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?
Quantum Mechanics – Condition for Operator on Quantum Hilbert Space to Behave Like Vector
angular momentumhilbert-spaceoperatorsquantum mechanicsvectors
Best Answer
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.