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?

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

1 Answer 1


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.


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