Writing Breit-Pauli spin-spin-coupling Hamiltonian as a sum of irreducible spin tensor operators

The spin-spin interaction part of the Breit-Pauli Hamiltonian for two electrons contains the term \begin{equation} 3(\mathbf S_1 \cdot \mathbf r)(\mathbf S_2 \cdot \mathbf r) - r^2 \mathbf S_1 \cdot \mathbf S_2 \qquad \qquad (1) \end{equation} where $\mathbf S_1$ and $\mathbf S_2$ are the spin operators for the two electrons and $\mathbf r$ is their distance vector. Now, due to the scalar products occurring in the expression, it is clear that this operator is a scalar under common rotations in spatial and spin space.

It is however advantageous to write the operator in a different form (see Roy McWeeny: Methods of Molecular Quantum Mechanics):

\begin{equation} 3\sum_{m=-2}^2 (-1)^m [\mathbf r \otimes \mathbf r]_{-m}^{(2)} [\mathbf S_1 \otimes \mathbf S_2]_{m}^{(2)} \qquad \qquad (2) \end{equation} Here $[\mathbf S_1 \otimes \mathbf S_2]_{m}^{(2)}$ (and analogous for the other term) are the rank 2 irreducible tensor operators which form the 5-dimensional irreducible representation of the rotation group present in the direct product of the vector representation with itself.

A short motivation for this form of the spin-spin operator: Due to the separation in a sum of terms which transform separately as irreducible tensors under spatial and spin space rotations respectively, one can use the Wigner-Eckart theorem for the determination of matrix elements in a perturbational treatment when the zeroth-order states are nonrelativistic spin eigenstates.

Now for my question: It is a pain in the ass to show by hand that the two above forms of the spin-spin operator are in fact identical. I would therefore like to learn in more abstract ways why the operator can be written as in equation (2). I'm especially wondering about the following:

• Why (and under what conditions in general) can an operator be written as a sum of terms which transform as spatial and spin tensor operators separately?
• Is it possible to derive the exact form of this separation without the need to go into lengthy detail?
• Why occur only rank 2 irreducible tensor operators? Rank 1 does not exist for the symmetric direct product $\mathbf r \otimes \mathbf r$, but what about rank 0?

Update: I've already understood a little bit and want to share it. In the following I denote irreps of SO(3) by the symbols $s$, $p$, $d$ etc. (like atomic orbitals).

We know that $\mathbf r \otimes \mathbf r = s \oplus d$ (spatial tensors) and $\mathbf S_1 \otimes \mathbf S_2 = s \oplus p \oplus d$ (spin tensors). Now we want to find elements from $(\mathbf r \otimes \mathbf r)\otimes(\mathbf S_1 \otimes \mathbf S_2)$ which transform as scalars under complete rotations. One possibility is $s\otimes s$, which leads to the operator $r^2 \mathbf S_1 \cdot \mathbf S_2$. The second possibility is $d \otimes d$. From the formula for the Clebsch-Gordan coefficients \begin{equation} \langle j, m, j, -m | J=0, M=0\rangle = \frac{(-1)^{j+m}}{\sqrt{2j+1}} \end{equation} we know that this operator must be proportional to equation (2).

Now still two questions remain:

• I only know that equation (1) is a scalar, so it could be an arbitrary linear combination of the two scalar operators I just found. How do I know that it is proportional to the $d\otimes d$ scalar operator, and does not contain any $r^2 \mathbf S_1 \cdot \mathbf S_2$?
• How do I know the proportionality constant?