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Adding angular momenta using levi-civta symbol

  1. Nov 8, 2013 #1
    Suppose you want to add 2 spin-1 particles.

    I understand you can get the j=1 triplet by $$e_{ijk}|j\rangle |k\rangle $$ where i, j, k run from -1, 0 , 1.

    The idea is that levi-civita symbol is a tensor under SO(3) rotations, so the contraction with the $$|j\rangle |k\rangle$$ tensor gives a vector under rotation, which is the triplet.

    However, isn't kronecker delta also a tensor under rotation SO(3)? So why can't you get the singlet from:

    $$\delta_{jk} |j\rangle |k\rangle = |1\rangle |1\rangle +|0\rangle |0\rangle+|-1\rangle |-1\rangle $$
    Looking at the textbooks, the middle |00> term is with a negative sign.

    Also, is there a group theoretic way to get the quintuplet?

    Thanks.
     
  2. jcsd
  3. Nov 8, 2013 #2

    Bill_K

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    Science Advisor

    Aren't you confusing the Cartesian basis with the spherical basis? When you say |k>, k = ±1, 0, that's the spherical basis. The relationship between the two bases is

    e+1 = - (ex + i ey)/√2
    e0 = ez
    e-1 = (ex - i ey)/√2

    Take the Kronecker delta in the form of the identity matrix,

    I = exex + eyey + ezez

    rewrite it in terms of e±1 and e0, and you'll see where the minus sign comes from..
     
  4. Nov 8, 2013 #3
    Thanks! I'm the grader for a graduate QM class, this will help the whole class.
     
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