This paper states that
However, the multiphonon relaxation bridging the $^5\!D_1$ and $^5\!D_0$ levels is a well-known exception because it is formally forbidden by the triangle rule and occurs due to $J$-mixing.
What is the triangle rule? Examples?
[Physics] the ‘triangle selection rule’ in spectroscopy
angular momentumspectroscopy
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these are selection rules for the electric dipole radiation. The transition from the excited state to a lower state of an atom is governed by the following matrix element: $$C \cdot \langle f | \hat{p} | i\rangle$$ You could also write the sum of positions of electrons $\Sigma\hat{r}$ instead of their total momentum $\hat{p}$ in the middle. This simple form of the operator in the middle is because at low enough frequencies, i.e. long enough wavelength, the atom simply finds itself in a uniform field, anf the electric potential $\phi$ for a uniform field is linear in $\hat{r}$. Equivalently, the matrix element may be converted from $\hat{r}$ to $\hat{p}$ and vice versa by realizing that the commutator of the Hamiltonian with $\hat{x}$ is proportional to $\hat{p}$.
At any rate, the operator in the middle is a 3-dimensional vector that only acts on the positions or momenta of the electrons, not their spins. So it has to commute with the total spin operator $\hat{S}$ of the electrons, and $\Delta S=0$ as a consequence.
On the other hand, it is a vector, i.e. a $j=1$ object as far as the SO(3) transformations go, and by combining its $j=1$ angular momentum with the orbital angular momentum $L_i$ of the initial state, you may get the final $L_f$ being $L_i\pm 1$ or $L_i$ according to the basic rules of the addition of angular momentum. You may imagine that you're just adding two vectors of specified lengths, $L_i$ and $1$, and depending on their relative angle, the length of the sum may go from $L_i-1$ to $L_i+1$.
Your list didn't allow $L_f=L_i$ and I think you are right that it is typically forbidden as well because it violates parity. The angular momentum to parity goes like $(-1)^L$, I guess, but because the vector operator is parity-odd, the initial and final states have to differ by their parity as well, which means that only $\Delta L=\pm 1$ is allowed. Everyone, please correct me if I am saying something incorrectly.
Best wishes Lubos
A magnetic dipole transition can be modelled as a time-dependent perturbation $V_{\text{md}}(t) = {e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}e^{-i \omega t}$. Fermi's Golden Rule tells us that the transition rate for $b-X,1$ is proportional to the matrix element of the perturbation between the initial and final states,
$$W \propto \langle \psi_b|{e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}|\psi_{X}\rangle,$$
where $|\psi_b\rangle$ is the excited state and $|\psi_{X}\rangle$ is the ground state (with three possible $M_S$ values.)
The effect of $\vec L$ and $\vec S$ will be to turn the final state into some combination of the triplet states, but it won't change $J$. Therefore we might expect the transition to be 'spin-forbidden':
$$W \propto \langle b^1\Sigma_g^+ |X^3\Sigma_{g,M_S=0,\pm1}^-\rangle = \langle J=0 | J=1 \rangle = 0.$$
This is where the spin-orbit coupling comes into play. Spin-orbit coupling is the reason why the singlet $(b)$ state has a higher energy than the triplet $(X)$ states. It is a perturbation of the form $V_\text{SO}={\mu\over\hbar}\vec{L}\cdot\vec{S}$, which can be rewritten as $V_\text{SO}={\mu \over 2\hbar}(J^2-L^2-S^2)$. In a spherically symmetric system like the helium atom, this perturbation commutes with the Hamiltonian, so all you get is a shift in the energy of the triplet (L=1) and singlet (L=0) states. However, in a linear molecule like $O_2$ you lose the spherical symmetry, so $[L^2,H]\neq0$ and in addition to an energy shift, you also get some mixing of the unperturbed eigenstates, so that the excited state is not exactly $|b^1\Sigma_g^+\rangle$, but rather $|\psi_b\rangle = c_1|b^1\Sigma_g^+\rangle + c_2|X^3\Sigma_{g,M_S=0}^-\rangle$. This mixing of J=0 and J=1 states is what allows $W$ to have a nonzero value. Since we can write $S_x = S_+ + S_-$, there will be a term in the transition rate like
$$W\propto c_2^*\langle X^3\Sigma_{g,M_S=0}^-|S_{\pm}|X^3\Sigma_{g,M_S=\mp1}^-\rangle+\cdots \neq 0.$$
Does this help? I know this is a bit hand-wavy so let me know if I can clarify anything.
Best Answer
The 'triangle rule' seems to be (as used, e.g. in Molecular Spectroscopy by D.J. Millen) the application of the triangle inequality to the addition of angular momentum. More specifically, if you are adding momenta with magnitudes $j_1$ and $j_2$ and magnetic quantum numbers $m_1$ and $m_2$, then the result will only have nonzero amplitude along the state $|j_3,m_3⟩$ (i.e. the Clebsch-Gordan coefficient will be nonzero) if the conditions $$ |j_k-j_l|\leq j_i \text{ for all distinct }i,k,l,\quad \text{and}\quad m_1+m_2=m_3. $$ These are justified in depth in any angular-momentum-in-QM book.
It is the first of those conditions that would be understood to be the triangle rule, as it directly embodies the triangle inequality. This is the rule that's at play here: a ${}^5\! D_1\to{}^5 \!D_0$ transition involves a change in $J$ from 1 to 0. However, since phonons have zero spin, the triangle rule forbids any phononic transitions which change the total angular momentum.
It's important to note, though, that for atoms as big as europium, the assumptions behind term symbols tend to break down, and instead of having a bunch of electrons with a well-defined total orbital and total spin angular momentum, the spin-orbit coupling on each electron is so big that you're best off by thinking of each electron as having a well-defined total $j$, and then adding those together. In the middle ground, this simply means that a term symbol like ${}^5\! D_1$ is only an approximate description of the true eigenstate, which includes a nonnegligible amplitude of ${}^5\! D_0$ character. This is what's known as $J$-mixing and it does enable phononic transitions between such states.