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Spin Angular Momentum and Operators

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Attached
    question.JPG

    2. Relevant equations

    ##J_+|j,m⟩ = ћ\sqrt{j(j+1)-m(m+1)} |j,m+1⟩##
    ##J_-|j,m⟩ = ћ\sqrt{j(j+1)-m(m-1)} |j,m-1⟩##
    ##J_z|j,m⟩ = mћ |j,m⟩##
    ##J^2|j,m⟩ = ћ^2j(j+1) |j,m⟩##

    3. The attempt at a solution

    Hi there,

    For part a, the expression we're looking for is given, but I'm unsure as to how to derive it from the equation for ##J_+## given above. Any help would be very much appreciated.
     
  2. jcsd
  3. Jan 7, 2016 #2

    TeethWhitener

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    Try applying the expression you're given for [itex]J_+[/itex] in part a) to each of the eigenstates of [itex]J_z[/itex] individually.
     
  4. Jan 7, 2016 #3

    blue_leaf77

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    You can use the completeness relation for the space in question, the space of angular momentum ##j=3/2##, which is ##\sum_m |m\rangle \langle m| = \mathbf{1}## with ##m=-3/2,-1/2,1/2,3/2##. Since this relation is equal to an identity operator, you can place it wherever and how many times you want around an operator without altering the operator itself.
     
  5. Jan 7, 2016 #4

    TeethWhitener

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    This is unnecessary. Simply apply [itex]J_+[/itex] to the [itex]J_z[/itex] eigenstates and use the fact that the eigenstates are orthogonal (i.e., [itex]\langle m | m' \rangle = \delta_{mm'}[/itex]).
     
  6. Jan 8, 2016 #5

    vela

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    It's just another and, in my opinion, more elegant way of doing the problem.
     
  7. Jan 8, 2016 #6

    PeroK

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    You can also look at this from a more general Linear Algebra point of view. In general, if you have a basis ##(e_i)## then ##|e_i> <e_j|## is an operator that picks out the ##j^{th}## component of a vector (##v_j##) and outputs a vector ##v_j e_i##. And, in fact, the set of all such operators is a basis for your operator space.

    You can, therefore, express any operator as a linear combination of these ##|e_i> <e_j|## operators. If you think about what the raising and lowering opertors do, you may see that they are well suited to being expressed in this basis. In this case, the ##(e_i)## are the states with ##m = \pm \frac{1}{2}, \pm \frac{3}{2}##
     
  8. Jan 8, 2016 #7

    TeethWhitener

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    Ok I'm thoroughly confused. How the heck do you solve part a of this problem using the completeness relation? I guess it's a little too elegant for me.
     
  9. Jan 8, 2016 #8

    nrqed

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    The simplest way is to start with TeetWhitener's advice. First thing first: do you know how to apply the operator given in the answer to each of the ##J_z## states? For example, what do you get if you apply it to ##|1/2\rangle ## ?
     
  10. Jan 8, 2016 #9

    vela

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    I think TeethWhitener knows how he would solve it using his own advice. ;)
     
  11. Jan 8, 2016 #10

    nrqed

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    AH! My bad! Hehehe...
     
  12. Jan 8, 2016 #11

    nrqed

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    Fo any operator A, you may write
    ## A = \sum_{n,m} |n\rangle\langle n| A |m\rangle \langle m| ##
    Knowing the effect of the operator on each basis state then yields the result.
     
  13. Jan 8, 2016 #12

    TSny

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    ##J_+ = J_+ \cdot \mathbb{1}## where ##\mathbb{1}## is the identity operator. Use blue_leaf's suggestion for the identity operator.
     
  14. Jan 16, 2016 #13
    Many thanks for all the replies. Hearing of both methods was very helpful.

    Just on a final note, for the last part of the question, part d), I'm applying the expression given for ##J_+## to ##|ψ⟩## as it's given in d). However 'm ending up with:

    ##J_+|ψ⟩= \frac{h}{2\sqrt{2}} (\sqrt{3}|3/2⟩ - 2|1/2⟩ - 3|-1/2⟩ ) ##

    Not sure if I'm approaching the question in the right way.
     
  15. Jan 16, 2016 #14

    vela

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    ##\hat{J}_x##, not ##\hat{J}_+##.
     
  16. Jan 16, 2016 #15
    D'oh!
     
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