Do i need to calculate the expectation value of the Hamiltonian?

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The discussion centers on the application of the generalized Ehrenfest Theorem to derive expressions for the time derivatives of expectation values of spin operators Sx and Sy. It clarifies that while one must compute the expectation value of the commutator, it is not necessary to find the expectation value of the Hamiltonian operator directly. The Hamiltonian provided is identified as H = ωSx, leading to the conclusion that Sx and H commute, resulting in a zero right-hand side for the Ehrenfest equation. However, a participant expresses confusion when their calculated time derivative of <Sx> does not equal zero, prompting questions about the correctness of their earlier calculations. The conversation emphasizes the importance of correctly identifying the Hamiltonian, which should be H = ωSz, given the context of a magnetic field in the z-direction.
  • #31
Zero1010 said:
I wasn't to sure about that. Any suggestions?
Look up the definition of the expected value ##\langle \hat X \rangle##.
 
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  • #32
Ok thanks
 

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