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In their book Akhiezer et al. give a definition of vector spherical harmonics (p.18 of Russian Edition) as

$$\pmb{Y}_{j\ell m}(\pmb \Omega) = \sum_{m' \lambda} \langle \ell m' 1\lambda| jm \rangle Y_{\ell m'} (\pmb \Omega) \hat{e}_{\lambda}$$

where

$$\hat{e}_{\pm1} = \mp \frac{1}{\sqrt{2}} (\hat e_x \pm i \hat e_y),\; \hat e_0 = \hat e_z$$

Then they state that $\hat J_z \pmb{Y}_{j\ell m}(\pmb \Omega) = m \pmb{Y}_{j\ell m}(\pmb \Omega)$, which is expected. Assuming $J_z = L_z + S_z$, I do not get how this can be true for the definition above. I would agree with $\hat L_z \pmb{Y}_{j\ell m}(\pmb \Omega) = m \pmb{Y}_{j\ell m}(\pmb \Omega)$ though. Is there some weird cancellation going on? What am I missing?

Also, if one knows more systematic and reliable reference for vector spherical harmonics, please, share. I am having troubles with relating the definitions to the polarization bases choice.


I have assumed $$J_z = L_z + S_z = -\frac{\hbar}{i} \Big( \frac{\partial}{\partial \phi} - \sigma_z \Big)$$

where $\sigma_z = \{\{1,0\},\{0,-1\}\}$ - Pauli matrix.

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This is wrong:

$$\hat L_z \pmb{Y}_{j\ell m}(\pmb \Omega) = m \pmb{Y}_{j\ell m}(\pmb \Omega)$$

Your definition of the vector spherical harmonics $$ \pmb{Y}_{j\ell m}(\pmb \Omega) = \sum_{m' \lambda} \langle \ell m' 1\lambda| jm \rangle Y_{\ell m'} (\pmb \Omega) \hat{e}_{\lambda} $$ has support over a bunch of scalar spherical harmonics $Y_{\ell m'} (\pmb \Omega)$ with different $\hat L_z$ eigenvalues $m'$. As in the general case, the superposition of eigenstates of an operator $\hat A$ with different eigenvalues will not be an eigenstate of that operator.

If you want to see it that way, then it is accurate to say that there are indeed 'cancellations': you have the same thing with the $\hat{e}_\lambda$ (which are eigenstates of $\hat S_z$, but with different eigenvalues), and the actions of $\hat L_z$ and $\hat S_z$ in directions 'away' from $\pmb{Y}_{j\ell m}(\pmb \Omega)$ will cancel out, with only a multiple of $\pmb{Y}_{j\ell m}(\pmb \Omega)$ remaining. However, those 'cancellations' are pretty indistinguishable from 'linear algebra happens' - they're not a particularly useful way to understand the structure.

As to what's "really" going on, it's exactly the same thing that you have in the quantum mechanical addition of angular momenta. Your vector spherical harmonics are functions of in the vector space $$ \pmb{Y}_{j\ell m} \in V=\left\{ \mathbf f:\mathbb S^2 \to \mathbb C^3 : \int_{\mathbb S^2} |\mathbf f(\pmb\Omega)|^2 \mathrm d \pmb\Omega <\infty \right\}, $$ and this vector space can be decomposed as the tensor product $$ V = L_2(\mathbb S^2,\mathbb C) \otimes \mathbb C^3. $$ You then have two representations of the rotation group $\mathrm{SO}(3)$, where the generators are denoted $\hat L_z$ and $\hat S_z$ respectively, in a tensor product, and you want to find the irreducible representations that make up that tensor space in terms of the irreducible representations on the individual tensor factors (i.e. $V_\ell =\mathrm{span}\{Y_{\ell m}: -\ell \leq m \leq \ell \}$ on $L_2(\mathbb S^2,\mathbb C)$, with $\mathbb C^3$ naturally forming an irreducible $\ell=1$ representation).

That is precisely the same mathematical problem as finding the eigenstates of the total angular momentum of a particle with orbital angular momentum $\ell$ and spin $1$, and it therefore has exactly the same solutions in terms of Clebsch-Gordan coefficients.

You can ask for a more detailed, answer, but (because the mathematical structures are identical) the only thing that you will get is a re-workthrough of the Clebsch-Gordan procedure that's been specialized to the spin-one case.

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  • $\begingroup$ Thanks a lot for your answer! Yes, I am terribly uncomfortable with symbolic representations that you are using in your explanation, but I have seen them a ton of times in the literature. Can you suggest a good reference? $\endgroup$ – MsTais Aug 30 '18 at 14:31
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    $\begingroup$ I don't have a good one to point to, I'm afraid. It would make a decent resource-recommendation question, though. $\endgroup$ – Emilio Pisanty Aug 30 '18 at 14:36

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