How to derive the selection rules $\Delta L= \pm 1$, $\Delta S=0$ for electron spectroscopy?
[Physics] Selection Rules in electron spectroscopy
quantum mechanicsspectroscopy
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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.
There are several different issues conflated together here: selection rules, separation between energy levels, and energy level population (which you didn't mention).
For vibrational spectroscopy, in the approximation that a vibrational mode behaves like a quantum harmonic oscillator, the energy levels are equally spaced and the selection rule is $\Delta n=\pm 1$, where $n$ is the quantum number. The reason for this is explained here. So you expect to see (and do see) an absorption transition from $n=0$ to $n=1$. You might also expect to see a transition from $n=1$ to $n=2$ etc. This would occur at the same frequency since the gap between successive energy levels is the same. However, it relies on there being a thermal equilibrium population of molecules already in the $n=1$ state. For most molecules, at normal temperatures, the population of $n=1$ and higher levels (determined by the Boltzmann factor) is rather low. At elevated temperatures, you might see such transitions; also the frequency won't be exactly at the same frequency as the $n=0\rightarrow 1$ transition, because of anharmonicity effects.
For a linear rotor, the quantum levels are at $BJ(J+1)$ where $B$ is a constant and $J$ is the quantum number. These are not evenly spaced. The selection rule is $\Delta J=\pm 1$ (angular momentum conservation). So you expect to see (and do see) transitions between successive levels: $J=0\rightarrow 1$, $J=1\rightarrow 2$ etc. In this case, at normal temperatures, the spacing between rotational levels is typically small compared with the available thermal energy. So those higher states are populated, at least for $J$ not too high. The frequencies are not all the same, but the energy level spacings change linearly with $J$: $$\Delta E_{J\rightarrow J+1}=B(J+1)(J+2)-BJ(J+1)=2B(J+1).$$ So you see a spectrum with equally spaced lines for $J=0,1,2\ldots$ (in this rigid rotor approximation).
EDIT following OP comment.
- The first line, $J=0\rightarrow 1$ is at a frequency $\nu$ given by $h\nu=\Delta E=2B$ where $h$ is Planck's constant.
- The second line, $J=1\rightarrow 2$ is at $h\nu=\Delta E=4B$.
- The third line, $J=2\rightarrow 3$ is at $h\nu=\Delta E=6B$.
And so on. Hence the lines in the spectrum are equally spaced, $2B$ apart (in energy units) or $2B/h$ in frequency units. You can also see a diagram of this in the Linear Molecules section of the Rotational Spectroscopy Wikipedia page (reproduced below under the terms of the CC BY-SA 3.0 licence). The diagram shows the link between the energy levels and the lines in the spectrum (the only difference is that the transitions on the energy level diagram on that page are drawn for emission lines, $J\leftarrow J+1$, but exactly the same frequencies occur for the corresponding absorption lines $J\rightarrow J+1$).
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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