I'm quite familiar with rotation in quantum/classical mechanics. I know rotation for an operator $O$ or state $|\psi \rangle$ acts like:
$$O \rightarrow R O R^{-1} \\ |\psi \rangle \rightarrow R |\psi \rangle $$
However, I don't understand how to apply this to the second quantized operator for example $c_{i \sigma} c_{j \sigma}^\dagger$ where $\sigma$ is spin index. I mean naively I can rotate this like an operator and so acts like:
$$c_{i \sigma} c_{j \sigma}^\dagger \rightarrow R c_{i \sigma} R^{-1} R c_{j \sigma}^\dagger R^{-1}$$ where $R \in SU(2)$, but then $c$ is in a different space from $R$. Maybe I'm supposed to do this? $$R c_{i \sigma} R^{-1} = c_{i R\sigma R^{-1}}$$
And if the above is true, then I have another confusion: in quantum mechanics, say two particle state, the only rotation invariant state is the singlet $\frac{1}{\sqrt{2}} |\uparrow \downarrow \rangle - | \downarrow \uparrow \rangle$ with $S(S+1) = 0$, and so analogously I think this operator is also rotational invariant
$$c_{i \uparrow} c_{j \downarrow} -c_{i \downarrow} c_{j \uparrow} $$
but this is not the only rotational invariant operator? Because it seems like $c_{i \sigma} c_{j \sigma}^\dagger$ is also invariant.. Is my thinking right?
In summary, I may be confused about how rotation acts in Fock space (for second quantized operator) versus Hilbert space (for spin state).