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For orthonormal complex spherical harmonics, $P_{\ell }(\hat{\mathbf{n}}\cdot \hat{\mathbf{m}})=4\pi {\sum }_{m=-\ell }^{\ell }{Y}_{\ell m}(\hat{\mathbf{n}})Y_{\ell m}^{\ast }(\hat{\mathbf{m}})/(2\ell +1)$. This converts an angle between two directions into a separable sum. Together with Orthogonality of Legendre polynomials, it projects rotationally invariant scattering kernels onto angular multipoles.

Here $\mu =\cos \theta$ and $a_{\ell ,m}=\sqrt{({\ell }^2-m^2)/[(2\ell +1)(2\ell -1)]}$. This follows by combining the associated Legendre function recurrence $(2\ell +1)\mu P_{\ell }^m=(\ell -m+1)P_{\ell +1}^m+(\ell +m)P_{\ell -1}^m$ with the normalized spherical harmonic constants. It gives the adjacent-multipole couplings in a photon Boltzmann hierarchy.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.

The angular basis ${}_s{Y}_{\ell m}$ for a spin-weight-$s$ field generalizes ordinary spherical harmonics. It has $\ell \ge |s|$. For scalar-source CMB polarization about a wavevector, only $m=0$ survives and ${}_{\pm 2}{Y}_{20}\propto {\sin }^2\theta$.

In the sudden approximation, changing the nuclear charge of a one-electron hydrogen atom from $Z_{i}e$ to $Z_{f}e$ leaves its electronic state unchanged during the change. Its subsequent probability to occupy a final state is the squared overlap with that state. Using normalized $1s$ wavefunctions ${\psi }_Z=({\beta }_Z^3/\pi {)}^{1/2}{e}^{-{\beta }_{Z}r}$, ${\beta }_Z=Z/a_0$, givesThis follows by integrating $4({\beta }_i{\beta }_f{)}^{3/2}{r}^2{e}^{-({\beta }_i+{\beta }_f)r}$ over positive $r$. A spherically symmetric initial state has zero overlap with every final state of nonzero orbital angular momentum, by the orthogonality of the spherical harmonics. Nuclear recoil and interactions with emitted particles can invalidate the perfectly spherical sudden-change model.

For a uniform-density sphere with incompressible flow of radius $R$ with one surface spherical harmonic displacement ${\xi }_r(R)=lA{R}^{l-1}{Y}_l^m$, the bulk density perturbation vanishes but the surface contributes $\delta \rho =\rho {\xi }_r(R)\delta (r-R)$. Continuity of the Newtonian gravitational potential and the Poisson equation give $[{\partial }_r\delta \Phi {]}_{\mathrm{out}-\mathrm{in}}=4\pi G\rho {\xi }_r(R)$. The regular interior and decaying exterior solutions areThe coefficient follows because the derivative jump is $-(2l+1)C{R}^{l-1}{Y}_l^m$. This surface source is why neglecting the bulk density perturbation does not justify neglecting the perturbed gravity.

The surface Laplacian is the Laplace-Beltrami operator of a surface with its induced Riemannian metric. It is ${\Delta }_{s}f={\nabla }_s\cdot {\nabla }_{s}f$, using the surface gradient and surface divergence. On a sphere of radius $a$, a degree-$l$ spherical harmonic has eigenvalue $-l(l+1)/a^2$.

A sphere with constant electric polarization $\mathbf{P}=P\hat{\mathbf{z}}$ has zero volume bound charge and surface bound charge density ${\sigma }_b=P\cos \theta$. Integrating $\mathbf{r}{\sigma }_{b}dA$ gives the displayed electric dipole moment. The surface distribution is purely degree one in spherical harmonics, so its external static electric potential is exactly dipolar. Its interior field is finite and is not that of a point singularity.

For an angular spherical harmonic $Y$ of degree $n\ge 1$, the radial, tangential gradient and tangential curl families can be normalized as $\mathbf{P}=\hat{\mathbf{r}}Y$, $\mathbf{B}=r\nabla Y/\sqrt{n(n+1)}$ and $\mathbf{C}=-r\hat{\mathbf{r}}\times \nabla Y/\sqrt{n(n+1)}$. They depend only on direction. At $n=0$ only the radial family is defined by these formulas.