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Spin base transformation

  1. Jul 18, 2013 #1
    There is something I'm struggling with and I can't seem to find the problem.


    We have the Z spinbase with:
    z = (1/sqrt(2))² <BRA|*(|s_z,+> + |s_z,->)
    which gives following z matrix:

    1 0
    0 1


    and we have for X:

    |s_x, +> = 1/sqrt(2) |s_z,+> + |s_z,->)
    |s_x, -> = 1/sqrt(2) |s_z,+> - |s_z,->)

    Now I have a problem with making the x matrix.
    this one is equal to

    0 1
    1 0

    but this doesn't fit with the base above?
    for example the first component:
    <s_x,+|s_x,+> = 1/2 {<s_z,+|s_z,+> + <s_z,+|s_z,-> + <s_z,-|s_z,+> + <s_z,-|s_z,-> }

    <s_z,+|s_z,+> = <s_z,-|s_z,-> = 1
    <s_z,+|s_z,-> = <s_z,-|s_z,+> = 0 because of orthogonality,


    so we get that <s_x,+|s_x,+> = 1 instead of 0?

    What do I do wrong?
     
  2. jcsd
  3. Jul 18, 2013 #2
    It's not that simple...

    To transform the operator to a different basis, you need to use an "S matrix." I'll represent it by [itex]\mathbb{S}[/itex] so it's less confusing because the spin operator normally use S.

    [itex]S_{z} \stackrel{→}{_{x}} \mathbb{S}^{\dagger}S_z\mathbb{S}[/itex]

    where [itex]\mathbb{S} → \bigl(\begin{smallmatrix} \langle +z\:|+x \rangle&\langle +z\:|-x \rangle\\ \langle -z\:|+x \rangle&\langle -z\:|-x \rangle \end{smallmatrix} \bigr)[/itex]

    But this won't get you the [itex]S_{x}[/itex] operator... To do this you need to determine the operator from its eigenvalues and eigenvectors, then rotate it to the z basis.
     
    Last edited: Jul 18, 2013
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