- Vector Operator vs. Spherical Tensor

1. Definitional Properties #

1.1 Definitional Property of Vector Operator #

$$ [ J_i , V_j ] = \sum_k i\hbar \epsilon_{ijk} V_k $$

or equivalently,

$$ [ V_i , J_j ] = \sum_k i\hbar \epsilon_{ijk} V_k $$

1.2 Corresponding Property of Spherical Tensor #

$$ [ J_0 , T_q^{(k)} ] = \hbar q T_q^{(k)} $$

and

$$ [J_{\pm} , T_q^{(k)}] = \hbar \sqrt{(k \mp q)(k \pm q +1)} T_{q \pm 1}^{(k)} $$

2. Rotation Transformations #

2.1 Similar Transformation as Rotation of Vector Operator in the Euclidean Space #

$$ \langle\psi| D^{\dagger}(R) V_i D(R) |\psi\rangle = \langle\psi| (R^T V_i R) |\psi\rangle $$

which implies that the vector operator transforms as:

$$ V_i \to V’_i = D^{\dagger}(R) V_i D(R) $$

Setting

$$ \mathbf{V} = \begin{pmatrix} V_1 \ V_2 \ V_3 \end{pmatrix}, \quad \mathbf{V’} = \begin{pmatrix} V’_1 \ V’_2 \ V’_3 \end{pmatrix} $$

we obtain

$$ \mathbf{V’} = R \mathbf{V}, $$

or

$$\sub{V’}{i} = D^{\dagger}(R) V_i D(R) = \sum_j R_{ij} V_j$$

2.2 Corresponding Property of Spherical Tensor in Complex Inner Space #

$$ D^{\dagger}(R) \sub{T}{q}^{(k)} D(R) = \sum_{q’} \sub{D}{qq’}^{(k)*}(R) \sub{T}{q’}^{(k)} $$