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For a spin-1/2 degree of freedom, we all know the Hilbert space of states is two-dimensional. The space of linear operators on that Hilbert space has dimension $2^{2}=4$, a basis of which is $S_{x},S_{y}, S_{z}$, and the identity.

These operators are orthogonal with respect to the infinite temperature inner product, meaning that $\text{Tr}\left[S_{\alpha}^{\dagger}S_{\beta}\right]\propto \delta_{\alpha\beta}$. I would like to generalize this to arbitrary spin $S$ - that is, I want to build up an orthonormal basis of operators for arbitrary spin. In general, the dimension of that space is given by $\left(2S+1\right)^{2}$.

By messing around, I have generated an orthonormal basis for the case $S=1$, which is given by $S_{x},S_{y},S_{z},S_{x}S_{y}+S_{y}S_{x},S_{y}S_{z}+S_{z}S_{y},S_{z}S_{x}+S_{x}S_{z},S_{x}^{2}-S_{y}^{2},2S_{z}^{2}-S_{x}^{2}-S_{y}^{2},$ and the identity. Clearly this has a resemblance to multipole expansions, but I am not sure how this generalizes to higher spin. I guess it's something like a higher-rank tensor decomposition, but this gets complicated since there are (presumably) index structures which are more complicated than just pairwise symmetrization/antisymmetrization.

If this has been worked out somewhere I'd appreciate a reference, or any suggestions about how to think about decomposing higher rank tensors to build up the basis myself.

Qmechanic
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miggle
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  • https://archive.org/details/landau-and-lifshitz-physics-textbooks-series/Vol%203%20-%20Landau%2C%20Lifshitz%20-%20Quantum%20mechanics%20-%20non-relativistic%20theory%20%283ed.%2C%201991%29/page/206/mode/1up?view=theater – The Tiler Jun 14 '23 at 14:43

2 Answers2

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There's a canonical set of tensor operators constructed from $\vert Sm\rangle$. Explicitly the operators

$$ T^S_{LM}=\sqrt{\frac{2L+1}{2S+1}}\sum_{mm'} C^{Sm'}_{Sm;LM} \vert Sm'\rangle\langle Sm\vert $$ where $C^{ Sm'}_{Sm;LM}$ is a Clebsch-Gordan coefficient, satisfy the orthogonality relation $$ \hbox{Tr}\left((T^S_{LM})^\dagger T^S_{L'M'}\right)=\delta_{LL'}\delta_{MM'}\, . $$

The $2L+1$ operators in the set $\{T^S_{LM}, M=-L,\ldots, L\}$ transform amongst themselves under rotation. $L$ runs from $0$ to $2S$. There are $(2S+1)^2$ operators in total and any $(2S+1)\times (2S+1)$ hermitian operator can be uniquely expanded in this set of tensor operators.

You can find additional details, for instance, in

Klimov, A.B. and Chumakov, S.M., 2009. A group-theoretical approach to quantum optics: models of atom-field interactions. John Wiley & Sons.

ZeroTheHero
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The generic S spin matrices are detailed in item 4 here. I am not sure I fully understand your question, frankly, and especially your odd claims about "higher tensor decomposition"... Anyway, let me summarize basic facts about N×N hermitian matrix parameterizations where the basis is orthonormal and $N=2S+1$; evidently you wish to construct the basis out of polynomials in the universal enveloping algebra of SU(2), except I don't know why. (The Lie algebra here is irrelevant, since commutators of the three $\vec S$ close into themselves, so we have to limit ourselves to the famously messier anticommutators.)

The number of independent N×N hermitian matrices is $N^2$, so N=2 is the last case where the identity and the three generators $\vec S$ suffice to provide the basis. The rest will have to be provided by the 6 anticommutators thereof, as you got for N=3; but, past this, you need to anticommute those, an uncontrolled process, and orthonormalize. (In your example, the orthogonality is automatic, by the vanishing of the "anomaly" for SU(2), lacking d coefficients, as explained in the linked question.)

Mathematical physicists normally opt for Generalizations of Gell-Mann matrices.

Perhaps it is appropriate to indicate what you will do with this basis: The Pauli matrices are rotation generators, so they have an obvious rotation group action, but the spin 1 case bilinears, beyond the quadratic Casimir (the identity) do not!

Cosmas Zachos
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