)/2$. When $D=3$, we have the coincidence $D(D-1)/2 = D$. For arbitrary $D$, consider these two definitions:

  • First definition: In relativistic quantum physics, the kinetic term for a spin-$1/2$ fermion involves partial derivatives in the combination $\sum_\mu\gamma^\mu\partial_\mu$ where the $\gamma^\mu$ are Dirac matrices and $\mu\in\{0,1,2,..,$D$\}$. The Hamiltonian involves the combination $\gamma^0\sum_k\gamma^k\partial_k$ with $k\in\{1,2,..,$D$\}$. We could define the Pauli matrix $\sigma_k$ to be proportional to the blocks of $\gamma^0\gamma^k$ in a basis where $\gamma^0$ is block-diagonal and $\gamma^k$ is block-off-diagonal. With this definition, the term describing how the spin-$1/2$ field interacts with the magnetic vector potential $A_k$ involves $\sum_k \sigma_k A_k$, where the index takes $D$ values.

  • Second definition: For any $D$, we could define the Pauli matrices to be the generators of rotations in the canonical planes. Then there are $D(D-1)/2$ Pauli matrices. With this definition, the term describing how the spin-$1/2$ field interacts with the magnetic field involves $\sum_k \sigma_k B_k$, where the index takes $D(D-1)/2$ values.

Whichever definition we use, the answer to the question is no:

  • With the first definition there are $D$ Pauli matrices, and the anticommutation relation shown in the OP follows from the anticommutation relation of the Dirac matrices. This works for any $D$, but it only makes sense for spin-$1/2$, because the Dirac matrices are specific to spin-$1/2$. So with this definition, the answer is no: a similar relation does not hold for spin $\geq 1$, because these Pauli matrices aren't even defined for spin $\geq 1$.

  • The second definition works for any spin $\geq 1/2$, but then the anticommutation relation shown in the OP does not hold for spin $\geq 1$ in $D=3$. If it did, then we would have both $\{\sigma_1,\sigma_2\}=0$ and $[\sigma_1,\sigma_2]\propto \sigma_3$, which together imply $\sigma_1\sigma_2\propto\sigma_3$, and similarly for cyclic permutations of the indices. This implies that the algebra generated by the $\sigma_k$ is linearly spanned by only $4$ matrices (the $\sigma_k$ and the identity matrix). This is only consistent with spin $1/2$, in which case the matrices have size $2\times 2$. So with this definition, the answer is still no: a similar relation does not hold for spin $\geq 1$, because the generators of rotations don't satisfy such a relation for spin $\geq 1$.

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