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:
-
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.
-
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.
Best Answer
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}
Thus, the CGs allow you to go in both direction: from the uncoupled to the coupled basis as in (1), and from the coupled to the uncoupled basis as in (2).