# Deriving spin rotation with the commutator of function relation

I'm trying to derive the rotation effect of the spin operator from the commutation relation (from here): $$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$

I started by doing: $$\left[e^{i\theta S_y}, S_z\right] = \left[S_y, S_z \right]i\theta e^{i\theta S_y}=-S_x\theta e^{i\theta S_y}$$ Multiplying by $$e^{-i\theta S_y}$$ from the right we get: $$e^{i\theta S_y} S_z e^{-i\theta S_y}-S_z e^{i\theta S_y} e^{-i\theta S_y}=-\theta S_x e^{i\theta S_y} e^{-i\theta S_y}$$ $$\Rightarrow e^{i\theta S_y} S_z e^{-i\theta S_y}=S_z-\theta S_x$$ Which is good for small angle approximation, but in other derivation I saw they got: $$e^{i\theta S_y} S_z e^{-i\theta S_y}=S_z cos(\theta)- S_x sin(\theta)$$

What am I doing wrong?

## 1 Answer

Yes. You are doing something wrong. The formula $$[f(A),B] =[A,B]\frac{\partial f(A)}{\partial A}$$ only holds if $$[A,[A,B]]=0$$. This condition is indeed stated in the link you gave to the formula.