Unit vector and commutator $\left[ S_zB_z,S_x \hat{x} \right]-\left[S_y \hat{y},S_zB_z \right]+\left[ S_zB_z,S_z \hat{z} \right]$

Question

I have a doubt about properties of commutators. I'm considering the operator: $$H=g\,S \cdot B_z \,\,\,\,\,\,\,\,\ g \in \mathbb{R}$$ I want to calcolate $\frac{d S}{dt}$: $$\frac{d S}{dt}=\frac{\partial S}{\partial t}-\frac{i g}{\hbar} \left[ S \cdot B_z,S \right]= \left[ S_zB_z,S_x \hat{x} \right]-\left[S_y \hat{y},S_zB_z \right]+\left[ S_zB_z,S_z \hat{z} \right]= \\ =\hat{x} \left[ S_zB_z,S_x \right]-\hat{y} \left[S_y \hat{y},S_zB_z \right]+\hat{z} \left[ S_zB_z,S_z \hat{z} \right]$$ I'm not sure that it's formally correct to take out the unit vector. I thought that a versor commutes with an observable if the respective coordinate commutes with it. But I'm not sure at all

Answer 1

The only non-commuting operators appear to be the three spin components of $\vec{S}=\hat{x} S_x+ \hat{y} S_y+ \hat{z} S_z$, so, for $H=g B_z S_z$, one simply has $$ \frac {d\vec{S}}{dt}= -igB_z [S_z,\vec{S}]~/\hbar= gB_z ~~\hat{z}\times \vec{S}. $$