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Matrix Elements

  1. Mar 10, 2010 #1
    Well, [itex]\langle j,m|S^2_{-}|j,m\rangle=\langle j,m|(S^2_{-}|j,m\rangle)[/itex] and [itex]S^2_{-}|j,m\rangle=S_{-}(S_{-}|j,m\rangle)=[/itex]___?

    Normally J_+*J_- is substituted as J^2 - J_z^2+hbarJ_z

    As described above, J_+(J_-J_+), wouldn't the content in parenthesese just cancel out to 1 ?
     
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  3. Mar 10, 2010 #2

    SpectraCat

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    I am going to need you to define some terms there. What is S-, the lowering operator for spins? What is j? Is it the generic total angular momentum quantum number, or is it for a specific case, such as an atom, where you are considering the coupling of orbital angular momentum (l) to spin (s)? Are you dealing specifically with states of spin 1/2 particles, or is it a generic case?
     
  4. Mar 10, 2010 #3
    What is S-, the lowering operator for spins?
    Yes.

    What is j? Is it the generic total angular momentum quantum number, or is it for a specific case, such as an atom, where you are considering the coupling of orbital angular momentum (l) to spin (s)?
    j is the generic total ang mom

    Are you dealing specifically with states of spin 1/2 particles, or is it a generic case?

    Spin 1 particle
     
    Last edited: Mar 10, 2010
  5. Mar 10, 2010 #4

    SpectraCat

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    So what happens in general when the spin lowering operator is applied to a state |j,m>? What are the properties of the scalar product of two angular momentum states?

    Answer those questions and you will have your answer.
     
  6. Mar 10, 2010 #5
    S_-|j,m>=|j,m-1>

    S_+|j,m-1> = |j,m> Is that right? Back to my question, J_+*J_- is substituted as J^2 - J_z^2+hbarJ_z which isn't 1. why is this?



    <j',m'|S_+^3|j,m> = c_jm^+c_jm+1^+c_jm+2^+<j',m'|j,m+3> delta_j',j delta_m',m+3
     
  7. Mar 10, 2010 #6
    MY secnod question is the following
    S_-S_+|+-> = h^2 |-+>

    Where does the eigenvalue h^2 come from?
     
  8. Mar 10, 2010 #7

    SpectraCat

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    Close, there should be a constant multiplying the state on the rhs of each equation above. But anyway, that isn't what you posted originally .. your expression from your OP has two applications of the lowering operator ...

    just to clarify, I guess you mean:
    [tex]J_{+}J{-}=J^{2}-J_{z}^{2} + \hbarJ_{z}[/tex]

    That is correct, but why would you think it equals one in the general case? That expression is an operator, so you should look at what it does to a state ... since any angular momentum eigenstate of each of the operators in that expression, this is pretty trivial:

    [tex][J^{2}-J_{z}^{2} + \hbarJ_{z}]|j,m> = J^{2}|j,m>-J_{z}^{2}|j,m> + \hbarJ_{z}|j,m> = \hbar^{2}[j(j+1) - m^{2} + m]|j,m>[/tex]

    Not sure what that last expression has to do with anything, but FWIW it looks basically correct (although it could do with some tex formatting :wink:.
     
  9. Mar 10, 2010 #8
    What constant are you referring to?
     
  10. Mar 10, 2010 #9

    SpectraCat

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    I don't really understand that ... what is |+->?

    In any case, hbar is the quantum mechanical unit of angular momentum. So the result of applying any angular momentum operator to a quantum state will always be multiplied by hbar. In the case above, the operator is a product of two angular momentum operators, so you get hbar*hbar from the application of that operator to a state.
     
  11. Mar 10, 2010 #10

    SpectraCat

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  12. Mar 10, 2010 #11
    + spin up

    - spin down



    Do you know how S_z would act on such a state given what I said earlier?

    S_-S_+|+-> = h^2 |-+>

    In addition

    S_+S_- |+-> = 0

    S_z^-*S_z^+|+-> = ?

    it supposed to be -hbar^2/4 not sure why.
     
  13. Mar 10, 2010 #12
    And just to clarify, S_-*S_+ = 1 without constants?
     
  14. Mar 11, 2010 #13
    ALso:

    S_+S_- + S_-S_+ = S^2-S_z^2 but this book I found says its also equal to 1 for a spin 1/2 particle?? Why?

    They also have

    S_+S_z = -1/2 S_+

    why is that one true?
     
  15. Mar 12, 2010 #14

    SpectraCat

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    No, ... you gave the correct expression for this *operator* earlier ... I showed you how to apply it to an eigenstate. It most certainly does not equal one.

    EDIT: However, since I now know you are talking about spin 1/2 eigenstates, I think you will find that the result of applying this to the eigenstate will give you 1, (well, actually hbar2, as I have said).

    Did you check out the website I linked?
     
    Last edited: Mar 12, 2010
  16. Mar 12, 2010 #15

    SpectraCat

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    Again, those are operators on the right hand side .. a spin 1/2 particle is an eigenstate of both of those operators ... plug in the state and evaluate the expression. However, it does not evaluate to one when applied to a spin 1/2 eigenstate .. work it out and see what you get. You are missing some factors of hbar in there as well.

    again, missing some factors of hbar ... and that is a possible result, but not the only one.
     
  17. Mar 12, 2010 #16
     
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