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.
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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."
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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.
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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.
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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.
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What does it mean for a transition to be spin-forbidden?
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What is a singlet-triplet intercombination band? Does this change the selection rules? How is this related to being spin-forbidden?
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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.
Best Answer
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.