Let $\mathcal H_1, \mathcal H_1$ be two Hilbert spaces, $\mathcal H=\mathcal H_1\otimes\mathcal H_2$ is their tensor product. Let $\mathcal L(\mathcal H_1),\mathcal L(\mathcal H_2),\mathcal L(\mathcal H)$ denote operators on the three Hilbert spaces, respectively. Then $$ \dim\mathcal L(\mathcal H)=(\dim\mathcal H)^2, $$ and similar formulae for $\mathcal H_1,\mathcal H_2$. Besides $\dim\mathcal H=\dim\mathcal H_1\cdot\dim\mathcal H_2$, so $$ \dim\mathcal L(\mathcal H)=\dim\mathcal L(\mathcal H_1)\cdot\dim\mathcal L(\mathcal H_2), $$ which inspires me to check if $$\mathcal L(\mathcal H)\stackrel?=\mathcal L(\mathcal H_1)\otimes\mathcal L(\mathcal H_2).$$
In $\mathcal L(\mathcal H_1)$, a set of tensor operators $\{\hat T_{1,q}^{(k)}\}$ can be chosen as a set of bases. And $\{\hat T^{(k)}_{2,q}\}$ for $\mathcal L(\mathcal H_2)$. Define
$$ \hat T_{qq'}^{(kk')} =\hat T_{1,q}^{(k)}\hat T_{2,q'}^{(k')}. $$
In this answer, a simple case is discussed, where $\mathcal H_1,\mathcal H_2$ are irreducible and $\mathcal L(\mathcal H)$ only contains scalar operators. But I'm not able to go further. Can it be shown that $\hat T_{qq'}^{(kk')}$ are linearly independent for different $(kk'qq')$? And what's the relationship between $\hat T_{qq'}^{(kk')}$ and $k$-order tensor operator $\hat T_q^{(k)}$ in $\mathcal H$?