# What symmetry has to be broken to result in the hyperfine structure splitting?

In Quantum Mechanics, symmetry operators $$\mathcal{S}$$ are associated with degeneracies if the Hamiltonian $$H$$ is conserved under the symmetry operation in question

$$\mathcal{S}^{\dagger}H\mathcal{S} = H. \tag{1}$$

This implies that $$[G,H]=0$$, where $$G$$ is the generator of $$\mathcal{S}$$, which in turn means that $$G$$ and $$H$$ have simultaneous eigenstates. If $$H$$ is changed such that Eq. $$(1)$$ is no longer valid, then the degeneracy is lifted.

Now, the hyperfine structure in atoms arises from the interaction of spin of the angular momentum of electrons $$\mathbf{J}$$ with the spin of the nucleus $$\mathbf{I}$$, which leads to the splitting of energy levels. What symmetry is broken when this interaction is accounted for in $$H$$ that lifts this degenercy? Knowing that angular momenta are generators of rotations, my initial thought is to say that some sort of rotational symmetry is broken here, but I can't see how.

• Write down the interaction term. Is J conserved? Is I? Review all quantum numbers here. Which ones are good, and which bad? Commented Aug 23, 2023 at 21:22

• $$\mathbf{J}$$ is the total (i.e. orbital + spin) angular momentum of the electrons, without the nuclear spin. The operator $$\mathbf{J}$$ generates the transformations $$U_e(\boldsymbol\phi)=\exp\left(\frac{i}{\hbar}\boldsymbol{\phi}\cdot\mathbf{J}\right)$$ where $$\boldsymbol{\phi}$$ denotes the rotation axis and angle. Such a transformation rotates the electrons (orbitals and spins) without rotating the nucleus. $$\mathbf{J}$$ is not conserved, and hence (due to Noether's theorem) $$U_e(\boldsymbol{\phi})$$ is not a symmetry transformation. Or in other words: rotating only the electrons is a broken symmetry.
• $$\mathbf{I}$$ is the nuclear spin, without the electrons' angular momentum. The operator $$\mathbf{I}$$ generates the transformations $$U_n(\boldsymbol{\phi})=\exp\left(\frac{i}{\hbar}\boldsymbol{\phi}\cdot\mathbf{I}\right)$$ where $$\boldsymbol{\phi}$$ denotes the rotation axis and angle. Such a transformation rotates the nucleus, without rotating the electrons. $$\mathbf{I}$$ is not conserved, and hence (due to Noether's theorem) $$U_n(\boldsymbol{\phi})$$ is not a symmetry transformation. Or in other words: rotating only the nucleus is a broken symmetry.
• On the other hand: $$\mathbf{J}+\mathbf{I}$$ (the total angular momentum of electrons and nucleus) is conserved. And therefore the transformations generated by it $$U_{e+n}(\boldsymbol{\phi})=\exp\left(\frac{i}{\hbar}\boldsymbol{\phi}\cdot(\mathbf{J}+\mathbf{I})\right)$$ (rotating electrons and nucleus together) are symmetry transformations.