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Change of Basis For Pauli Matrix From Z Diagonal to X Diagonal Basis

  1. Oct 14, 2014 #1
    I want to find a matrix such that it takes a spin z ket in the z basis,

    [itex] | \; S_z + >_z [/itex]

    and operates on it, giving me a spin z ket in the x basis,

    [itex] U \; | \; S_z + >_z = | \; S_z + >_x [/itex]

    I would have thought that I could find this transformation operator matrix simply by using the following argument:

    [itex] U \; | \; S_z >_z = | \; S_z >_x [/itex]

    [itex] _z< S_z \; | U \; | \; S_z >_z = _z< S_z \; | \; S_z >_x [/itex]

    Therefore, elements of U are given by the inner product [itex] _z< S_z \; | \; S_z >_x [/itex]

    However, to compute the inner product [itex] _z< S_z \; | \; S_z >_x [/itex] , I need to know [itex] | \; S_z >_x [/itex], which is exactly what I am trying to find.

    Where is my misunderstanding?

    I have shown that the matrix given by the inner products [itex] _z< S_z \; | \; S_x >_z [/itex] gives the matrix:

    [itex] \begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix} [/itex],

    which cannot be correct, since [itex] _z< S_z \; | \; S_x >_z \neq _z< S_z \; | \; S_z >_x [/itex].

    However, starting in the x basis and calculating [itex] | \; S_z >_x [/itex] shows that the above matrix works. I imagine this is not a coincidence, but it seems to be implying that [itex] | \; S_z >_x = | \; S_x + >_z [/itex].

    This is not a homework question.
  2. jcsd
  3. Oct 14, 2014 #2
    I also just realized that this should have been posted in the Advanced Homework section.
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