In QM we work with $H=L_2(\mathbb{R}^3)$ as a Hilbert space of square-integrable complex-valued functions. Now we define a special set of three operators $L_x, L_y, L_z$ by $L_i = \varepsilon_{ijk} \hat{x}_j \hat{p}_k$, where $\hat{x}_j$ is the operator that multiplies by $x_j$ and $\hat{p}_k$ is the operator $-i \hbar \partial_k$. We say that a set/tuple of three hermitian operators $A_i: H \to H$ is a vector $\vec A=(A_1, A_2, A_3)$/behaves like a vector/ is a vector operator, if $[L_i, A_j]=\varepsilon_{ijk} i \hbar A_k$. My question is, how many of such $\vec{A}$ are there? If I know for example $A_1$, I guess under some conditions there are unique $A_2, A_3$ to make $(A_1, A_2, A_3)$ a vector operator by the above definition? Is this true and what are such conditions?
2 Answers
It depends on the precise rigorous definition of a vector of selfadjoint operators. Usually, there is a dense invariant subspace for the operators and the generators of rotations, where all linear combinations of the three operators are essentially selfadjoint. In that domain, the exponentials of the angular momenta can be integrated obtaining that $$U(R) A_k U(R)^\dagger = \sum_{j=1}^3 R_{kj}A_j\tag{1}$$ for every 3 rotation $R\in SO(3)$.
Since every unit vector in $\mathbb{R}^3$ can be mapped to any other such vector by a rotation, it is clear that, e.g., from $A_1$ we can reconstruct $A_k$ for $k=2,3$ using $(1)$.
In summary, the condition for $A_1$ to be part of a vector of operators is that there are other two operators $A_2, A_3$ satisfying the commutation relations you wrote (plus some technical hypothesis).
However, if $A_k$ is a component of a vector of observables under the action of the rotation group, that component determines the other two components uniquely.
If you multiply $A_k$ by the same scalar function $f(r)$, then $f(r)\vec A$ remains a vector so since $f(r)$ is otherwise pretty arbitrary there are (up to some technical stuff like domains etc.) infinitely many of those.
In addition you can also have combinations of vector operators coupled to yield another vector operators. Whereas $\vec r=(\hat x,\hat y,\hat z)$ is clearly a vector, so is $\vec p=(\hat p_x,\hat p_y,\hat p_z)$ and so is $\vec L=\vec r\times \vec p$. More generally, if $\vec A$ and $\vec B$ are vectors, so will be $\vec C$ with (spherical) components $$ C_k=\sum_{mm'}A_{1m}B_{2m'}C^{1k}_{1m;1m'} $$ where $C^{Lk}_{\ell_1m;\ell_2m'}$ is a Clebsch. Of course you can do triple coupling of vector operators as well, or you could couple an $\ell_1=2$ operator with an $\ell_2=1$ operator to get another $L=1$ operator.
Obviously the combinations are infinite. All these tensor operators will just differ by their reduced matrix elements, which will explicitly depend on this $f(r)$ or on the reduced matrix elements of the various components of the composite tensor.
