Vector operators in spherical basis

Question

This question is about the components of a vector operator in the spherical basis. In 3D real Euclidean space, a vector $\mathbf{v}$ can be expanded in the standard Cartesian components as $$ \mathbf{v} = v^x \,\mathbf{e}_x + v^y \,\mathbf{e}_y + v^z \,\mathbf{e}_z = v^i \,\mathbf{e}_i $$ where the summation convention is assumed. If one now instead considers a complex space, one may introduce a different basis (spherical basis), which I always see defined as $$ \mathbf{e}_{\pm 1} = \mp \frac{1}{\sqrt{2}} \,(\mathbf{e}_x \pm i \mathbf{e}_y) \\ \mathbf{e}_0 = \mathbf{e}_z \tag{1} \label{1} $$ The vector $\mathbf{v}$ can now be expanded in this spherical basis as $\mathbf{v} = v^q \,\mathbf{e}_q$. By substituting the relations between spherical and Cartesian basis vectors, one finds that these components are given by $$ v^{\pm 1} = \frac{1}{\sqrt{2}} \,(\mp v^x + i v^y) \\ v^0 = v^z \tag{2} \label{2} $$ In quantum mechanics, a vector operator $\mathbf{V}$ is defined as a set of operators that satisfy the commutation relation $[J_i, V_j] = i\hbar \epsilon_{ijk} V_k$ with the angular momentum operator $\mathbf{J}$. It would therefore seem reasonable to define the spherical components of such a vector operator analogously to those of a 3D vector via \eqref{2}. However, with this definition one would find that these components, call them $V^q$, would not form a spherical tensor operator $V_q^1$, defined by the relations $$ [J_z, V_q^k] = \hbar \,q \,V_q^k \\ [J_{\pm}, V_q^k] = \hbar \sqrt{k(k+1) - q(q\pm 1)} \,V_{q\pm 1}^k $$ Instead, the spherical components of a vector operator are usually defined as (see Biedenharn and Louck "Angular momentum in quantum physics" or Sobel'man "Introduction to the theory of atomic spectra", for example) $$ V_{\pm 1}^1 = \mp \frac{1}{\sqrt{2}} \,(V_x \pm i V_y) \\ V_0^1 = V_z \tag{3} \label{3} $$ and they form a spherical tensor operator with rank k=1.

So why is the spherical basis chosen as in \eqref{1} if the components resulting from it in \eqref{2} are not taken as "the spherical components of a vector operator"? Rather, the components defined via \eqref{3} are used.

Answer 1

It allows you to write the commutation relations using Clebsch-Gordan technology $$ [\hat J_\mu,{\cal M}_{JM}]= \sqrt{J(J+1)} C^{J,M+\mu}_{JM 1\mu} {\cal M}_{JM+\mu}\, , $$ construct composite tensors using Clebsch-Gordan technology, and factor matrix elements into a reduced part and a Clebsch-Gordan coefficient. The additional signs can be deduced by looking at the different spherical harmonics expressed in terms of Cartesian coordinates, v.g. \begin{align} Y^{-1}_1(\theta,\varphi)&=+\frac{1}{2}\sqrt{\frac{3}{2\pi}} \frac{x-iy}{r}\, ,\\ Y^{1}_1(\theta,\varphi)&=-\frac{1}{2}\sqrt{\frac{3}{2\pi}} \frac{x+iy}{r}\, . \end{align}

(Indeed the normalizationfactor $\sqrt{J(J+1)}$ is just the reduced matrix element of the tensor operator.)