# Swapping spin operators when commuting

I'm working on a problem at the moment and its asking me to use ehrenfast's theorum to show that my previous answers were correct, but I'm out by a minus sign.

I assumed that if $$[S_z, S_x] = ihS_y$$, then $$[S_x, S_z]$$ must be $$-ihS_y$$.

Is this really stupid?

If I know that $$\hat{H} = \omega S_z$$, then my expectation value for $$S_x$$ from Eherenfest's theorem becomes $$-\omega \langle S_y \rangle$$ right?

For some reason the previous sections sort of requires me to find it as just $$\omega \langle S_y \rangle$$ but I can't see anything wrong with the way I did my integrals for that part so I'm hoping I've done something something stupid here with Ehrenfast.

EDIT to clarify:

I'm basically just asking if there is any difference between $$[S_z, S_x]$$ and $$[S_x, S_z]$$. We know that for spin $$[S_z, S_x] = ihS_y$$, it's a standard thing. If I swap the operators and commute, is it the same or is it negative?

• You should add more context and details. I really don't understand the question. Mar 2, 2022 at 11:05
• Regarding the edit: Hint: Check the definition of $[A,B]$. Mar 2, 2022 at 11:25

Indeed it holds for any two operators A and B: $$[A,B]=AB-BA=-(BA-AB)=-[B,A]$$ You need to provide more details in order to get a more detailed answer to your question.