I know the transition rule $\Delta S = 0$. But where does that rule come from?

Is it just very unlikely that an absorbed/emitted $\gamma$ will carry the energy necessary for a spin flip?

Or is there another, deeper reason?


2 Answers 2


$\let\D=\Delta \let\om=\omega \let\lam=\lambda \def\hH{\hat H}$ $\def\bB{{\bf B}} \def\bk{{\bf k}} \def\br{{\bf r}}$ The reason for $\D S=0$ is deeper than angular momentum conservation and stronger than electric dipole approximation.

Consider an em plane wave impinging over the atom. If spin magnetic moments are neglected, the only possible interaction is between the wave's electric field and the electrons charges. Even if no approximation is done (electric dipole or other) the corresponding term in the hamiltonian will be symmetric in the electrons' positions operators.

Now we know that parahelium states are symmetric under interchange of space coordinates, antisymmetric under spins interchange, where for orthohelium the opposite is true: space antisymmetric, spin symmetric. Given the different symmetries wrt space coordinates, no matrix element may exist for the interaction hamiltonian (symmetric) between parahelium and orthohelium states.

We are thus obliged to take recourse to interaction of the em wave with spin magnetic moments. The general expression of the wave's magnetic field is $\bB=\bB_0 \exp[i(\bk\cdot\br-\om t)]$. The lowest approximation in its expansion in powers of $\bk$ is $\bB_0$, giving rise to an interaction hamiltonian $-\bB\cdot(\mu_1+\mu_2)$, where $\mu_1$, $\mu_2$ are the spin magnetic moments of the electrons. This term is symmetric in the spins, so the same selection rule $\D S=0$ applies.

In order to violate that selection rule the ensuing term must be taken into account: $i\,\bB_0\,(\bk\cdot\br)$. This would work, but let's try to compute its magnitude. It will be something like $B_0\,k\,a_0\,\mu_0$ ($a_0=$ Bohr's radius; electron magnetic moment is one Bohr's magneton; Gauss units are used). For comparison consider an electric dipole transition: we will have $E_0\,e\,a_0$. Recalling that in a plane wave $E_0=B_0$ the ratio is $${k\mu_0 \over e} = {2\pi \hbar \over \lam m c} = {h \over \lam m c}.$$

Inserting numerical values ($\lam = 0.5\,\rm nm$ in the visible range, $h/mc=2.4\,\rm pm$ we find a ratio $5\cdot10^{-3}$ for the matrix element, $2.5\cdot10^{-5}$ for transition probability.


The requirement that $\Delta S=0$ comes from the conservation of angular momentum when the radiation is taken in the dipole approximation (i.e. as an oscillating uniform electric field, with no magnetic field. This is the first (and main) step in a hierarchy of transition types which is described in Wikipedia here. There are higher rungs in that ladder, starting with magnetic dipole transitions, which do allow for spin flips.


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