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Homework Statement
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.
Homework Equations
[tex]\left[ S_{z},S_{y}\right] =-i\hbar S_{x}[/tex]
[tex]\{ S_{x},S_{z}\}=0[/tex]
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:
[tex]=S_{z}^{n-2}S_{y}S_{z}S_{z}[/tex]
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?