I am reading Peter Woit's book Quantum Theory, Groups and Representations, section 9.42 (a similar version can be found here). In the book he proved a version of the Clebsch-Gordan decomposition theorem for $G=SU(2)$:

Theorem 9.1. The tensor product $(\Pi_{V^{n_1}\otimes V^{n_2}},V^{n_1}\otimes V^{n_2})$ decomposes into irreducible representations as $$(\Pi_{n_1+n_2},V^{n_1+n_2})\oplus (\Pi_{n_1+n_2-2},V^{n_1+n_2-2})\oplus\cdots\oplus (\Pi_{|n_1-n_2|},V^{|n_1-n_2|}).$$ Here, $(\Pi_k,V^k)$ is the $k+1$-dimensional irreducible representation of $SU(2).$

And his proof of this theorem goes roughly like this (this can be found in section 9.4.2):

proof: The group $U(1)=\{e^{i\theta}\in\mathbb{C}:0\leq \theta<2\pi\}$ can be embedded naturally into $SU(2)$ by sending $e^{i\theta}$ to the matrix $ \begin{pmatrix} e^{i\theta} & 0\\ 0& e^{-i\theta} \end{pmatrix}. $ And it can be shown that the action of $\Pi_n$ on $U(1)\subset SU(2)$ is given by $$ \Pi_n\begin{pmatrix} e^{i\theta} & 0\\ 0& e^{-i\theta} \end{pmatrix}= \begin{pmatrix} e^{in\theta}&&&\\ & e^{i(n-2)\theta}&&\\ &&\ddots&&\\ &&& e^{i(-n+2)\theta}\\ &&&& e^{i(-n)\theta} \end{pmatrix}. $$ Hence one can show that the character function on this element is given by $$ \chi_{V^n}\begin{pmatrix} e^{i\theta} & 0\\ 0& e^{-i\theta} \end{pmatrix}=e^{in\theta}+e^{i(n-2)\theta}+\cdots+e^{-in\theta}= \frac{e^{i(n+1) \theta}-e^{-i(n+1) \theta}}{e^{i \theta}-e^{-i \theta}}=\frac{\sin ~(n+1) \theta}{\sin \theta} $$ using the Weyl character formula. Now since the character on a tensor product is multiplicative, one can show that on the group $U(1)$, we have $$ \begin{align*} \chi_{V^{n_1}\otimes V^{n_2}}&=\chi_{V^{n_1}}\chi_{V^{n_2}}\\ &=(e^{in_1\theta}+e^{i(n_1-2)\theta}+\cdots+e^{-in_1\theta})\frac{e^{i(n_2+1) \theta}-e^{-i(n_2+1) \theta}}{e^{i \theta}-e^{-i \theta}}\\ &=\frac{(e^{i(n_1+n_2+1)\theta}-e^{-i(n_1+n_2+1)\theta})+\cdots+(e^{i(n_2-n_1+1)\theta}-e^{-i(n_2-n_1+1)\theta})}{e^{i\theta}-e^{-i\theta}}\\ &=\chi_{V^{n_1+n_2}}+\chi_{V^{n_1+n_2-2}}+\cdots +\chi_{V^{n_2-n_1}} \end{align*} $$ where we are assuming $n_2>n_1$.* So, when we decompose the tensor product of irreducibles into a direct sum of irreducibles, the ones that must occur are exactly those of Theorem 9.1.*

Remarks: Thanks to Meng Cheng's answer, I now see that since every element of $SU(2)$ is a conjugate of some element in $U(1)$, and since the character is a class function, it suffices to consider $\chi|_{U(1)}.$ Then using the fact that $SU(2)$ is compact, hence completely reducible, two representations of $SU(2)$ are equivalent iff they have the same character. This completes the proof. (Although I am not entirely sure how to write $S\in SU(2)$ as $QUQ^{-1}$ for some $Q\in SU(2), U\in U(1)$. I will appreciate it if somebody can explain it to me).

My Question:

I know that the Clebsch-Gordan decomposition theorem has something to do with the Clebsch-Gordan coefficients and the addition of angular momentum. But from this purely abstract treatment, I cannot see how they are related. Could anyone enlighten me?

  • $\begingroup$ For Q 1, characters are functions of conjugacy classes, and for SU(2) up to conjugacy every element is in a U(1) subgroup. So it's enough to consider U(1). For 2, it's a standard result in finite-dimensional representations of (finite or compact) groups. $\endgroup$
    – Meng Cheng
    Commented Sep 2, 2022 at 4:28
  • $\begingroup$ To reopen this post (v1) consider to only ask 1 question per post. $\endgroup$
    – Qmechanic
    Commented Sep 2, 2022 at 5:12
  • $\begingroup$ Woit as in most pure math treatments of the question only care about the abstract decomposition, whereas physicists require a decomposition with explicit intertwiners, i.e., as in ACuriousMind's answer one needs to explain precisely what $| J_im_{J_i}\rangle$ means as a vector inside $V_{j_1}\otimes V_{j_2}$. You can find more info in my MO answer to mathoverflow.net/questions/439481/… $\endgroup$ Commented Oct 10, 2023 at 19:04

1 Answer 1


In the decomposition $$ V_{j_1}\otimes V_{j_2} = \oplus_i V_{J_i}$$ you have on the one hand the natural basis $\lvert j_1m_1j_2m_2\rangle = \lvert j_1m_1\rangle \otimes \lvert j_2m_2\rangle$ of the tensor product and on the other hand the natural bases $\lvert J_i m_{J_i}\rangle$ of each of the summands on the r.h.s.

The Clebsch-Gordan coefficients are the inner products $\langle j_1m_1j_2m_2\vert J_i m_{J_i}\rangle$ between these two bases (defined to be zero when $J_i$ does not occur on the r.h.s. at all).

  • $\begingroup$ Thanks! But in the case of angular momentum, is the group still SU(2)? Or is it SO(3) instead? $\endgroup$ Commented Sep 2, 2022 at 21:18
  • $\begingroup$ @TianyiWang That...probably doesn't matter as much as you think, since we usually look at projective representations of $\mathrm{SO}(3)$, which are the same as the linear representations of $\mathrm{SU}(2)$, see also this Q&A of mine. $\endgroup$
    – ACuriousMind
    Commented Sep 2, 2022 at 22:03
  • $\begingroup$ Thanks! I'll take a look! $\endgroup$ Commented Sep 2, 2022 at 22:52

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