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🌀 Rotational-Vibrational Spectrum: P & R Branches

Interactive rovibrational spectroscopy simulator: a spinning, vibrating diatomic molecule splits a single vibrational line into a P branch and an R branch of rotational lines. Tune temperature and the rotational constant B, click any line to watch that exact J to J' transition happen in 3D.

Chemistry & Materials3DModerate60 FPS

How it Works

A vibrating diatomic molecule almost never sits still rotationally while it absorbs an infrared photon — in the gas phase it is always tumbling, and that rotation persists through the vibrational transition. So instead of one sharp line at the vibrational frequency ν₀, real gas-phase spectra show a forest of closely spaced lines fanning out on either side. Each line is a combined rotation-vibration transition: the molecule jumps from vibrational level v=0 to v=1 and its rotational quantum number J changes by exactly ±1 (ΔJ=0, the Q branch, is forbidden for a simple Σ-state diatomic like these). Lines with ΔJ=+1 land above ν₀ and form the R branch; lines with ΔJ=−1 land below ν₀ and form the P branch. No line appears exactly at ν₀ itself — that gap is the spectroscopic fingerprint of a diatomic's Q-branch-forbidden rotation-vibration spectrum.

E(v,J) = (v+1/2)hcν₀ + B·hc·J(J+1) ν_R(J) = ν₀ + 2B(J+1) J = 0,1,2,… ν_P(J) = ν₀ − 2B·J J = 1,2,3,… N(J) ∝ (2J+1)·exp[−B·J(J+1)·hc/kT]

Each line's intensity is set by how many molecules actually sit in that starting J state before the photon arrives. The Boltzmann population N(J) grows with the (2J+1) rotational degeneracy but shrinks exponentially with rotational energy, so it peaks at an intermediate J and falls off on both sides — that is why the branches bulge outward from ν₀ and then fade, rather than simply getting weaker line by line. Raising the temperature slider pushes that peak toward higher J, spreading the branches wider and shifting the envelope's maximum away from the band center; a small, light molecule (large B) spaces its lines far apart, while a heavier one packs them close together.

Pick a molecule to change its rotational constant B and vibrational origin ν₀, drag the temperature slider to watch the intensity envelope reshape itself, and click any line in the spectrum to select its exact J → J' transition — the energy-level ladder highlights the jump and the 3D molecule spins and stretches at that transition's own rotational speed.

Frequently Asked Questions

Why is there a gap exactly at ν₀ where no line appears?

That gap is the missing Q branch. A Q-branch line would require ΔJ=0 — a vibrational transition with no change in rotational state at all. For a diatomic molecule in a Σ electronic state (zero orbital angular momentum along the bond axis), that transition is forbidden by the selection rules, so nothing absorbs right at the bare vibrational frequency. Only P (ΔJ=−1) and R (ΔJ=+1) lines are allowed, leaving a clean gap of width 4B centered on ν₀.

Why do the branches bulge outward and then fade instead of monotonically weakening?

Line intensity tracks the population of the starting J level, N(J) ∝ (2J+1)e^(−BJ(J+1)hc/kT). The (2J+1) degeneracy factor grows with J (more M_J sublevels to populate) while the Boltzmann exponential shrinks with J. Their product rises, peaks near J_max ≈ √(kT/2Bhc) − 1/2, then falls — so the strongest lines sit a few steps away from the band center on each side, not immediately next to it.

Why does raising temperature spread the branches wider?

Higher temperature populates higher rotational states more heavily, pushing J_max up. Since each branch line sits at ν₀ ± 2B(J+1) or ν₀ − 2BJ, a larger populated J range means lines extend further from the center, and the peak of the intensity envelope moves outward with it — a hotter gas has a visibly broader rovibrational band.

Why does HCl have more widely spaced lines than CO?

Line spacing is set by 2B, and B = h/(8π²cI) is inversely proportional to the molecule's moment of inertia I = μr². HCl's reduced mass μ is small (hydrogen is light), giving it a small I and therefore a large B — its lines sit roughly 20 cm⁻¹ apart. CO's two atoms are both heavier, giving a much larger I, a smaller B, and lines packed close enough to nearly blur together at low resolution.

What physically changes in the molecule between a P-branch and an R-branch transition?

In an R-branch transition (ΔJ=+1) the molecule absorbs the photon's energy into both a vibrational quantum and extra rotational energy — it ends up spinning faster after absorption. In a P-branch transition (ΔJ=−1) part of the photon's energy pays for the vibrational jump while the molecule actually gives up rotational energy, ending up spinning slower. That is why R-branch lines sit above ν₀ (photon carries more energy) and P-branch lines sit below it (photon carries less).

Is this the same as the discrete IR "fingerprint" peaks in a normal infrared spectrum?

Not quite. A standard low-resolution IR spectrum (like a benchtop FTIR of a liquid or solid sample) shows one broadened peak per vibrational mode because rotation is damped out by molecular collisions and the resolution is too coarse to separate individual J lines. This fine rotational structure only becomes visible for a dilute gas measured at high spectral resolution, which is exactly the regime this simulator models.

⚙ Under the hood

Watch a single vibrational absorption line of a spinning diatomic molecule split into a P branch and an R branch of rotational lines. Tune temperature and pick HCl, CO or NO, then click any line to watch that exact J to J' transition play out on a 3D molecule and an energy-level ladder.

Rovibrational SpectroscopyP and R BranchesRigid RotorRotational ConstantDiatomic MoleculeBoltzmann DistributionSelection RulesChemistry

3D · Three.js / WebGL renderer · 60 FPS target · runs fully client-side, no install