How to write exchange interaction as Spin operators
I have been reading about quantum magnetism from <PERSON> book, and Equation (2.9) reads
$$J^F \sum_{s s'} c^{\dagger}{is} c^{\dagger}{i's'}c_{is'}c_{i's} = -2 J^F\left (\mathbf{S}i \cdot \mathbf{S}{i'} + \frac{1}{4} n_i n_{i'} \right)$$
Here $S^{\alpha}{i}$, $\alpha = x,y,z$ are components of spin one half operators: $$\mathbf{S}_i \equiv \frac{1}{2} \sum{s s'} c^{\dagger}{is} \vec{\sigma}{ss'}c_{is'}$$
Here $\vec{\sigma}$ are the usual Pauli matrices. And $n_i = \sum_{s} c^{\dagger}{is} c{is}$
For $s=2$, I can explicitly write the right hand side and obtain the left hand side with some simplification. 281
But that does not seem satisfactory to me. I should be able to algebraically obtain this result at a higher abstraction level.
Please give me ideas on how to start.
Edit:
Given the information by NowIGetToLearnWhatAHeadIs, I do the following $$\mathbf{S}i \cdot \mathbf{S}{i'} = \frac{1}{4} \sum_{ss'} c^{\dagger}{is} c{is'} \left( \vec{\sigma}{ss'} \cdot \vec{\sigma}{ss'} \right) c^{\dagger}{i's} c{i's'} \ = \frac{1}{4} \sum_{ss'} c^{\dagger}{is} c{is'} \delta_{s s'}^2 c^{\dagger}{i's} c{i's'} \ = - \frac{1}{4} \sum_{s} c^{\dagger}{is} c^{\dagger}{i's} c_{is} c_{i's}$$
Also $$\frac{1}{4} n_i n_{i'} = \frac{1}{4} \sum_{s} c^{\dagger}{is} c{is} \sum_{s'} c^{\dagger}{i's'} c{i's'} \ = - \frac{1}{4} \sum_{ss'} c^{\dagger}{is} c^{\dagger}{i's'} c_{is} c_{i's'}$$
For $s = s'$, this adds up with the previous expression to give $$- \frac{1}{2} \sum_{s} c^{\dagger}{is} c^{\dagger}{i's} c_{is} c_{i's}$$
which describe the diagonal terms for our complete expression. But for $s \neq s'$, we have $$- \frac{1}{4} \sum_{ss'} c^{\dagger}{is} c^{\dagger}{i's'} c_{is} c_{i's'}$$
but what I need is $$- \frac{1}{2} \sum_{ss'} c^{\dagger}{is} c^{\dagger}{i's'} c_{is'} c_{i's}$$
How do I do this transformation? 262