Consider the three \(\ell=1\) spherical harmonics:
\[Y_{1}^{-1}(\theta,\phi)=\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi}\]
\[Y_0^0(\theta)=\sqrt{\frac{3}{4\pi}}\cos\theta\]
\[Y_{1}^{1}(\theta,\phi)=-\sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi}\]
Although the spherical harmonics are just functions of \(\theta,\phi\) on the sphere \(S^2\), one can introduce an \(r\)-dependence simply by multiplying each of them by \(r\). The resulting functions are then best written in Cartesian coordinates \((x,y,z)\):
\[rY_{1}^{-1}(x,y,z)=\sqrt{\frac{3}{4\pi}}\frac{x-iy}{\sqrt{2}}\]
\[rY_0^0(x,y,z)=\sqrt{\frac{3}{4\pi}} z\]
\[rY_{1}^{1}(x,y,z)=-\sqrt{\frac{3}{4\pi}}\frac{x+iy}{\sqrt{2}}\]
But this suggests something quite interesting. It means if we have any vector \(\textbf x\in\textbf R^3\) (i.e. a rank \(\ell=1\) tensor) with contravariant components \(\textbf x=x\hat{\textbf i}+y\hat{\textbf j}+z\hat{\textbf k}\) in a Cartesian (aka orthonormal) basis \(\hat{\textbf i},\hat{\textbf j},\hat{\textbf k}\), then by rotating into the so-called spherical basis \(\hat{\textbf e}_{-1}=\frac{\hat{\textbf i}-i\hat{\textbf j}}{\sqrt{2}},\hat{\textbf e}_{0}=\hat{\textbf k},\hat{\textbf e}_{1}=-\frac{\hat{\textbf i}+i\hat{\textbf j}}{\sqrt{2}}\), the contravariant components \(\textbf x=x^{\mu}\hat{\textbf e}_{\mu}\) of the same vector \(\textbf x\) in this spherical basis are now proportional to the spherical harmonics! But this gives us a handle on how such components must transform under rotations because we know how spherical harmonics transform under rotations (and the \(r\)-prefactor is constant with respect to rotations)! To wit, recall that \(Y_{\ell}^m(\hat{\textbf n})=\langle\hat{\textbf n}|\ell, m\rangle\) are the (angular part of the) position space wavefunctions of orbital angular momentum eigenstates, and for each fixed total angular momentum \(j\in\textbf N/2\), the \(2j+1\)-dimensional subspace \(\mathcal H_{j}:=\text{span}_{m=-j,…,j}|j,m\rangle\) is \(e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}\)-invariant for all \(\Delta \boldsymbol{\phi}\) in \(SO(3)\). More precisely, these states transform according to the components of the Wigner D-matrix \(D^j(\boldsymbol{\phi})\in U(2j+1)\):
\[e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}|j,m\rangle=\sum_{m’=-j}^jD_{m’m}^j(\Delta{\boldsymbol{\phi}})|j,m’\rangle\]
with \(D_{m’m}^j(\Delta{\boldsymbol{\phi}})=\langle j,m’|e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}|j,m\rangle\).
Example: Consider the vector \(\textbf x=\hat{\textbf i}-2\hat{\textbf j}+3\hat{\textbf k}\) in some Cartesian basis. Suppose we wished to rotate \(\textbf x\) by an angle of \(\pi/5\) about the \(z\)-axis. Ordinarily, we could just achieve this by acting on the Cartesian components of \(\textbf x\) with a suitable rotation matrix:
\[\begin{pmatrix}\cos \pi/5 & -\sin\pi/5 & 0 \\ \sin\pi/5 & \cos\pi/5 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ -2 \\ 3\end{pmatrix}=\begin{pmatrix}\cos\pi/5+2\sin\pi/5 \\ \sin\pi/5-2\cos\pi/5 \\ 3\end{pmatrix}\]
Alternatively, in the spherical basis we instead have \(\textbf x=\frac{1+2i}{\sqrt{2}}\hat{\textbf e}_{-1}+3\hat{\textbf e}_{0}-\frac{1-2i}{\sqrt 2}\hat{\textbf e}_{1}\). We then note that the associated Wigner \(D\)-matrix has components \(D^1_{m’m}(\pi/5)=\langle 1,m’|e^{-i\pi/5 J_3}|1,m\rangle=\delta_{m’m}e^{-im\pi/5}\) and so \(D^1(\pi/5)=\text{diag}(e^{i\pi/5},1,e^{-i\pi/5})\). Acting with \(D^1(\pi/5)\) on the components of the vector \(\textbf x\) in the spherical basis then yields:
\[\begin{pmatrix}e^{i\pi/5} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{-i\pi/5}\end{pmatrix}\begin{pmatrix}(1+2i)/\sqrt{2} \\ 3 \\ -(1-2i)/\sqrt{2}\end{pmatrix}=\begin{pmatrix}e^{i\pi/5}(1+2i)/\sqrt{2}\\ 3\\ -e^{-i\pi/5}(1-2i)/\sqrt{2}\end{pmatrix}\]
Taking the earlier Cartesian result and converting it into the spherical basis shows that the two calculations match as claimed.
So far we have been talking about a vector \(\textbf x\in\textbf R^3\). Of course though, all of this discussion applies just as well to vector operators on \(\mathcal H=L^2(\textbf R^3\to\textbf C)\) in quantum mechanics such as \(\textbf X,\textbf P\), etc. To reiterate, to rotate a vector in \(\textbf R^3\), one can always first perform a unitary change of basis from Cartesian to spherical, rotate the spherical components of the vector operator as if they were orbital angular momentum eigenstates (or spherical harmonics if you like) using a suitable Wigner \(D\)-matrix, and the result will be the rotated vector operator in the spherical basis.
Even more generally, one need not limit oneself to scalar operators or vector operators; any tensor operator can be converted from Cartesian components to spherical components (with the latter dubbed a “spherical tensor operator” although I find it to be misnomer as it’s not a different kind of tensor operator, just one whose components happen to expressed in the spherical basis). More precisely, if \(T^j_m\) are the spherical components of a rank \(j\in\textbf N\) tensor operator \(T\) with \(2j+1\) component operators \(T^j_m\) for \(m=-j,…,j\), then these components transform under rotations via the Wigner \(D\)-matrix:
\[e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}T^j_m e^{i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}=\sum_{m’=-j}^jD_{m’m}^j(\Delta{\boldsymbol{\phi}})T^j_{m’}\]
As already suggested, rank \(j=0\) scalar operators have \(2j+1=1\) spherical component operator that transforms under rotations as \(e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}T^0_0e^{i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}=D_{00}^0(\Delta{\boldsymbol{\phi}})T_0^0=T_0^0\) which has the logically equivalent infinitesimal formulation \([\textbf J,T_0^0]=0\). Meanwhile, a rank \(j=1\) vector operator with \(2j+1=3\) spherical component operators \(T^1_{-1},T^1_{0},T^1_1\) transforms under rotations as \(e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}T^1_me^{i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}=D_{-1m}^1(\Delta{\boldsymbol{\phi}})T_{-1}^1+D_{0m}^1(\Delta{\boldsymbol{\phi}})T_0^1+D_{1m}^1(\Delta{\boldsymbol{\phi}})T_1^1\), or infinitesimally \(\). This should be compared with the analogous rotation behavior statement for the Cartesian components \(e^{-i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}T_ie^{i\Delta{\boldsymbol{\phi}}\cdot\textbf J/\hbar}=R_{ij}T_j\), or infinitesimally \([J_i,T_j]=i\hbar\varepsilon_{ijk}T_k\). More generally then, one can also look at the infinitesimal limit of the above defining equation for spherical components of a tensor operator to obtain logically equivalent commutation relations:
\[[J_3,T^{\ell}_m]=m\hbar T^{\ell}_m\]
\[[J_{\pm},T^{\ell}_m]=\hbar\sqrt{(\ell\mp m)(\ell\pm m +1)}T^{\ell}_{m\pm 1}\]
\[\sum_{i=1}^3[J_i,[J_i,T^{\ell}_m]]=\ell(\ell+1)\hbar^2 T^{\ell}_m\]
Hopefully this all makes sense! Finally then, we arrive at:
Theorem (Wigner-Eckart Theorem): Let \(T:\mathcal H\to\mathcal H\) be a spherical tensor operator of rank \(j\) with \(2j+1\) components \(T_{(j)}^{m}\) for \(m=-j,…,j\). Then there exists a so-called reduced matrix element \(\langle j_2||T^{(j)}||j_1\rangle\) independent of \(m_1,m_2,m\) which acts as a proportionality constant between the matrix elements of \(T\) in the angular momentum eigenbasis and Clebsch-Gordan coefficients that can be easily computed or looked up in tables:
\[\langle j_2,m_2|T_j^m|j_1,m_1\rangle=\langle j_2||T^{(j)}||j_1\rangle\langle j_1,m_1|\otimes\langle j_2,m_2|j,m\rangle\]
In particular, recall that the Clebsch-Gordan coefficients \langle j_1,m_1|\otimes\langle j_2,m_2|j,m\rangle have the standard selection rules \(m=m_1+m_2\) and \(|j_1-j_2|\leq j\leq j_1+j_2\). By virtue of the Wigner-Eckart theorem, these selection rules are also directly inherited by the matrix elements \(\langle j_2,m_2|T_j^m|j_1,m_1\rangle\).