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Quantum Numbers of Spin 1/2

  1. Jan 30, 2012 #1
    this sounds like a question that should be answered somewhere, but i cant find it.

    how many quantum numbers does it take to specify the state of a spin 1/2?

    1. if it points along z, then just 1: the eigenvalue of Sz will do (up to global phase)
    A. whether or not in a B field
    B. if there is some more complex Hamiltonian, then this is probably no longer enough? or as long as there is 1 dof with 2 possible outcomes it is?

    2. What if the spin points along Sx, then if we knew exactly where it pointed, only 1 quantum number and its operator will give us all the information.
    A. but if we didnt know where it pointed we wouldn't know which operators would be "good."
    so if we have this spin in a B filed in some superposition of Sz up down e-states, it would take more than the eigenvalue of Sz to specify the state with the amount of superposition and the phase between the up and down.

    3. What if the was no B field, and the two states were degenerate, how many quantum numbers would it take to specify the state? or is this a question that cannot be asked, for the two are indistinguishable? (use to erase information by turning off B field?)
  2. jcsd
  3. Jan 30, 2012 #2


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    To specify the state, you need only know that it's spin 1/2 and one quantum number along any of these axes.

    The maximal set of commuting operators is simply S^2 and one of Sx, Sy, or Sz. As Sx, and Sy, and Xz do not commute with each other, you cannot specify them all simultaneously.
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