Graduate What happens when you commute Sx and Sz in spin operators?

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The discussion centers on the commutation relation between spin operators, specifically examining the relationship between [Sz, Sx] and [Sx, Sz]. It is established that [Sz, Sx] = ihbar Sy, leading to the inquiry about the result of [Sx, Sz]. A participant points out the general property of commutators, stating that [A, B] = -[B, A], which clarifies the confusion. The conversation highlights the importance of understanding these fundamental properties in quantum mechanics. Taking a break is suggested to alleviate the confusion experienced by one participant.
Dennmac
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So we know [Sz, Sx] = ihbar Sy (S with hats on) so what happens if you get [Sx, Sz]? Is it the same result? Just trying to work out if I've gone wrong somewhere
 
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Dennmac said:
So we know [Sz, Sx] = ihbar Sy (S with hats on) so what happens if you get [Sx, Sz]? Is it the same result? Just trying to work out if I've gone wrong somewhere
In general ##[A, B] = - [B, A]##. The proof is as easy as they come.
 
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Facepalm...thank you! Letters are swimming in front my eyes, I think it's time for a break!
 
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I am slowly going through the book 'What Is a Quantum Field Theory?' by Michel Talagrand. I came across the following quote: One does not" prove” the basic principles of Quantum Mechanics. The ultimate test for a model is the agreement of its predictions with experiments. Although it may seem trite, it does fit in with my modelling view of QM. The more I think about it, the more I believe it could be saying something quite profound. For example, precisely what is the justification of...

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