I am trying to understand the relation between Wigner's $3j$-symbols (or Clebsch-Gordan coefficients) and matrix coefficients of intertwiners. I am new to this topic and need some help to understand it properly.

So, let us start with the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which is the complexification of $\mathfrak{su}(2)$. Now, as usual, all the irreducible representations of $\mathfrak{sl}(2,\mathbb{C})$ can be described by spins $j\in\mathbb{N}_{0}/2$ and have dimension $2j+1$. Now it is a general fact that for a simple and complex Lie algebra, ever finite-dimensional irreducible representation can be labeled with a heighest weight $\Lambda$, which in the case of $\mathfrak{sl}(2,\mathbb{C})$ are given by $2j$. In general, the tensor product of two such heighest weight modules is fully reducible and hence we can write $$V_{\Lambda}\otimes V_{\Lambda^{\prime}}=\bigoplus_{i}C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}V_{\Lambda_{i}}$$ with some coefficients, usually called multiplicities or Littlewood-Richardson coefficients. Now in order to relate this to intertwiners, we first of all know, according to the Lemma of Schur, that the space of intertwiners between two irreducible finite-dimensional representations of a complex Lie algebra $\mathfrak{g}$ is either $0$-dimensional (if they are not isomorphic) or $1$-dimensional (if they are isomorphic). As a consequence, we get that the Littlewood-Richardson coefficients are given by the dimension of the space of intertwiners from $V_{\Lambda_{i}}$ to $V_{\Lambda}\otimes V_{\Lambda^{\prime}}$, i.e.

$$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{\Lambda_{i}},V_{\Lambda}\otimes V_{\Lambda^{\prime}}))$$

So in the case of $\mathfrak{sl}(2,\mathbb{C})$ (or equivalently $\mathfrak{su}(2)$), the coefficients are hence given by

$$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}}))$$

So far so good. In a textbook (Fuch, Schweigert — Symmetries, Lie algebras and representations), they then say that using this correspondence, it is clear that the Clebsch-Gordan coefficients for fixed $J$ are the matrix coefficients of intertwiners in $\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}})$ for "a definite choice of basis". Can anyone explain this step to me in more detail? Furthermore, what does "a definite choice of basis" mean in this context? Also, what is then the correspondence with the $3j$-symbols?

Thank you very much!

EDIT: In order to clarify the conventions I use, let me add some more details about the Clebsch-Gordan coefficients: Let $J_{\pm},J_{0}$ be the three generators of $\mathfrak{sl}(2,\mathbb{C})$, i.e. $$[J_{0},J_{\pm}]=\pm J_{\pm}\hspace{1cm}\text{and}\hspace{1cm}[J_{+},J_{-}]=2J_{0}$$ Then there is a basis $\{\mid j,m\rangle\}_{-j\leq m\leq j}$ of $V_{2j}$ satisfying $$J_{0}\mid j,m\rangle=m\mid j,m\rangle$$ $$J_{\pm}\mid j,m\rangle=\sqrt{j(j+1)-m(m\pm 1)}\mid j,m\pm 1\rangle$$ In order to define the Clebsch-Gordan coefficients, one usually defines to different bases of the tensor product $V_{2j}\otimes V_{2j^{\prime}}$:

(1) The tensor product of the bases described above, i.e. $\{\mid j_{1},j_{2},m_{1},m_{2}\rangle\}_{-j_{1}\leq m_{1}\leq j_{1},-j_{2}\leq m_{2}\leq j_{2}}$ where $\mid j_{1},j_{2},m_{1},m_{2}\rangle:=\mid j_{1},m_{1}\rangle\otimes\mid j_{2},m_{2}\rangle$.

(2) Define the total spin $\vec{J}:=\vec{J}_{1}+\vec{J}_{2}$. Then there is a basis $\{\mid J,M\rangle:=\mid j_{1},j_{2},J,M\rangle\}$ satisfying $$\vec{J}^{2}\mid J,M\rangle=J(J+1)\mid J,M\rangle$$

Then the Clebsch-Gordan coefficients are the coefficients of the change of basis matrix, i.e. $\langle j_{1},j_{2},m_{1},m_{2}\mid J,M\rangle$

  • $\begingroup$ Sorry, maybe I should add the definition of the Clebsch-Gordon coefficients...$J$ are the eigenvalues of the "total spin" appearing in the definition of the Clebsch-Gordon coefficients... $\endgroup$ Commented Jul 3, 2021 at 17:16

2 Answers 2


There is a standard basis for finite-dimensional $SO(3)$ (and hence $\mathfrak{su}(2)$) representations, denoted by $\left| j, m\right\rangle$; the total spin $j\ge 0$ labels the representation, the magnetic quantum number $m=-j,\ldots,j$ labels the basis. If you look at the first few equations in the Wikipedia article on $3j$-symbols, one can immediately realize that they give the following formula for the (unique up to a scalar multiple) intertwiners in this basis: $$ I_{j_1,j_2}^{j_3} \colon \left| j_3, m_3\right\rangle \mapsto \sum_{m_1=-j_1}^{j_1} \sum_{m_2=-j_2}^{j_2} (-1)^{-j_1+j_2+m_3} \sqrt{2j_3+1} \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & -m_3 \end{pmatrix} \left|j_1, m_1\right\rangle \otimes \left|j_2, m_2\right\rangle.$$

  • 1
    $\begingroup$ Okay, thank you very much for your answer. So, to understand correctly, $I_{j_{1}j_{2}}^{j_{3}}$ in your notation is the (unique up to multiple) intertwiner of the form $I_{j_{1}j_{2}}^{j_{3}}:V_{j_{3}}\to V_{j_{2}}\otimes V_{j_{2}}$ and hence, using your formula, the matrix coefficients of this intertwiner are exactly given by $(-1)^{-j_{1}+j_{2}+m_{3}}\sqrt{2j_{3}+1}$ times the $3j$-symbol? $\endgroup$ Commented Jul 3, 2021 at 17:34
  • $\begingroup$ @G.Blaickner You got it! $\endgroup$ Commented Jul 3, 2021 at 17:37
  • $\begingroup$ Okay, great! Thank you again! $\endgroup$ Commented Jul 3, 2021 at 17:37

You can find a quick review of the relevant definitions of $3j$, $6j$, $9j$, symbold in Section 7 of my article with Chipalkatti "The higher transvectants are redundant", Annales de l'Institut Fourier, Tome 59 (2009) no. 5, pp. 1671-1713.

I think the best best way to see the vector space corresponding to the spin $j$ representation (instead of spherical harmonics etc.) is the space of homogeneous polynomials with complex coefficients in two variables and of degree $2j$. The group $SU(2)$ obviously acts on $\mathbb{C}^2$ and thus has a natural action on functions on $\mathbb{C}^2$, in particular polynomial functions, and even more in particular, homogeneous polynomial functions. These are called binary forms.

Finally, please correct the use of "Clebsch-Gordon" to "Clebsch-Gordan" with an "a". Half a century before physicists got interested in these coefficients etc. Paul Gordan and Alfred Clebsch figured out the decomposition into irreducibles of a tensor product of irreducibles for $SU(2)$, with explicit intertwiners. The analogous result in higher dimension is pretty much an open problem. For $SU(3)$, this was done recently by Böhning and Graf von Bothmer in this article. For higher dimension, and for tensor products of symmetric powers only, there are such explicit results, most notably by Jerzy Weyman and Peter Olver. See this article and references therein.

  • $\begingroup$ +1 for the correct spelling of "Gordan" $\endgroup$
    – Buzz
    Commented Jul 4, 2021 at 0:38
  • $\begingroup$ Thank you very much for your answer and for providing some references! $\endgroup$ Commented Jul 4, 2021 at 11:42

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