1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Addition of Angular Momenta - Degeneracy

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

    Show that when angular momenta ##j_1## and ##j_2## are combined, the total number of states is ##(2j_1 +1)(2j_1+2)##

    1z3p2dt.png

    2. Relevant equations



    3. The attempt at a solution

    For the two gyros in the box, there are ##2j_1 +1## possible orientations of the first gyro, and for each of these orientations, the second gyro can be oriented in ##2j_2+1## ways.

    So shouldn't the total number of ways be ##(2j_1+1)(2j_2+1)##?
     
  2. jcsd
  3. Apr 10, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Yes, there is an error in the question. While your reasoning is valid, I think the setter wants you to actually compute the sum on the LHS explicitly.
     
  4. Apr 10, 2014 #3
    How would I compute it explicitly?
     
  5. Apr 11, 2014 #4

    CAF123

    User Avatar
    Gold Member

    Rewrite the sum like $$\sum_{J=j_1 - j_2}^{j_1 + j_2} 2J+1 = 2\sum_{J=j_1 - j_2}^{j_1 + j_2}J + \sum_{J=j_1 - j_2}^{j_1 + j_2} 1$$

    Try to relabel the indices on each of these sums on the RHS so that you can use the well-known sum formula for the sum of the first ##n## positive integers. To use that, relabel the indices such that the sum starts at J=0.

    The supposition is that ##j_1 \geq j_2## so ##j_1 - j_2## is non-negative which means rewriting the sums is a bit easier.
     
  6. Apr 12, 2014 #5
    [tex]\sum_{J=j_1 - j_2}^{j_1 + j_2} 2J+1[/tex]
    [tex] = 2\sum_{J=j_1 - j_2}^{j_1 + j_2}J + \sum_{J=j_1 - j_2}^{j_1 + j_2} 1[/tex]
    [tex]= 2(\frac{j_2+1}{2})(2j_1) + 2j_2+1[/tex]
    [tex] = (2j_1+1)(2j_2+1)[/tex]

    I suppose the sum of states is a constant? Because degeneracy for each particle is constant, and degeneracy of a combined state depends on degeneracy of each particle. Hence sum of states before mixing = sum of states after mixing.
     
    Last edited: Apr 12, 2014
  7. Apr 12, 2014 #6

    CAF123

    User Avatar
    Gold Member

    Did you not expect ##(2j_1 + 1)(2j_2+1)##? Rewrite, for example, $$\sum_{J=j_1-j_2}^{j_1+j_2}\,J = \sum_{J=0}^{j_1+j_2}\,J - \sum_{J=0}^{(j_1-j_2)-1}\,J $$ and apply the sum of n integers formula.
     
  8. Apr 12, 2014 #7
    I did, and I got ##(2j_1+1)(2j_2+1)##.

    Why would the sum of states before and after mixing be equal? Is my explanation correct?
     
    Last edited: Apr 12, 2014
  9. Apr 13, 2014 #8

    CAF123

    User Avatar
    Gold Member

    Yes. Effectively you can express any of your quantum states as a linear combination of states in the coupled (or in your language, mixed) basis or those in the uncoupled basis. Since both of these bases describe the same physical space of states, then the number of basis elements must be the same.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted