Why are there no transitions between orthohelium and parahelium? 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?
 A: $\let\D=\Delta \let\om=\omega \let\lam=\lambda \def\hH{\hat H}$
$\def\bB{{\bf B}} \def\bk{{\bf k}} \def\br{{\bf r}}$
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.
A: 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. 
