Energy Levels of the OH Molecule

The OH transitions^G.1 that occur in the FIR are due to spin-rotation transitions in the ground electronic state. This has an electron configuration of $1s\sigma^{2}2s\sigma^{2}2p\sigma^{2}2p\pi^{3}$ where the letters identify the various electron orbitals in a similar way to the description of atoms. The following discussion of molecular energy levels is given in more detail by Herzberg (1950); Herzberg (1971). The letters $s$, $p$, $d$...denote electrons with orbital angular momentum quantum number, $l=0$, 1, 2...The component of orbital angular momentum in the electric field direction has a quantum number, $m_{l}$, where $m_{l}=l$, $l-1$, $l-2$, ... $-l$. Molecular orbitals are represented by $\sigma$, $\pi$, $\delta$...denoting the magnitude of the orbital angular momentum in the electric field direction (i.e. the internuclear axis), $\lambda=0$, 1, 2..., where $\lambda=\vert m_{l}\vert=l$, $l-1$, ... 0. Molecular orbitals with $\lambda\neq0$ are doubly degenerate as they can have $m_{l}=\pm\lambda$. The total electron orbital angular momentum quantum number is given by,

$$\Lambda=\sum \lambda_{i}$$ (G.2)

This sum must take into account the direction of $\lambda_{i}$ in the internuclear axis (i.e. $\pm\lambda_{i}$). The magnitude of the resultant defines the electronic state, where $\vert\Lambda\vert$=0, 1, 2...are denoted by $\Sigma$, $\Pi$, $\Delta$...The resultant electron spin is given by the vector sum,

$$S=\sum{s_{i}}$$ (G.3)

where $s_{i}$=$\frac{1}{2}$, the spin of each individual electron. Spin-orbit coupling causes states with $\Lambda\neq0$ to be split into $2S+1$ components. For OH, the ground electronic state is split into two as $\Lambda=1$ and $S=\frac{1}{2}$. In this case there are two ladders of rotational levels, labelled $^{2}\Pi _{1/2}$ and $^{2}\Pi _{3/2}$. The subscript denotes the total (orbital + spin) angular momentum quantum number of the electrons, given by $\vert\Lambda+\Sigma_{s}\vert$, where $\Sigma_{s}$ is the component of the total electron spin, $S$, along the internuclear axis. The superscript denotes number of spin-components, given by $2S+1$.

The effect of rotational motion on the electron spin-orbit coupling leads to several coupling cases known as Hund's cases. In Hund's case (a) the electronic motion is closely coupled to the line joining the nuclei and the spin-orbit coupling is large. This has the effect that the coupling of rotational motion with electron motion is weak. The total angular momentum quantum number including rotation is given by $J$. The lowest rotational level occurs with $J=\vert\Lambda+\Sigma_{s}\vert$.

In Hund's case (b) the electron spin is only weakly coupled to the internuclear axis and there is much stronger coupling with rotational motion. In this case the component of the total electron spin along the internuclear axis, $\Sigma_{s}$, is not well defined. The total angular momentum quantum number apart from spin is denoted by $N$ and rotational levels with the same value of $N$ lie close together in energy. The total angular momentum quantum number is $J=N+S$ and consequently the lowest rotational level has $J=\vert\Lambda+S\vert$.

For multiplet states with $\Lambda\neq0$ there is a further splitting in addition to the spin-splitting described above. This is due to the two directions possible for the electron orbital angular momentum ($\Lambda$) along the internuclear axis. In the presence of rotation the degeneracy between the two directions is broken and each rotational level is split into components with positive or negative parity. This is known as $\Lambda$-type (or parity) doubling and is due to the coupling of the rotation to the orbital motion of the electrons.

For the OH molecule the spin-splitting is not large. At low values of $J$ the coupling is intermediate between Hund's case (a) and (b). As $J$ increases the spin-splitting decreases and the coupling goes towards Hund's case (b). Figure 7.1 shows a representation of the energy levels of OH up to $J=9/2$.

In reality the OH molecule energy levels are more complicated as the nuclear spin of the $^{1}$H atom causes an extra contribution to the total angular momentum. Each energy level is split into two hyperfine components and the total angular momentum has a quantum number given by, $F=J+I$, where $I$ is the nuclear spin quantum number. For $^{1}$H, $I$=1/2.

Figure 7.1: Low lying rotational energy levels in the electronic and vibrational ground state of the OH molecule, adapted from Brown et al. (1982). The $\Lambda$-doubling of each level is shown but the splitting has been exaggerated for clarity. The relative spacing of the levels in energy is only approximate. The $J$ values for each level are shown and the wavelength of each transition is given in $\mu$m.