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Two spin-1 objects: Addition of Angular Momenta

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose in a container there are two spin-1 objects, A and B. It is found that ##J=j_1+j_22## and ##M=m_1+m_2=1##. What is the probability when ##J_z## is measured for A, values of m=0, m=-1 and m=+1 will be obtained?

    mbid09.png


    2. Relevant equations



    3. The attempt at a solution

    [tex]|j,j> = |2,2> = |(1,1)_A>|(1,1)_B>[/tex]

    Starting:
    [tex]J_{-}|2,2> = \left(J_{-}^A + J_{-}^B\right)|(1,1)_A>|(1,1)_B>[/tex]
    [tex]\sqrt{2(2+1)-2(2-1)}|2,1> = \sqrt{1(1+1)-1(1-1)}\left(|(1,0)_A>|(1,1)_B> + |(1,1)_A>|(1,0)_B>\right)[/tex]
    [tex]|2,1> = \frac{1}{\sqrt{2}}\left(|(1,0)_A>|(1,1)_B> + |(1,1)_A>|(1,0)_B>\right)[/tex]

    For m=0, probability = 1/2
    For m=1, probability = 1/2
    For m=-1, probability = 0

    Is this right?
     
  2. jcsd
  3. Apr 10, 2014 #2

    CAF123

    User Avatar
    Gold Member

    I would say yes, that is correct. So essentially you are writing the coupled state |2,1> in the uncoupled basis. Since j1+j2=1+1 = 2 = J and m1+m2=1, you know that the only possible values for the mi are 0 and 1, since any other combination does not give a sum of 1.

    So the coupled state |J,M> decomposed in the uncoupled basis |j1j2m1m2> → |m1m2> would usually consist of 9 terms for fixed j1 and j2 both equal to 1. But by the above restriction, the only possibilities are |0,1> and |1,0>, as you found.

    So the decomposition must be |2,1> = c1|0,1> + c2|1,0>. For normalized states, |c1|2+ |c2|2 = 1 => c1 = c2 = 1/√2.

    Thought I would give a more descriptive solution, which helps in understanding what you are actually doing.
     
    Last edited: Apr 10, 2014
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