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A spherical tensor operator of rank $k$ is defined such that under a rotation $\mathcal R(\alpha,\beta,\gamma) \in \mathrm{SO}(3)$, it transforms as: $$\hat U(\mathcal R) \hat T_q^k \hat U^\dagger(\mathcal R) \equiv \sum_{q' = -k}^k\mathcal D_{q'q}^{(k)}(\mathcal R) \hat{T}_{q'}^k$$ where $\hat U(\mathcal R)$ is the unitary representation of $\mathcal R$ in our Hilbert space, and $\mathcal D_{q'q}^{(k)}$ are Wigner D-matrices.

I understand the significance of this definition, since it means that the subspace associated with a specific rank $k$, i.e. $V_k := \mathrm{span}\{\hat T^k_q\}_{q=-k}^k$ , is invariant under rotations. However, let's take two arbitrary elements of $V_k$, which I'll call ${\hat A}$ and $\hat B$. These elements can be written in terms of our spherical tensor basis by definition: $${\hat A} = \sum_{q=-k}^k a_q \hat T_q^k$$ $${\hat B} = \sum_{q=-k}^k b_q \hat T_q^k$$ My question is: Can I always find some rotation $\mathcal R \in \mathrm{SO}(3)$ that transforms $\hat A$ into $\hat B$? In other words, are such elements of $V_k$ generally connected through rotations?

For example, does there exist some rotation represented by $\hat{U}(\mathcal R)$, such that $\hat U (\hat T_2^2+\hat T_{-2}^2) \hat U^{\dagger} = \hat T_0^2$ ?

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Observe that since the D matrices are unitary, $\hat{A}$ and $\hat{B}$ must have the same norm in order to be connected by a rotation.

If we consider only normalized tensors your question turns out to be if there is a rotation $R$ such that \begin{equation} U\hat{T}_q^kU^\dagger=\sum_{q'=-k}^k D^{(k)}_{q'q}\hat{T}^k_{q'}=\sum_{q'=-k}^k a_{q'q}\hat{T}^k_{q'} \end{equation} for any set $(a_{-kq},...,a_{kq})$ with $\sum_{q'=-k}^k |a_{q'q}|^2=1$. This question is equivalent to ask if the set of Wigner matrices of rank $k$ contains $U(2k+1)$ (and, since the D's are unitary, if the sets coincide). To see that this cannot be the case it's sufficient to notice that the D's (if $k>0$) are parameterized by 3 real numbers while the set $U(2k+1)$ has $(2k+1)^2$ independent components, so the statement is true only for the trivial case $k=0$.

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