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Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.

The ruby R-line shift compares the same emission feature under two conditions. It is distinct from the Stokes shift, which compares absorption and emission features. In ruby, chromium ions in the corundum crystal provide the electronic transitions; compression changes their local environment and transition energies.^[ref]

For consistently defined vacuum wavelengths, the reciprocal relation givesThus a positive wavelength shift means a negative reciprocal shift, and $E=hc/\lambda$ gives lower photon energy; $h$ is the Planck constant and $c$ the speed of light. For spectroscopic wavenumber, use vacuum wavelength and keep that convention consistent. A calibration specified in air wavelength must retain its convention or be converted consistently. The reciprocal relation is nonlinear: a line exactly linear in $\Delta \lambda$ is not exactly linear in $\Delta (1/\lambda )$.

For a purely mathematical check with chosen vacuum wavelengths ${\lambda }_0=700.0\mathrm{nm}$ and $\Delta \lambda =2.0\mathrm{nm}$, the reciprocal shift is $-40.7000\mathrm{c}{\mathrm{m}}^{-1}$. The small-shift approximation $-\Delta \lambda /{\lambda }_0^2$ gives $-40.8163\mathrm{c}{\mathrm{m}}^{-1}$, about $0.286\%$ too large in magnitude. These chosen values illustrate the conversion; they are not measured ruby lines or an additional pressure calibration.

In the quadratic vibrational term expression $E_v/(hc)={\tilde{\omega }}_e(v+1/2)-{\tilde{\omega }}_e{x}_e(v+1/2{)}^2$, the constant $x_e$ is dimensionless, while ${\tilde{\omega }}_e{x}_e$ has units of spectroscopic wavenumber. A positive value makes adjacent level spacings decrease. For an exact Morse oscillator, $D_e/(hc)={\tilde{\omega }}_e/(4x_e)$. Using only two low-lying transitions to extrapolate this identity to a real molecule assumes the higher-order anharmonic terms are negligible all the way to dissociation.

A spectroscopic wavenumber commonly means the reciprocal of the vacuum wavelength ${\lambda }_0$ of electromagnetic radiation, often in $\mathrm{c}{\mathrm{m}}^{-1}$. It is an ordinary spatial frequency, so the vacuum angular wavenumber is $K_0=2\pi \tilde{\nu }$. Specify vacuum or in-medium wavelength: the same frequency can have different wavelengths in different media. IUPAC's “wavenumber in medium” entry defines the in-medium reciprocal $1/\lambda$ and notes that unqualified spectroscopic usage often means the vacuum reciprocal.^[ref]

For example, a vacuum wavelength of $500\mathrm{nm}=5\times 10^{-5}\mathrm{cm}$ corresponds to $\tilde{\nu }=20000\mathrm{c}{\mathrm{m}}^{-1}=2\times 10^6{\mathrm{m}}^{-1}$. Multiplication by $2\pi$ gives $K_0=4\pi \times 10^6\mathrm{rad}{\mathrm{m}}^{-1}$. Converting centimetres to metres and converting cycles to radians are separate operations.

The Stokes shift compares the lowest absorption-band maximum with the fluorescence emission spectrum maximum. Relaxation often makes $\Delta \lambda >0$. This comparison differs from using an arbitrary excitation wavelength. A spectral-density maximum can move when changing from per unit wavelength to per unit spectroscopic wavenumber; specify the spectral convention.^[ref]

An original illustrative energy budget makes the distinction concrete. Excitation at $450\mathrm{nm}$ and emission at $520\mathrm{nm}$ correspond to about $2.76$ and $2.38\mathrm{eV}$ per photon, using $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$ in vacuum. The difference, about $0.37\mathrm{eV}$, must go elsewhere in the complete energy balance. These chosen wavelengths are not measured band maxima for a particular fluorophore.

Anti-Stokes emission has shorter wavelength than excitation. Additional energy, for example from a thermally populated level of molecular vibration or multiple excitation photons, makes it possible. The one-photon, initially relaxed drawing does not cover these cases.^[ref]

The University of Florida's Rotation Vibration Spectrum of the HCl Molecule, University of Florida laboratory protocol, printed pp. 2–3 and 10–11, uses a scanning Michelson interferometer for Fourier transform spectroscopy. An air-filled silicon-window cell supplies the background. Supervised preparation adds concentrated hydrogen chloride liquid; the cleaned cell subsequently receives DCl liquid. Acquisition settings are identical.

The readout is transmission or absorbance against spectroscopic wavenumber, resolving molecular rotation alongside stretching. Background addresses instrument and cell contributions; cleaning addresses carryover. Excessive absorption impairs analysis. Assigning isotopes and rotational branches precedes fitting band parameters to test isotope scaling of diatomic vibration. These are prescribed measurements, not numerical observations supplied here.^[ref]

For a transparent ideal model, add the linear rigid rotor to the harmonic vibrational energy levels and assume the same frequency-valued rotational constant $B$ in both vibrational states:For a spin-free diatomic molecule with no electronic angular momentum, electric-dipole selection rules give $\Delta J=\pm 1$ for this vibrational absorption. Subtracting initial from final energy eigenvalues yieldsThese are two rotational branches about the vibrational band origin, not a single stretch line. For stipulated dimensionless values ${\tilde{\nu }}_{\mathrm{harm}}=100$ and $\tilde{B}=1$, the first three R-branch positions are $102,104,106$, and the first three P-branch positions are $98,96,94$. There is no $J=0\to -1$ transition. The numbers illustrate level subtraction; they are not HCl measurements. Real upper and lower rotational constants differ, and centrifugal distortion shifts lines further.

A wavenumber measures how rapidly a wave phase changes with distance. Two conventions coexist: angular wavenumber counts radians per unit length, while ordinary spatial frequency counts cycles per unit length. For a monochromatic wave with wavelength $\lambda >0$,Here $K\ge 0$ is the magnitude of the angular wavenumber. The symbol $k$ can instead denote a signed one-dimensional angular wavenumber: a factor $e^{ikx}$ has wavelength $2\pi /|k|$ for real $k\ne 0$. Both conventions have dimensions of inverse length; writing $\mathrm{rad}{\mathrm{m}}^{-1}$ or “cycles per metre” exposes the convention. A quoted value in ${\mathrm{m}}^{-1}$ alone does not resolve the factor of $2\pi$. Spectroscopic wavenumber usually uses the cycles-per-length convention.^[ref]

For a real monochromatic plane wave, fix the conventionThe wavevector $\mathbf{k}$ supplies both the spatial wave phase slope and the normal to the wavefronts; its dot product with position makes the wave phase constant on parallel planes. The angular frequency is $\omega =2\pi f$, with $f>0$ the ordinary frequency. Moving a distance $\lambda$ along $\mathbf{k}$ changes the wave phase by $2\pi$. Holding a wave crest's wave phase fixed gives its phase velocity $\omega \mathbf{k}/K^2$ and phase speedIn one dimension the signed phase velocity is $\omega /k$. This identifies a moving pattern; it does not identify the motion of the medium's particles or a wave packet's group velocity.

For the synthetic plane wave below, choose $\lambda =0.40\mathrm{m}$ and $f=5\mathrm{Hz}$. Then $K=5\pi \mathrm{rad}{\mathrm{m}}^{-1}$, ${\nu }_x=2.5$ cycles per metre, $\omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}$, and $v_p=2\mathrm{m}{\mathrm{s}}^{-1}$. The wave crests are $0.40\mathrm{m}$ apart at one instant; after a quarter oscillation period, each has advanced $0.10\mathrm{m}$. The lower panel retains whole turns of wave phase, making its slope visible instead of reducing it modulo $2\pi$.