# 9j symbol expansion in 6j symbols

1. Apr 10, 2013

### Yoran91

Hello everyone,

I'm going through Edmond's Angular momentum in quantum mechanics and I found a particular expansion of the 9j symbol in terms of three 6j symbols which I didn't quite understand.
The problem lies in equation 6.4.1 which I simply don't understand.

On page 100/101, (section 6.4), Edmond states that the unitary transformation
$\langle (j_1 j_2) j_{12}, (j_3 j_4) j_{34},jm | (j_1 j_3) j_{13},(j_2 j_4)j_{24},jm \rangle$ can be performed in three steps. In each of these steps the coupling or addition of only three angular momentum operators is carried out.

He states :
$\langle (j_1 j_2) j_{12}, (j_3 j_4) j_{34},jm | (j_1 j_3) j_{13},(j_2 j_4)j_{24},jm \rangle = \\ \Sigma \langle (j_1 j_2)j_{12},j_{34},j | j_1,(j_2 j_{34}) j',j \rangle \\ \times \langle j_2,(j_3 j_4)j_{34},j' | j_3,(j_2 j_4)j_{24},j' \rangle\\ \times \langle j_1,(j_3 j_24)j',j|(j_1 j_3)j_{13},j_{24},j \rangle$.

However, I have no idea how he precisely reduces the addition of four angular momentum operator to three angular momentum operator in these steps. Can anyone explain this?

2. Apr 10, 2013

### fzero

It's worthwhile to recall that the 3j (or Clebsch-Gordan), 6j and 9j-symbols all arise from decomposing a product of angular momentum representations into a sum of irreducible representations. For a product of 2 irreps, we find 3j symbols. If we take a product of 3 irreps, we are lead to 6j-symbols. However, we can take a pair of the 3 spins and use the 3j-symbols to decompose the pairwise-product, and then take a product with the 3rd spin. By this we mean a distribution of the form:

$$\mathbf{j}^{(1)} \otimes \mathbf{j}^{(2)} \otimes \mathbf{j}^{(3)} = \left( \mathbf{j}^{(12)}_1 \oplus \cdots \oplus \mathbf{j}^{(12)}_n \right) \otimes \mathbf{j}^{(3)}.$$

In this way, we will find expressions for the 6j-symbols in terms of the 3j-symbols. Similarly, we can use the 6j-symbols to decompose the product of a triplet of 3 spins in a product of 4 spins,

$$\mathbf{j}^{(1)} \otimes \mathbf{j}^{(2)} \otimes \mathbf{j}^{(3)}\otimes \mathbf{j}^{(4)} = \left( \mathbf{j}^{(123)}_1 \oplus \cdots \oplus \mathbf{j}^{(123)}_n \right) \otimes \mathbf{j}^{(4)}.$$

I wanted to explain the idea, rather than the details, which are tedious. But the mechanics of arriving at the explicit expressions is analogous to the steps needed to work out the C-G or 3j-symbols in the product of 2 spins. One lists all of the states on each side of the decomposition correspondence. Then one uses the SU(2) generators in the ladder combinations to find the highest and lowest-weight states. We can then use the ladder operators to act on these to fill out the rest of the multiplets.

3. Apr 11, 2013

### Yoran91

I don't really see how you would get 9j symbols in terms of 6j symbols.

I understand that if we consider the triple product of Hilbert spaces $H_1 \otimes H_2 \otimes H_3$,that we can decompose it in two ways:
$H_1 \otimes \left( H_2 \otimes H_3 \right)$ and $\left( H_1 \otimes H_2 \right) \otimes H_3$.

Then, decomposing into irreducible subspaces, we would find eigenvectors of the form $|j_1 j_{23} jm \rangle$ in the first scheme and $|j_{12} j_3 jm\rangle$ in the second.

Then the 6j symbols relate these two adding schemes, i.e. they constitute a unitary transformation between the two sets of basis vectors for $H_1 \otimes H_2 \otimes H_3$.

I see that these basis vectors can be decomposed in terms of 3j symbols (or CG-coefficients), so one could find a relation between 6j and 3j symbols.

How does this work in the product $H_1 \otimes H_2 \otimes H_3 \otimes H_4$?
I see we would consider
$H_1 \otimes \left( H_2 \otimes H_3 \otimes H_4 \right)$,

and apply our knowledge from the first part. We would then have eigenvectors of the from
$|j_2 j_{34} j' m' \rangle$ and $|j_{23} j_4 j' m' \rangle$ and there would be a unitary transformation between these vectors.

However, I don't see how you would go from the basis vectors of $\left( H_1 \otimes H_2 \right) \otimes \left( H_3 \otimes H_4 \right)$ to these eigenvectors and then to basis vectors of $\left( H_1 \otimes H_3 \right) \otimes \left( H_2 \otimes H_4 \right)$.

4. Apr 11, 2013

### fzero

Decomposing

$$H_1 \otimes H_2 \otimes H_3 \otimes H_4 = \oplus_a H_a$$

gives the 9j-symbols.

Decomposing

$$H_2 \otimes H_3 \otimes H_4 = \oplus_\alpha H'_\alpha$$

gives 6j-symbols. If we now equate

$$\oplus_a H_a = H_1 \otimes \left( \oplus_\alpha H'_\alpha\right),$$

we'll start getting relationships betwen 9j and 6j-symbols.

You could do these splits and write 9j-symbols in terms of 3j-symbols. But this is equivalent to writing 9j-symbols in terms of 6j-symbols and then rewriting those 6j-symbols in terms of 3j-symbols.

5. Apr 12, 2013

### Yoran91

I'm sorry but I still don't understand.

I don't see how the tensor product decomposition gives 9j symbols, or how the tensor product of three spaces gives 6j symbols, at least not in the case you've given.

All I know about the 6j symbols arising in the product is through rearranging the tensor product and then finding a unitary transformation between the two bases one obtains, as I've shown in the previous post.

I don't see what the 'a' in
$H_1 \otimes H_2 \otimes H_3 = \oplus_a H_a$ is. Thus I don't understand what the 6j symbols would look like.

Last edited: Apr 12, 2013