# Product of spherical tensors

Consider a spin $$j$$ system. A spin $$j$$ spherical tensor $$T^k_q(j)$$ of rank $$k$$ is a $$(2j+1) \times (2j+1)$$ matrix.

Given two spherical tensors of spin $$j$$, say $$T^{k_1}_{q_1}(j)$$ and $$T^{k_2}_{q_2}(j)$$, we can multiply them, since they are just matrices.

It is an interesting fact that the product $$T^{k_1}_{q_1}(j) T^{k_2}_{q_2}(j)$$ is always a (real) linear combination of $$T^k_q$$ for $$k$$ in the range $$|k_1-k_2| \leq k \leq min \{ 2j,k_1+k_2 \}$$.

A reference for this fact is equation 16 of 2.4.4. page 45 in Quantum Theory of Angular Momentum by D A Varshalovich, A N Moskalev and V K Khersonskii. However no derivation, explanation or proof is provided in this reference.

Does anyone have a derivation, explanation or proof for the fact $$T^{k_1}_{q_1}(j) T^{k_2}_{q_2}(j)=\sum_{k=|k_1-k_2|}^{min \{ 2j,k_1+k_2 \}} C(k_1,q_1,k_2,q_2,k,q,j) T^k_q(j)$$ here $$C(k_1,q_1,k_2,q_2,k,q,j)$$ is a coefficient with an explicit form in terms of the Clebsch-Gordan coefficients and 6j symbols, for details see the reference.

It is known that the $$\{T_q^k(j): 0 \leq k \leq 2j, -k \leq q \leq k \}$$ are an orthonormal basis for the space of $$(2j+1) \times (2j+1)$$ matrices. In particular, the $$T^k_q(j)$$ form a spanning set. Thus we expect the product $$T^{k_1}_{q_1}(j) T^{k_2}_{q_2}(j)$$ to be in the span of some $$T^k_q(j)$$ for $$0 \leq k \leq 2j$$. My question is really about the bounds on $$k$$ given in the product formula. Why is $$|k_1-k_2|\leq k \leq k_1+k_2$$?

I'm especially interested in understand the upper bound $$k \leq k_1+k_2$$

For example, for $$j=\tfrac{3}{2}$$ and $$k_1=1=k_2$$ it is already not obvious/ not intuitive to me why the product $$T^{1}_{q_1}(\tfrac{3}{2}) T^{1}_{q_2}(\tfrac{3}{2})$$ can be expressed as a linear combination of spherical tensors without using any of the seven rank $$3$$ spherical tensors $$T^3_q(\tfrac{3}{2})$$.