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Vector operators $\vec{V}$ in quantum mechanics are usually defined as those that commute in a particular way with the spatial Angular Momentum $\vec{L}$: $[L_i,V_j]=i\hbar\varepsilon_{ijk}V_k$. I am not aware of a more general definition, but can we define some sort of "vectoriality" with respect to spin angular momentum (or any set of operators such that $[J_i,J_j]=i\hbar\varepsilon_{ijk}J_k$)?

Is it correct to think that the "vectorial"(or even "tensorial") property of operators is valid for each one of the tensor spaces ($\mathcal{E}_{spatial}\otimes\mathcal{E}_{spin}\otimes \cdots$) available once "spin" degrees of freedom have been added? Maybe there is even some sort of "tensoriality" with respect to a more general algebra of commutation.

DanielSank
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crimson_sprite
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1 Answers1

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The notion of vector operator makes sense when the Hilbert space admits a preferred strongly continuous unitary representation of $SU(2)$ which leaves invariant the space. Every such representation admits 3 selfadjoint generators $J_i$ with $i=1,2,3$. Then the condition you wrote, that a triple of operators $V_i$, defined on a common subspace invariant under the representation, is equivalent to $$U_u V_i U_u^{-1} = \sum_{j=1}^3R_{ij}(u) V_j$$ where $R(u)$ is the standard orthogonal matrix associated to every element $u\in SU(2)$. In that sense the triple of operators $V_i$ transforms how components of vectors in the physical space. This definition can be applied once you have a representation of $SU(2)$ acting on a Hilbert space. It does not matter if the generators represent the angular momentum or the spin etc...

Valter Moretti
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