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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.

Qmechanic
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1 Answers1

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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)

ZeroTheHero
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