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I Deuteron Spin

  1. Jan 18, 2016 #1
    My notes say:

    http://i.imgur.com/Z0v7Psi.png

    The particular part I don't understand is that we know I = 1, and the text says:

    for S = 1, I can be equal to 1 for L = 0, 1, 2.


    If we know I = 1 and S = 1, how can L be anything other than 0?

    Thanks as always!
     
  2. jcsd
  3. Jan 18, 2016 #2

    blue_leaf77

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    Addition of quantum angular momentum ##\mathbf{I} = \mathbf{L} + \mathbf{S}## yields, for the possible value of ##I##,
    $$I = |L-S|,|L-S|+1,\ldots,L+S-1,L+S$$.
    If you plug in ##S=1## to the above equation, you should see that ##I## can be unity only if ##L = 0,1,2##.
     
  4. Jan 18, 2016 #3
    Thank you!
     
  5. Jul 6, 2016 #4
    Can you elaborate how to look at this kind of problem when solving?
    How exactly do you plug in ##S=1## in this equation and get only 0,1,2 as solutions for L?
     
  6. Jul 7, 2016 #5

    blue_leaf77

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    Given ##L = S = 1##, the possible total angular momenta are
    $$
    I = |1-1|,|1-1|+1,1+1 = 0,1,2
    $$
    To put it in a simple way, the min total angular momenta is ##|L-S|## while the maximum is ##L+S##. Between these values the angular momenta are spaced uniformly at unit increment.
     
  7. Jul 8, 2016 #6

    ChrisVer

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    In general I don't like what these notes say... For S=0 indeed you have antiparallel spins... but that is too vague since you can have antiparallel spin configuration for S=1 too....
    In particular the spin states are (I give it as [itex]|S, S_z>[/itex] in the LHS and [itex]|pn>[/itex] spins in the RHS ):
    S=1
    [itex]|1,1> = |\uparrow_p \uparrow_n>[/itex]
    [itex]|1,0> =\frac{1}{\sqrt{2}}\Big( |\uparrow_p \downarrow_n> +|\downarrow_p \uparrow_n> \Big) [/itex]
    [itex]|1,-1> = |\downarrow_p \downarrow_n>[/itex]
    S=0
    [itex]|0,0> =\frac{1}{\sqrt{2}}\Big( |\uparrow_p \downarrow_n> -|\downarrow_p \uparrow_n> \Big) [/itex]

    As you can see, with adding two spins 1/2 you can either get either total spin 1 (triplet) or 0 (singlet)...in an extension of your question, how did the addition of two 1/2s resulted to something that is 0?
    What will happen if you add three spins 1/2?
     
  8. Jul 10, 2016 #7

    vanhees71

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