Symmetry reason why magnetic dipole transitions are suppressed

Question

In the theory of light-matter interaction, electric dipole transitions between two atomic states of same parity are forbidden. This is because the Hamiltonian conserves parity. Is there a symmetry reason why magnetic dipole transitions happen but their amplitudes are extremely suppressed?

Answer 1

This comes from the heirarchy of length scales in the atomic problem. Very briefly, the EM-atom transition matrix element has a term like

$$\int e^{i\vec{k} \cdot \vec{x} }\psi'(\vec{x})\left[\vec{A_0} \cdot (-i\hbar \vec{\nabla} )\right]\psi(\vec{x}) d^3x $$

A standard approximation to this is the dipole approximation, in which we say that $$e^{i\vec{k} \cdot \vec{x} }\approx 1 \text{ .}$$ This is justified by the fact that $\psi$ typically has a term like $e^{-r/a_0}$, so the spatial integral is over a volume about the size of the atom. This is usually much smaller than the wavelength of light considered, so that the EM field is basically constant spatially over the atom. It turns out that when you take this approximation you end up with an electric dipole coupling, because the integral reduces to something like $$\hat{\epsilon}\cdot\int \psi'(\vec{x})\left[ e\vec{x}\right]\psi(\vec{x}) d^3x $$ where $\hat{\epsilon}$ is the photon polarization. Taking the next order in the expansion of $e^{i\vec{k} \cdot \vec{x} }$ gives the magnetic dipole transition, as well as the electric quadrupole. Lots of books give up to the dipole approximation, like Sakurai and Shankar, but I'm not sure of a good treatment of the higher terms.

For atoms, one can say from dimensional arguments that the ratio of a generic magnetic to electric dipole matrix element is $(Z\alpha)^2$, with $Z$ the atomic mass (which determines the typical atom size) and $\alpha$ the fine structure constant (1).

Regarding Ben's point, for the first nucleus I found searching ($^{12}$C), it looks like $r/\lambda\approx 0.01.$ So it might naively appear that the dipole approximation is not too bad, but this is still two orders of magnitude larger than a typical value for atoms.

(1): Foot, Atomic physics.

Answer 2

No. As a counterexample, magnetic dipole transitions in nuclei are often quite strong, competing against E1 and E2 transitions on fairly even terms. Since nuclei and the relevant forces (strong and electromagnetic) don't break any symmetry that is unbroken for atoms, any relative weakness of magnetic dipole transitions in atoms can't be due to symmetry.