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Raising and lowering operators for spin

  1. Dec 6, 2007 #1
    When we set the raising and lowering operators for spin to be [tex]S_{\pm} = S_x \pm i S_y[/tex], what convention are we following (i.e. why is the first term taken to be S_x and the second taken to be S_y)?
  2. jcsd
  3. Dec 6, 2007 #2


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    If we label the eigenstates of [itex]S_z[/itex] as [itex]|{+}\rangle[/itex] and [itex]|{-}\rangle[/itex], so that [itex]S_z|{\pm}\rangle=\pm\frac12 |{\pm}\rangle[/itex], then
    That is, [itex]S_+[/itex] raises the value of [itex]S_z[/itex], and [itex]S_-[/itex] lowers it. That is how the raising and lowering operators are defined, and [itex]S_x \pm i S_y[/itex] is just what they work out to be.
  4. Dec 6, 2007 #3
    It is the right hand convention.
    think about the angular momentum...
    it is defined mathematically L=R x P... it is a "right handed representation".
    We can of sure define that in left hand repr...
  5. Dec 7, 2007 #4
    I think that is a Math method.
    angular momentum: After "operate" with Math signals, Lz will be raised or lowered. (with operation L+=Lx+iLy)

    If define [tex]L \pm = L_x \pm iL_y [/tex]
    When comute them:
    [tex][L_z ,L \pm ] = ... = \pm \hbar (L_x \pm iL_y )[/tex]
    (Griffiths D.J_Quantum Mechanics...)
    If define "left hand" the signal will become to convert.
    +- become -+
    That is not good!
    Last edited: Dec 7, 2007
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