# Anticommutator of spin-1 matrices

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?

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.

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 $$\geq 1$$.