Clebsch-Gordan Decomposition for 6 x 3

[nigelscott]

Homework Statement

[/B]
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:

T^ij ⊗ T^k = Q^ijk = (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:

ε^ijl{ε_lmnQ^mnk + ε_lnmQ^nmk) = ε^ijl(Q^k_l + Q^k_l)

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 Q^k_l and should be writing Q^k_l - δ^k_lQ. However, If I do this I now have to add the singlet which shouldn't be there. What am I going wrong?

 

[stevendaryl]

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:

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

To figure out the coefficients, you start with:

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

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?

 

[nigelscott]

Thanks. Yes, I am familiar with using the ladder operators. I was more focused on the procedure outlined using the 3 ⊗ 8 by Georgi (LIe Algebras in Particle Physics page 143) and also here https://physics.stackexchange.com/questions/102554/tensor-decomposition-under-mathrmsu3 . I was trying to work this out for the 6 ⊗ 3.

 

[stevendaryl]

Thanks. Yes, I am familiar with using the ladder operators. I was more focused on the procedure outlined using the 3 ⊗ 8 by Georgi (LIe Algebras in Particle Physics page 143) and also here https://physics.stackexchange.com/questions/102554/tensor-decomposition-under-mathrmsu3 . I was trying to work this out for the 6 ⊗ 3.

Oh. Are you talking about angular-momentum, or SU(3) (or something else)?

 

[nigelscott]

SU(3).

 

[Orodruin]

The easiest way of doing this is to use Young tableaux. Are you familiar with them?

SU(3).

Note that this is vital information that was left out of your first post.

 

[nigelscott]

Yes, I am familiar and recognize that Young tableaux is easier for practical purposes. However, this approach should give the same result, yes?