# Hund's cases

In rotational-vibrational and electronic spectroscopy of diatomic molecules, Hund's coupling cases are idealized cases where specific terms appearing in the molecular Hamiltonian and involving couplings between angular momenta are assumed to dominate over all other terms. There are five cases, traditionally notated with the letters (a) through (e). Most diatomic molecules are somewhere between the idealized cases (a) and (b).[1]

## Angular momenta

To describe the Hund's coupling cases, we use the following angular momenta:

## Choosing the applicable Hund's case

Hund's coupling cases are idealizations. The appropriate case for a given situation can be found by comparing three strengths: the electrostatic coupling of ${\displaystyle \mathbf {L} }$ to the internuclear axis, the spin-orbit coupling, and the rotational coupling of ${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {S} }$ to the total angular momentum ${\displaystyle \mathbf {J} }$.

Hund's case Electrostatic Spin-orbit Rotational
(a) strong intermediate weak
(b) strong weak intermediate
(c) intermediate strong weak
(d) intermediate weak strong
(e) weak intermediate strong
strong intermediate

The last two rows are degenerate because they have the same good quantum numbers.[2]

## Case (a)

In case (a), ${\displaystyle \mathbf {L} }$ is electrostatically coupled to the internuclear axis, and ${\displaystyle \mathbf {S} }$ is coupled to ${\displaystyle \mathbf {L} }$ by spin-orbit coupling. Then both ${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {S} }$ have well-defined axial components ${\displaystyle \Lambda }$ and ${\displaystyle \Sigma }$, respectively. ${\displaystyle {\boldsymbol {\Omega }}}$ defines a vector of magnitude ${\displaystyle \Omega =\Lambda +\Sigma }$ pointing along the internuclear axis. Combined with the rotational angular momentum of the nuclei ${\displaystyle \mathbf {R} }$, we have ${\displaystyle \mathbf {J} ={\boldsymbol {\Omega }}+\mathbf {R} }$. In this case, the procession of ${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {S} }$ around the nuclear axis is assumed to be much faster than the nutation of ${\displaystyle {\boldsymbol {\Omega }}}$ and ${\displaystyle \mathbf {R} }$ around ${\displaystyle \mathbf {J} }$.

The good quantum numbers in case (a) are ${\displaystyle \Lambda }$, ${\displaystyle S}$, ${\displaystyle \Sigma }$, ${\displaystyle J}$ and ${\displaystyle \Omega }$. We express the rotational energy operator as ${\displaystyle H_{rot}=B\mathbf {R} ^{2}=B(\mathbf {J} -\mathbf {L} -\mathbf {S} )^{2}}$, where ${\displaystyle B}$ is a rotational constant. There are, ideally, ${\displaystyle 2S+1}$ fine-structure states, each with rotational levels having relative energies ${\displaystyle BJ(J+1)}$ starting with ${\displaystyle J=\Omega }$.[1]

## Case (b)

In case (b), the spin-orbit coupling is weak or non-existent (in the case ${\displaystyle \Lambda =0}$). In this case, we take ${\displaystyle \mathbf {N} ={\boldsymbol {\lambda }}+\mathbf {R} }$ and ${\displaystyle {\mathbf {J} }={\mathbf {N} }+{\mathbf {S} }}$ and assume ${\displaystyle \mathbf {L} }$ precesses quickly around the internuclear axis.

The good quantum numbers in case (b) are ${\displaystyle \Lambda }$, ${\displaystyle N}$, ${\displaystyle S}$, and ${\displaystyle J}$. We express the rotational energy operator as ${\displaystyle H_{rot}=B\mathbf {R} ^{2}=B(\mathbf {N} -\mathbf {L} )^{2}}$, where ${\displaystyle B}$ is a rotational constant. The rotational levels therefore have relative energies ${\displaystyle BN(N+1)}$ starting with ${\displaystyle N=\Lambda }$.[1]

## Case (c)

In case (c), the spin-orbit coupling is stronger than the coupling to the internuclear axis, and ${\displaystyle \lambda }$ and ${\displaystyle \sigma }$ from case (a) cannot be defined. Instead${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {S} }$ combine to form ${\displaystyle \mathbf {J} _{a}}$, which has a projection along the internuclear axis of magnitude ${\displaystyle \Omega }$. Then ${\displaystyle \mathbf {J} ={\boldsymbol {\Omega }}+\mathbf {R} }$, as in case (a).

The good quantum numbers in case (c) are ${\displaystyle J_{a}}$, ${\displaystyle J}$, and ${\displaystyle \Omega }$.[1]

## Case (d)

In case (d), the rotational coupling between ${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {R} }$ is much stronger than the electrostatic coupling of ${\displaystyle \mathbf {L} }$ to the internuclear axis. Thus we form ${\displaystyle \mathbf {N} }$ by coupling ${\displaystyle \mathbf {L} }$ and ${\displaystyle \mathbf {R} }$ and the form ${\displaystyle \mathbf {J} }$ by coupling ${\displaystyle \mathbf {N} }$ and ${\displaystyle \mathbf {S} }$.

The good quantum numbers in case (d) are ${\displaystyle L}$, ${\displaystyle R}$, ${\displaystyle N}$, ${\displaystyle S}$, and ${\displaystyle J}$. Because ${\displaystyle R}$ is a good quantum number, the rotational energy is simply ${\displaystyle H_{rot}=B\mathbf {R} ^{2}=BR(R+1)}$.[1]

## Case (e)

In case (e), we first form ${\displaystyle \mathbf {J} _{a}}$ and then form ${\displaystyle \mathbf {J} }$ by coupling ${\displaystyle \mathbf {J} _{a}}$ and ${\displaystyle \mathbf {R} }$. This case is rare but has been observed.[3]

The good quantum numbers in case (e) are ${\displaystyle J_{a}}$, ${\displaystyle R}$, and ${\displaystyle J}$. Because ${\displaystyle R}$ is once again a good quantum number, the rotational energy is ${\displaystyle H_{rot}=B\mathbf {R} ^{2}=BR(R+1)}$.[1]

## References

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