# Understanding a statement for the Clebsch-Gordan coefficients

In wikipedia the following statement is made,

"the tensor product representation decomposes as the direct sum of one copy of each of the irreducible representations of dimension $$2J+1$$, where $$J$$ takes values from $$|j_1-j_2|$$ to $$j_1+j_2$$."

I don't understand the statement. For as much as I understand, for the system we consider two types of Hilbert spaces:

1. The Hilbert space which is spanned by the eigenvectors of the uncoupled system, and these are tensor products of the eigenstates of each angular momentum.

2. The Hilbert space which is spanned by the eigenvectors of the coupled system, which are primarly eigenstates of the total angular momentum.

And I know that in order to transition from the uncoupled basis to the coupled, each ket in the coupled basis is expressed as a linear combination basis kets in the uncoupled basis, multiplied with a corresponding Clebsch-Gordan coefficient.

But the statement I do not understand. In fact the statement looks to me as if it is saying the reverse, meaning:

the tensor product representation i.e an eigenstate of the uncoupled Hilbert space such as $$|j_1,j_2;m_1,m_2\rangle$$ is decomposed (expressed or expanded) as a sum of the coupled basis kets corresponding to a value $$J$$. Which is the total opposite of what the Clebsch-Gordan formula does.

• It's hard to penetrate through verbiage. Can you provide an example of tensoring two doublets and Clebsching into a triplet and a singlet? It is one line with trivial Clebsches. It would best illustrate your question, instead of nebulous generalities, no? Commented Aug 8, 2023 at 18:29
• I mean, the Clebsch-Gordan coefficients "go" in both directions. You can use them to expand the uncoupled states in terms of the total-J states, or you can use them to express the total-J states in terms of the uncoupled states, because both are bases for the total angular momentum space. Commented Aug 8, 2023 at 18:30
• @CosmasZachos j_1=1 and j_2=1/2. So the uncoupled basis is spanned by 6 ket vectors, each expressed as a tensor product of the eigenstates of each angular momentum. For this case it says: Thus, the six-dimensional tensor product representation decomposes as the direct sum of a two-dimensional representation and a four-dimensional representation." Does it mean that some states in the uncoupled basis, are expressed as a linear combination of 4 kets in the coupled basis ? Commented Aug 8, 2023 at 18:34
• Illustrate your final question with an example. I have no clue which 4 states you might be possibly talking about. Choose an M. The Clebsch matrix is real and invertible, if that's what you are asking. Write the explicit expressions I mentioned. Unless you've done that, you are talking in circles. Commented Aug 8, 2023 at 19:08
If $$\vert j_1m_1;j_2 m_2\rangle =\sum_{J(M)} C^{JM}_{j_1m_1;j_2m_2} \vert JM\rangle \tag{1}$$ where the Clebsch-Gordan coefficients are the inner products $$C^{JM}_{j_1m_1;j_2m_2} =\langle JM\vert j_1m_1;j_2m_2\rangle = \langle j_1m_1;j_2m_2\vert JM\rangle$$ since the CG's are real. Next, start with $$\vert JM\rangle$$ and expand it in the uncoupled basis using the identity operator $$\hat{\mathbb{1}}=\sum_{j_1m_1;j_2m_2}\vert j_1m_1;j_2m_2\rangle \langle j_1m_1;j_2m_2 \vert$$ to get \begin{align} \vert JM\rangle &= \sum_{j_1m_1;j_2m_2}\vert j_1m_1;j_2m_2\rangle \langle j_1m_1;j_2m_2 \vert JM\rangle \\ &=\sum_{j_1m_1;j_2m_2}\vert j_1m_1;j_2m_2\rangle C^{JM}_{j_1m_1;j_2m_2}\, . \tag{2} \end{align}
The basis-changing Clebsch matrix is orthogonal, real, and thus invertible. In your particular example, choose an M, additively conserved in all equations. So, then, for M = 1/2, as per your comment question, $$|1~~0~~1/2~~1/2\rangle = \sqrt{2/3} |3/2~~1/2\rangle -1/\sqrt{3}|1/2~~1/2\rangle .$$ Can you explain which four states you are talking abut?