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Angular Momentum

  1. Apr 14, 2010 #1
    S_z = hbar/2 [ (|+><+|)-(|-><-|) ]

    Normally we're given these relations, how does one derive them?
  2. jcsd
  3. Apr 14, 2010 #2
    The general form of any operator in terms of its eigenvalues and eigenvectors is
    A = \sum_a |a\rangle a \langle a |
    where |a> is the eigenvector with eigenvalue a. This is simply like expressing an operator in its diagonal basis, and the operator [tex] |a\rangle \langle a |[/tex] is the projection operator for that eigenvector. That's what the S_z operator you've written is.
  4. Apr 14, 2010 #3
    Sorry. I don't see how that is derived. Can you show me explicitly?

    What does that say about S_x?

    S_x = hbar/2 [ (|+><-| + |-><+|)]
  5. Apr 14, 2010 #4


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    Do you understand how the resolution of the identity works? The expression you have in square brackets is equal to 1 (more specifically the 2x2 identity matrix) ... do you understand why? That is a good place to start here ...
  6. Apr 14, 2010 #5
    There's a negative sign in front of |-><-| so its not the identity.

    But (|+><+|)+(|-><-|) = 1 is.

    Actually for an arbitrary set [tex]

    \sum_a |a\rangle \langle a | = 1


    |+> = ( 1 0 )^T
    |-> = ( 0 1 )^T
    Last edited: Apr 14, 2010
  7. Apr 14, 2010 #6


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    Sorry ... you are of course correct (it's late and I am not paying close enough attention). However, the equation you gave was generated from the resolution of the identity as follows:

    [tex]S_z = S_z \{|+\rangle\langle+|\:+\:|-\rangle\langle-|\} [/tex]

    But of course,


    So we have (parentheses used to group terms for clarity):

    [tex]S_z = (S_z|+\rangle)\langle+|\:+\:(S_z|-\rangle)\langle-| = \frac{\hbar}{2}[\:|+\rangle\langle+|\:-\:|-\rangle\langle-|\:][/tex]

    As you correctly wrote. As for Sx, the treatment is basically the same, although you have to be a little bit careful at the beginning. I find it clearest to do the resolution of the identity in terms of the eigenstates of Sx (as opposed to |+> and |->, which are the eigenstates of Sz by convention). Then use the expressions for the Sx eigenstates in the basis of |+> and |-> to simplify the result.

    EDIT: Ok, that TeX code came out pretty ugly, but it is correct .. let me know if you have any more questions. Bedtime now.
  8. Apr 16, 2010 #7
    What's S_x |+> and S_x |->?
  9. Apr 16, 2010 #8


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    What do you think? Are you familiar with the Stern-Gerlach experiment? The |+> and |-> are eigenstates of Sz, so what should happen when the Sx operator is applied to them? Are Sz and Sx commuting operators?
    Last edited: Apr 16, 2010
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