The tensor product of two states with spin can be decomposed into irreducible representations via Clebsh-Gordan coefficients

$$|j_1, m_1, j_2, m_2 \rangle = \sum C^{JM}_{j_1, m_1, j_2, m_2} |JM\rangle\,.$$

Since spherical harmonics $Y_{\ell m}$ are representations of $SO(3)$, I would have expected a similar decomposition, i.e.

$$Y_{\ell_1 m_1} (\Omega) Y_{\ell_2 m_2}(\Omega) = \sum C^{L M}_{\ell_1 m_1 \ell_2 m_2} Y_{L M}(\Omega)\,.$$

However, the Wikipedia page on Clebsh-Gordan coefficients instead gives the expansion

$$Y_{\ell_1 m_1} (\Omega) Y_{\ell_2 m_2}(\Omega) = \sum_{L,M} \sqrt{\frac{(2\ell_1 + 1)(2\ell_2 + 1)}{4\pi (2 L+1)}} C^{L M}_{\ell_1 m_1 \ell_2 m_2}C^{L 0}_{\ell_1 0 \ell_2 0} Y_{L M}(\Omega)\,.$$

How can I understand where these additional terms come from? I've found some derivations of the expression in Sakurai's Modern Quantum Mechanics, and I can follow the derivation, but I don't understand where the discrepancy arises on the level of representation theory.

  • $\begingroup$ L could not slip outside the summation sign. Where do you find a "discrepancy", instead of an identity? $\endgroup$ Commented Nov 5, 2019 at 22:52
  • $\begingroup$ The discrepancy I have is that the Clebsh-Gordan coefficients is supposed to give the decomposition of the tensor product of two representations into a direct sum of irreducibles. As such, I would expect them to give the decomposition of the product of spherical harmonics (the second equation). Instead, it is the third equation that describes this. Perhaps the problem comes from identifying the product of two harmonics as a "tensor product," but since the decomposition still involves Clebsh-Gordan coefficients, it seems like it has some sort of interpretation in terms of representation theory. $\endgroup$ Commented Nov 6, 2019 at 0:14
  • $\begingroup$ The answer should explain this to you. The fallacy in your expected equation is blindly dotting two $|\omega\rangle$s on the l.h.s. and just one on the right. $\endgroup$ Commented Nov 6, 2019 at 0:25

1 Answer 1


The “missing” Clebsch is hidden by the nature of the spherical harmonics as coset functions, i.e. functions over $SU(2)/U(1)$.

The best way to understand the occurrence of this CG is by expressing the spherical harmonics in terms of full $SU(2)$ group functions: \begin{align} Y_{LM}(\beta,\alpha)=\sqrt{\frac{2L+1}{4\pi}}D^{L*}_{M_10}(\alpha,\beta,\gamma). \tag{1} \end{align} The $\gamma$ dependence (i.e. the $U(1)$ factors) drops out because the second projection $M_2=0$.

As a special case of combining full group functions, we thus have \begin{align} D^{L*}_{M_10}(\Omega)D^{\ell*}_{m_10}(\Omega)= \left[\langle L M_1\vert\langle \ell m_1\vert\right] R(\Omega)\left[ \vert L 0\rangle \vert \ell 0\rangle\right]^* \end{align} and so one CG is needed to combine the kets: \begin{align} \vert L 0\rangle \vert \ell 0\rangle = \sum_{J} C_{L0;\ell 0}^{J0}\vert J 0\rangle \tag{2} \end{align} and one is needed to combine the bras.

Note the proportionality factor in (1) is what produces the various $\sqrt{2L+1}$ factors in your expression.

FYI there’s quite a sneaky way of evaluating the CG of (2) in Claude Cohen-Tannoudji’s QM book (with Diu and Laloe)

  • $\begingroup$ Thanks a lot, I think I almost understand this. My one confusion is the following - I had always understood spherical harmonics to be the space under which the representation of $SO(3)$ acts on. i.e, the ket $| \ell m\rangle$ should be identified with $Y_{\ell,m}$, with the inner product defined as integration over the unit sphere. Is this not actually the case? $\endgroup$ Commented Nov 6, 2019 at 0:29
  • $\begingroup$ If you think of functions as $\psi(x)=\langle x\vert \psi\rangle$ then your spherical harmonic is basically $\langle \ell 0\vert R(\Omega)\vert \ell m\rangle$. and then it makes sense to integrate over $S^2$ since the function depends on the spherical angles. The point is: you still need an inner product, i.e. one bra (It is fixed in the irrep) and one ket, to define the function. (I’m being sloppy with the complex conjugate here.) I think this is the simplest way to see this CG but there are more convoluted explanations (v.g. Cohen-Tannoudji) which evaluate directly the missing CG. $\endgroup$ Commented Nov 6, 2019 at 0:35
  • $\begingroup$ Ah that makes sense, thanks a lot. $\endgroup$ Commented Nov 6, 2019 at 0:42

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