I am studying the transition from the second excited electronic state of molecular oxygen, $b^1\Sigma_g^+$ , to the ground state, $X^3\Sigma_g^-$. I know that the ground state has total angular momentum $J=1$, total spin $S=1$, and three spin sublevels $(m_s=-1,0,+1)$. The upper state has $J=0$, $S=0$, and one sublevel with $m_s=0$. I am specifically interested in the transition from the upper state to the ground state $m=\pm1$ level and I will refer to it as the $b-X,1$ transition.

I would like to understand the the selection rules that govern this transition and the language that is used to describe this transition.

Here is what I know so far about this transition.

  • Brecha, Pedrotti, Krause, "Magnetic rotation spectroscopy of molecular oxygen with a diode laser" JOSA 1998: They describe this transition as "doubly weak" and I quote, "First, it is a magnetic dipole transition and thus is roughly 5 orders of magnitude weaker than a normal electric dipole transition. Second, the transition is a singlet-triplet intercombination band, making it 3 orders of magnitude weaker still."

  • Minaev, Agren, 1997: They describe this transition as "magnetic dipolar" and "spin-forbidden." They also say that it is spin-orbit coupling that accounts for this "doubly weak" transition being as large as it is.

  • Wikipedia-Selection Rules: Provides the rules for Magnetic Dipole (M1) transitions and discusses Spin-Obrit (LS) Coupling but I do not understand where this came from.

  • Sannigrahi, "Derivation of Selection Rules for Magnetic Dipole Transitions," 1982: This is pretty clear but just seems to say that $\Delta m_s=\pm1$. This is true for the $b-X,1$ transition but how do I tie in $L$ and $J$ into the selection rules?

Most of what I have found on the internet and in textbooks relating to selection rules is directed to electric dipole transitions. Where I have found something discussing magnetic dipole transitions, it is either a summary of the rules or I do not have the required background to understand it.

  • What does it mean for a transition to be spin-forbidden?

  • What is a singlet-triplet intercombination band? Does this change the selection rules? How is this related to being spin-forbidden?

  • What does spin-orbit coupling have with this?

I look forward to any feedback on this topic.

Follow up Questions

Meaning of Spin-Forbidden

Does 'spin-forbidden' just mean that the transition from a $J=1$ to $J=0$ state is not allowed because the selection rules for magnetic dipole transitions say that $J$ cannot change? I expected 'spin-forbidden' to imply something about the change of the spin between the initial and final states.

For example, suppose I have a time-dependent perturbation like $V_{md}(t) = \frac{e}{m} \vec{S}\cdot \vec{B}(t)$ and I am interested in the transition rate between the initial state $\left| s m_s \right\rangle$ and the final state $\left| s' m_s' \right\rangle$ with the quantization axis in the z direction. As you pointed out, the transition rate will be proportional to $\left\langle s' m_s'\right| \vec{S} \cdot \vec{B}(t) \left| s m_s \right \rangle$. Now if $\vec{B}(t)$ is circularly polarized, the rules for an allowed transition will be $s'-s=0$ and $m_s'-m_s = \pm1$. For $\vec{B}(t)$ polarized in the z direction, $s'-s=0$ and $m_s'-m_s = 0$ so there is not a transition to other states. I would think that if $s'-s=0$ and $m_s'-m_s = \pm1$ are not true (like if $s'=1$ and $s=0$), then the magnetic dipole transition between $\left| s m_s \right\rangle$ and $\left| s' m_s' \right\rangle$ would be called 'spin-forbidden.'

The same arguments could be made for $\vec{L}$ or $\vec{J}$ as I did with $\vec{S}$. Would you also call a magnetic dipole transition between $\left| L=0 \right\rangle$ and $\left| L=1 \right\rangle$ 'spin-forbidden?'

Spin-Orbit Coupling

Now the difference in the energies of the singlet $(b)$ and triplet, ground state $(X)$ is $1.63\text{ eV}$. That seems too large to be due to spin-orbit coupling breaking a degeneracy. If I was pretending this was a hydrogen atom, I would say this was like a transition where the principle quantum number $n$ changed. I am not sure how to talk about this in a molecule.

This isn't key to the rest of your explanation but I did want to clarify to make sure we were on the page.

Singlet-Triplet Intercombination Band

Do you know what the expression 'singlet-triplet intercombination band' is referring to? After your explanation, it seems to refer to the mixing of the singlet and triplet states due to spin-orbit coupling (SOC). Is this true?

Mixing of unperturbed states

How did you know that the perturbed upper state could be written as a combination of the unperturbed upper state and the $M_s=0$ ground state? It makes sense that the states would get mixed up by SOC but I don't know how. If this is a messy explanation, don't worry about it.

I would like to reiterate that your (George G's) explanation has been incredibly helpful. Thank you.

  • 1
    $\begingroup$ Glad I could help! I've never done anything with molecular transitions before, so I'm not sure what energies to expect. I did forget when I wrote that response though that the different spin states will also affect the coulomb interaction of the electrons. This happens because the total wave function of a two-electron state needs to be anti-symmetric, so the symmetric (triplet) spin states have anti-symmetric spatial parts, which means they have a lower interaction energy. $\endgroup$
    – George G
    Commented Mar 21, 2014 at 0:30
  • $\begingroup$ I haven't heard the term 'singlet-triplet intercombination band' before, so I'm not sure what that means. As for the mixing of states, $M_S$ can't change because those molecular states are eigenstates of $S_z$ and $S^2$, just like in hydrogen because the spin part doesn't care about the geometry of the system. However, you do get mixing of the different J's because orbital angular momentum states do care about geometry. $\endgroup$
    – George G
    Commented Mar 21, 2014 at 0:51
  • $\begingroup$ That makes sense that spin does not care about geometry while $J$ will. Thanks! $\endgroup$ Commented Mar 21, 2014 at 1:09

1 Answer 1


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.


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