Applications to Molecular Rotations

Molecular rotational spectra connect the orientation wavefunction to a measurable pattern of microwave or Raman lines. We use an ideal linear rotor in a fixed nondegenerate electronic and vibrational state, with electronic and nuclear spin structure neglected. Open-shell molecules and states with electronic angular momentum need additional labels and can have different branch rules.

For a linear rigid rotor, the states are

$$|J,M\rangle, \qquad J=0,1,2,\ldots, \qquad M=-J,\ldots,J.$$

The ideal energies are

$$E_J=BJ(J+1).$$

Here $B=\hbar^2/(2I)$ is an energy and $I$ is the moment of inertia. The rigid-rotor model supplies these levels. The transition operator explains why the leading electric-dipole pure rotational rule is

$$\Delta J=\pm1, \qquad \Delta M=0,\pm1,$$

and why many molecules do not show a simple electric-dipole rotational spectrum.

Rotor States as Spherical Harmonics

For an ideal linear rotor, the orientation of the molecular axis is described by a unit vector $\hat{\mathbf n}$ with polar angles $(\theta,\phi)$. The rotor wavefunctions are spherical harmonics:

$$\langle \theta,\phi|J,M\rangle = Y_J^M(\theta,\phi).$$

The rotational angular momentum labels are ordinary integer angular-momentum labels. The $M$ quantum number is a projection on a laboratory quantization axis. In the absence of external fields, the energy depends on $J$ but not on $M$.

Permanent Dipole Operator

A polar linear molecule has a permanent electric dipole moment along its body-fixed molecular axis:

$$\boldsymbol\mu = \mu\,\hat{\mathbf n}.$$

In the laboratory frame, the spherical components of this dipole are proportional to rank-$1$ spherical harmonics:

$$\mu_q = \mu\,C_q^{(1)}(\hat{\mathbf n}), \qquad q=0,\pm1.$$

A permanent body-fixed dipole does not imply $\langle JM|\boldsymbol\mu|JM\rangle\ne0$ in zero field. A parity eigenstate has zero laboratory dipole expectation, yet can have nonzero dipole matrix elements to another rotational state. Thus the molecular dipole operator is a rank-$1$ tensor under space-fixed rotations. The same vector-operator logic used for atomic dipole transitions applies here, but the states are rotor states rather than electronic orbital states.

Matrix Element and Three-j Symbols

The angular matrix element has the standard form

$$\langle J'M'|\mu_q|JM\rangle = \mu (-1)^{M'} \sqrt{(2J'+1)(2J+1)} \begin{pmatrix} J' & 1 & J\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} J' & 1 & J\\ -M' & q & M \end{pmatrix}.$$

The exact phase convention follows the spherical-harmonic convention used for the $3j$ symbols. The zeros are convention-independent.

The second $3j$ symbol gives

$$-M'+q+M=0,$$

so

$$M'=M+q.$$

Since $q=0,\pm1$, the magnetic selection rule is

$$\Delta M=0,\pm1.$$

The triangle condition gives

$$J'=J-1,J,J+1,$$

where nonnegative angular momentum is understood. But the first $3j$ symbol,

$$\begin{pmatrix} J' & 1 & J\\ 0 & 0 & 0 \end{pmatrix},$$

vanishes unless $J'+1+J$ is even. This removes $J'=J$ and leaves

$$\Delta J=\pm1.$$

For absorption in the ideal rotor, the usual branch is

$$J\to J+1.$$

For emission,

$$J\to J-1.$$

Line Positions

The symmetry rule tells which adjacent levels can be connected by the electric dipole operator. The line positions then come from energy differences:

$$E_{J+1}-E_J = B(J+1)(J+2)-BJ(J+1) = 2B(J+1).$$

Thus the ideal absorption line ladder is evenly spaced:

$$2B,\quad4B,\quad6B,\ldots$$

in energy units. If $B$ is quoted in frequency or wavenumber units, the same arithmetic pattern appears in those units. The connection between line spacings and moment of inertia is developed in Rotational Spectra.

Why Homonuclear Diatomics Are Different

The electric-dipole pure rotational spectrum requires a permanent dipole moment:

$$\mu\ne0.$$

Ideal homonuclear diatomic molecules such as $\mathrm H_2$, $\mathrm N_2$, and $\mathrm O_2$ have no permanent electric dipole. Their body-fixed charge distribution has no polar vector pointing from one distinguishable end to the other.

Therefore the ordinary $E1$ pure rotational matrix element vanishes at the operator level. This does not mean the molecule lacks rotational levels. It means the leading microwave electric-dipole probe does not connect them.

Other mechanisms can still probe rotations:

• rotational Raman scattering through the polarizability tensor;
• weak magnetic or quadrupole mechanisms in special cases;
• rovibrational electric-quadrupole transitions in homonuclear diatomics;
• collision-induced absorption or environment-induced dipoles.

Ordinary stretching of an isolated homonuclear diatomic does not generate an electric dipole: inversion symmetry holds at every bond length, so it does not restore an E1 rovibrational spectrum. In other molecules, an infrared-active vibration can change the dipole. The interaction mechanism must therefore be specified as carefully as the rotor levels.

Raman Preview

Rotational Raman scattering is controlled by the molecular polarizability rather than the permanent dipole. For an ordinary linear molecule away from resonant or antisymmetric-polarizability effects, $\alpha_{ij}=\alpha_\perp\delta_{ij}+(\alpha_\parallel-\alpha_\perp)n_in_j$. It has scalar and rank-$2$ irreducible parts. Rotational Raman lines require nonzero anisotropy $\alpha_\parallel-\alpha_\perp$; an isotropic polarizability cannot change the orientation state.

The scalar rank-$0$ part gives no change in $J$. The rank-$2$ part gives the angular possibilities

$$\Delta J=0,\pm1,\pm2$$

from the triangle rule, but the $3j$ symbol with all three magnetic entries zero and the even parity of $n_in_j$ remove the usual pure rotational Raman transition with $\Delta J=\pm1$ for a linear rotor. The characteristic rotational Raman branches are

$$\Delta J=\pm2,$$

with $\Delta J=0$ unshifted when the $M$ sublevels are degenerate and no other internal state changes. The Stokes rotational shifts are $E_{J+2}-E_J=2B(2J+3)$: $6B,10B,14B,\ldots$, spaced by $4B$.

This preview is included only to clarify why nonpolar molecules can have rotational structure in Raman spectra even when the ordinary electric-dipole microwave spectrum is absent.

Polarization and M Components

The selection rule

$$\Delta M=q$$

depends on the spherical component of the driving field. Relative to a quantization axis:

Component	Rule
$q=0$	$\Delta M=0$
$q=+1$	$\Delta M=+1$
$q=-1$	$\Delta M=-1$

In zero field the $M$ sublevels are degenerate, so many spectra do not resolve these components. External electric or magnetic fields choose a physical axis, split or mix levels, and make the polarization dependence more explicit.

For a static electric field along $z$, the perturbation has the form

$$H_{\rm Stark} = -\mu E\,C_0^{(1)}(\hat{\mathbf n}).$$

It preserves axial symmetry about $z$, so $M$ remains a good label, but it mixes different $J$ values subject to the rank-$1$ rule. The first encounter with field effects is Rotor in External Fields.

Beyond the Linear Rotor

Real molecular rotational spectra can require more structure:

• symmetric tops have an additional body-fixed projection label;
• asymmetric tops do not have a single simple $J(J+1)$ energy formula;
• nuclear spin statistics can remove or reweight alternating rotational levels;
• vibration-rotation coupling changes the rotational constants;
• electronic angular momentum, spin-rotation coupling, hyperfine structure, and external fields split levels.

The tensor-operator habit remains the same. Identify the good angular-momentum labels, express the interaction as irreducible tensor components, apply angular and parity rules, and only then compute reduced matrix elements and rates.

Continue with Rotations of Molecules for symmetric and asymmetric tops, centrifugal distortion, and fitting microwave spectra. Rovibrational Coupling explains how changing the vibrational state adds P, Q, and R branches and changes the rotational constants.

Common Mistakes

• Explaining $\Delta J=\pm1$ only from adjacent line spacings. The rule comes from the dipole matrix element; the spacings come from the rotor energies.
• Forgetting the $3j$ symbol with zeros, which removes $\Delta J=0$ for pure rotational electric-dipole transitions of a linear rotor.
• Expecting homonuclear diatomic molecules to have ordinary electric-dipole microwave spectra.
• Treating the absence of an $E1$ rotational line as absence of rotational levels.
• Ignoring polarization and therefore losing the $\Delta M$ rule.
• Applying linear-rotor rules unchanged to symmetric or asymmetric tops.
• Forgetting nuclear exchange symmetry in homonuclear molecules.

Exercises

1. Use the $3j$ expression to show why $\Delta J=0$ is absent for an electric-dipole pure rotational transition of a linear rotor.

Solution

The relevant factor is

$$\begin{pmatrix} J' & 1 & J\\ 0 & 0 & 0 \end{pmatrix}.$$

A $3j$ symbol with all lower entries zero vanishes unless $J'+1+J$ is even. If $J'=J$, then

$$J'+1+J=2J+1,$$

which is odd. Therefore the factor vanishes and $\Delta J=0$ is absent.

2. Calculate $\langle1,0|\mu_0|0,0\rangle$ directly from spherical harmonics. Why can it be nonzero even though each state’s permanent laboratory dipole expectation is zero?

Solution

Since $Y_{00}=1/\sqrt{4\pi}$, $Y_{10}=\sqrt{3/(4\pi)}\cos\theta$, and $\mu_0=\mu\cos\theta$,

$$\langle1,0|\mu_0|0,0\rangle =\frac{\mu\sqrt3}{4\pi}\int\cos^2\theta\,d\Omega =\frac\mu{\sqrt3}.$$

Each diagonal expectation is zero by parity. This off-diagonal element instead connects opposite parities, so the electric field can drive the transition. A body-fixed dipole and a laboratory expectation value describe different quantities.

3. Why does an ideal homonuclear diatomic molecule lack an ordinary electric-dipole pure rotational spectrum?

Solution

The ordinary pure rotational microwave transition uses the permanent electric dipole operator. An ideal homonuclear diatomic molecule has no permanent body-fixed dipole moment, so the relevant operator coefficient is zero. The rotational levels still exist, but the leading $E1$ pure rotational matrix element is absent.

4. A polar linear rotor has rotational constant $B$ in energy units. What is the absorption energy for the $J=2\to3$ line?

Solution

For absorption,

$$\Delta E_J = E_{J+1}-E_J = 2B(J+1).$$

With $J=2$,

$$\Delta E=2B(3)=6B.$$

References

• P. F. Bernath, Spectra of Atoms and Molecules, 3rd ed., Oxford University Press, 2016.
• J. M. Brown and A. Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• G. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules, 2nd ed., Van Nostrand, 1950.
• C. H. Townes and A. L. Schawlow, Microwave Spectroscopy, Dover, 1975.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.