# Clebsch-Gordan Decomposition for 6 x 3

## Homework Statement

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I am trying to get the C-G Decomposition for 6 ⊗ 3.

2. Homework Equations

Neglecting coefficients a tensor can be decomposed into a symmetric part and an antisymmetric part. For the 6 ⊗ 3 = (2,0) ⊗ (1,0) this is:

Tij ⊗ Tk = Qijk = (Q{ij}k + Q{ji}k) + (Q[ij]k + Q[ji]k)

Where the focus is only on the interchange of the i and j indeces.

3. The Attempt at a Solution

Consider the antisymmetric term: (Q[ij]k + Q[ji]k). Use the invariant tensor to get:

εijllmnQmnk + εlnmQnmk) = εijl(Qkl + Qkl)

So the symmetric part is the '10' (3,0) and the antisymmetric part is the '8' (1,1). The symmetric part is traceless. However, I think I have neglected the trace of the antisymmetric term Qkl and should be writing Qkl - δklQ. However, If I do this I now have to add the singlet which shouldn't be there. What am I going wrong?

stevendaryl
Staff Emeritus
I'm not sure at what level you're trying to understand the C-G decomposition, but if you're just interested in relating two different representations of angular momentum states, then to me, the best approach is to repeatedly use raising/lowering operators.

You have two systems, one with angular momentum 5/2 (so it is a "6", in that there are 6 states: ##m=+5/2, +3/2, +1/2, -1/2, -3/2, -5/2##). You have a second system with angular momentum 1 (so it is a "3" with 3 states: ##m=1, 0, -1##). You can combine them to get a state that has angular momentum 7/2 (an "8"), 5/2 (a "6") or 3/2 (a "4"). So the equation is:

##6 \otimes 3 = 8 \oplus 6 \oplus 4##

##|j_1 = 5/2, m_1 = 5/2, j_2 = 1, m_2 = 1\rangle = |j = 7/2, m = 7/2\rangle##

We have ##J_1{-} + J_2{-} = J_{-}##, so act on the left side with ##J_1{-} + J_2{-} ## and act on the right side with ##J_{-}##. You use the fact that

##J_{-} |j\ m\rangle = \sqrt{j (j+1) - m (m-1)} |j\ m-1\rangle##

This gives you:

##\sqrt{5/2 7/2 - 5/2 3/2} |5/2, 3/2,1, 1\rangle + \sqrt{2} |5/2, 3/2, 1, 0\rangle ##
##= \sqrt{7/2 9/2 - 7/2 5/} | 7/2, 5/2\rangle##

So ##\sqrt{5/7} |5/2, 3/2, 1, 1\rangle + \sqrt{2/7} |5/2, 5/2, 1, 0\rangle = |7/2, 5/2\rangle##

You can repeatedly operate with ##J_{-1}## to find out all the coefficients relating ##j_1 = 5/2, j_2 = 1## to ##j=7/2##.

Then you can go back to
So ##\sqrt{5/7} |5/2, 3/2, 1, 1\rangle + \sqrt{2/7} |5/2, 5/2, 1, 0\rangle = |7/2, 5/2\rangle##

Choosing the orthogonal combination, you find:
So ##\sqrt{2/7} |5/2, 3/2, 1, 1\rangle - \sqrt{5/7} |5/2, 5/2, 1, 0\rangle = |5/2, 5/2\rangle##

This approach is a concrete way to get the coefficients. It looks like you're trying to understand things at a more abstract level?

SU(3).

Orodruin
Staff Emeritus