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Spin commutators, [Sz^n, Sy]

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    The entire problem is quite in depth. But what I am having trouble with is just a small part of it, and it boils down to finding the following commutator:
    [tex]\left[ S_{z}^{n},S_{y}\right][/tex]
    where S_{z} and S_{y} are the quantum mechanical spin matrices.

    The reason is that I have to commute S_{y} with an exponential that has S_{z} in it. So I expand the exponential as a series which contains S_{z}^{n}, so I need to find the above commutator.

    2. Relevant equations
    [tex]\left[ S_{z},S_{y}\right] =-i\hbar S_{x}[/tex]
    [tex]\{ S_{x},S_{z}\}=0[/tex]

    3. The attempt at a solution
    [tex]S_{z}^{n}S_{y}=S_{z}^{n-1}(S_{z}S_{y})=S_{z}^{n-1}(S_{y}S_{z}-i\hbar S_{x})[/tex]
    [tex]=S_{z}^{n-2}(S_{z}S_{y}S_{z}-i\hbar S_{z}S_{x})[/tex]
    [tex]=S_{z}^{n-2}((S_{y}S_{z}-i\hbar S_{x})S_{z}-i\hbar S_{z}S_{x})[/tex]
    [tex]=S_{z}^{n-2}(S_{y}S_{z}S_{z}-i\hbar (S_{x}S_{z}+S_{z}S_{x}))[/tex]
    By the anticommutator relation for X and Z given above, the inner parenthesis is zero:

    This seems very strange though, otherwise it appears that if I keep doing this then as long as "n" is an even number, then [itex]\left[ S_{z}^{n},S_{y}\right][/itex] will commute.
    and that the commutator will only be different from zero, with a value of [itex]-i\hbar S_{x}S_{z}^{n-1}[/itex] only if "n" is odd.

    Is this right?
  2. jcsd
  3. Feb 29, 2012 #2
    Try taking a few powers of the pauli matrices and notice what happens to even and odd powers of them, you should see why an even power of any pauli matrix commutes with any power of any pauli matrix
    With this knowledge in hand, notcie what [itex]-i \hbar S_x S_z^{n-1}[/itex] reduces to for odd n!

    You are right but you went on quite a convoluted path to get there :p
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