Molecular Physics Applications

A freely rotating molecule has continuous rotational symmetry in the laboratory even when its shape has only a finite point group in the body frame. Keeping those two actions distinct helps organize rotational levels, vibrational modes, and transition intensities. The approximation also matters: electronic spin, nonadiabatic motion, nuclear exchange, and external fields can change the useful labels.

Helpful background. The rigid rotor introduces rotational multiplets; molecular rotational transitions supplies the tensor rules. Born–Oppenheimer Berry phase explains the geometric connection between electronic and nuclear motion.

Laboratory rotations and the body-frame point group

In an isolated molecule, rotating the entire system in the laboratory does not change the energy. Its conserved total angular momentum includes the degrees of freedom retained in the model. In the simplest spin-free linear rotor on a nondegenerate electronic surface, the rotational Hamiltonian is

$$H_{\rm rot}=\frac{\mathbf J^2}{2I},\qquad E_J=B_EJ(J+1),\qquad B_E=\frac{\hbar^2}{2I}.$$

Here $I$ is the moment of inertia, $J=0,1,\ldots$, and $M=-J,\ldots,J$. The laboratory projection $M$ labels a $(2J+1)$-fold multiplet. Identical-nucleus exchange and nuclear-spin statistics can restrict which rotational states accompany a given nuclear-spin state; the naive rotor list does not settle those restrictions.

The point group instead describes transformations of the equilibrium nuclear framework in the body frame. It classifies electronic states, normal displacements, and operator components. It does not replace the full laboratory rotation group. A transition can be nonzero only if

$$\Gamma_f^*\otimes\Gamma_O\otimes\Gamma_i \supset\Gamma_{\rm sym},$$

where $\Gamma_{\rm sym}$ is the totally symmetric representation. This is a necessary condition; the corresponding matrix element can still vanish. Molecular symmetry gives explicit reductions, and character tables provides compact lookup values.

Rotational lines and corrections that preserve angular momentum

For a polar spin-free linear rotor in the leading electric-dipole approximation, the pure rotational rules are $\Delta J=\pm1$ and $\Delta M=0,\pm1$, with the component selected by polarization. Absorption $J\to J+1$ requires photon energy

$$\hbar\omega=2B_E(J+1).$$

An ideal homonuclear diatomic on a fixed nondegenerate electronic surface has no permanent electric dipole and no ordinary pure rotational electric-dipole lines. Its rotational levels still exist; Raman or higher-multipole processes can probe them.

Centrifugal distortion changes the energy formula without necessarily changing the symmetry. The rotationally invariant effective expression

$$E_J=B_EJ(J+1)-D_E[J(J+1)]^2$$

still has exact $J,M$ labels and $M$ degeneracy within this spin-free model. For small distortion, $D_EJ(J+1)\ll B_E$, it shifts the line energy by $-4D_E(J+1)^3$. Likewise, rotationally invariant vibration–rotation coupling need not destroy total angular-momentum labels. A laboratory field can break that invariance; internal corrections need not.

See molecular rotations for nonrigid models and rovibrational coupling for the full rotational and vibrational labeling problem.

Vibrational activity depends on the probe

Near a stable equilibrium, mass-weighted normal coordinates give the harmonic approximation

$$H_{\rm vib}\simeq\sum_a\left(\frac{P_a^2}{2}+\frac{\omega_a^2Q_a^2}{2}\right).$$

Translation and overall rotation have been removed before identifying the vibrational coordinates. A nonlinear molecule with $N$ nuclei has $3N-6$ vibrational modes; a linear one has $3N-5$.

Infrared electric-dipole activity depends on the derivative of the dipole with respect to a mode coordinate. Raman activity in the usual leading polarizability treatment depends on the corresponding polarizability derivative. The two probes transform differently, so a mode may be active in one and absent in the other.

For a centrosymmetric molecule, normal modes have even ($g$) or odd ($u$) inversion character. A dipole component is odd and a polarizability component is even. From the even vibrational ground state, a fundamental infrared transition requires a suitable odd mode, while a fundamental Raman transition requires a suitable even mode. This gives the familiar mutual exclusion in these approximations; not every odd or even mode is necessarily active, because the remaining point-group restrictions still apply.

Polyatomic normal modes and diatomic vibrations connect this symmetry reasoning to actual spectra and anharmonic corrections.

Electronic surfaces and geometric phase

At fixed nuclear positions $R$, the electronic equation is $H_e(R)|\phi_a(R)\rangle=E_a(R)|\phi_a(R)\rangle$. The nuclear potential surface includes this electronic energy and the nuclear repulsion, if the latter was not already included in $H_e(R)$. The variation of $|\phi_a(R)\rangle$ produces derivative couplings and, on an isolated surface, a Berry connection.

A loop around a suitable conical intersection can return the electronic eigenvector with a minus sign. The nuclear wavefunction must compensate so that the complete state has the required behavior. A one-surface geometric phase describes a loop avoiding the degeneracy; motion through a small-gap region can require coupled electronic surfaces.

The Born–Oppenheimer approximation, nonadiabatic coupling, and conical intersections develop the relevant dynamical tests. Slow nuclear motion alone is not a sufficient guarantee when the electronic gap becomes small.

A static field mixes rotor states

For a polar linear rotor in a $z$-directed electric field, $H_{\rm Stark}=-\mu E\cos\theta$. The field preserves rotations about $z$, so $M$ remains good in this model, while $J$ generally does not. The odd interaction mixes neighboring opposite-parity rotor states. Within a fixed-$J$ field-free multiplet, all $M$ states have the same parity, so the entire restricted first-order Stark matrix vanishes despite the $M$ degeneracy. For each fixed $M$, neighboring $J$ levels are separated in this ideal rotor. First-order state mixing can orient the molecule, while energy shifts are generally quadratic at weak field. See a rotor in external fields.

Exercises

1. Derive the distortion correction to the absorption line $J\to J+1$ from the energy expression above. Does it split the $M$ multiplet?

Solution

The difference in $J(J+1)$ is $2(J+1)$, and the difference of its squares is $4(J+1)^3$. Hence $\hbar\omega=2B_E(J+1)-4D_E(J+1)^3$. There is no $M$ dependence, so this correction does not split the multiplet. Extrapolating the truncated expression to arbitrarily high $J$ would exceed its regime of validity.

2. An inversion-even normal mode has zero first-order infrared dipole matrix element from the vibrational ground state. Must its Raman matrix element be nonzero?

Solution

No. Even parity passes the Raman parity restriction, but the mode must also transform as an allowed polarizability component under the full point group. Its polarizability derivative can additionally vanish or be very small. Symmetry permission alone is not an intensity prediction.

References

• J. M. Brown and A. Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003.
• P. R. Bunker and P. Jensen, Molecular Symmetry and Spectroscopy, 2nd ed., NRC Research Press, 1998.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• G. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules, 2nd ed., Van Nostrand, 1950.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.