Vibration-Rotation Spectra and P, Q, R Branches

Point an infrared spectrometer at a cell of hydrogen chloride gas and you do not see the single sharp line that a bare "the bond vibrates at one frequency" picture predicts. You see a picket fence: a couple of dozen nearly evenly spaced lines straddling a conspicuous hole where the naive line should have been — the fingerprint of two quantized motions at once. This article takes three results as given and assembles them into that fence. From aq-optics-born-oppenheimer-approximation comes the clean separation of electronic, vibrational, and rotational energy into additive ladders. From aq-optics-harmonic-oscillator-vibration comes the vibrational quantum $\Delta v=\pm1$ , worth a tenth of an electron-volt or so — an infrared photon. From aq-optics-rigid-rotor-model comes the rotational ladder $B\,J(J+1)$ , whose rungs are a thousand times finer. And from aq-optics-molecular-selection-rules comes the dipole rule that a vibrational jump must be accompanied by $\Delta J=\pm1$ . Stack these and the single vibrational line shatters into a band. Transitions that spin the molecule up ( $\Delta J=+1$ ) form the R branch on the high-frequency side; transitions that spin it down ( $\Delta J=-1$ ) form the P branch on the low side. The band centre, where the molecule would change vibration without touching its rotation ( $\Delta J=0$ , the Q branch), is empty for an ordinary closed-shell diatomic — the hole in the fence. Read it carefully and the fence hands you the molecule's dimensions: the line spacing gives the bond length to a few picometres, the slow asymmetry between the branches gives how the bond stretches when it vibrates harder, and the rise-and-fall of the line heights is a non-contact thermometer.

Key takeaways

Absorbing one infrared photon lets a diatomic molecule change vibration ( $v\to v+1$ ) and rotation ( $J\to J\pm1$ ) at the same instant, so a single vibrational line splits into a band of rotational lines.
$\Delta J=+1$ builds the R branch above the origin, $\Delta J=-1$ the P branch below it, and $\Delta J=0$ (the Q branch) is missing for closed-shell diatomics — leaving a gap at the band centre.
In the rigid limit the lines are spaced by $2B$ and the P and R branches are mirror images with a $4B$ hole between the innermost lines.
Real bonds stretch more in higher $v$ , so $B_1$ is a hair smaller than $B_0$ ; the R branch slowly converges (and can turn into a band head) while the P branch fans out.
Line heights follow the population $(2J+1)\,e^{-hcB J(J+1)/k_BT}$ , giving each branch a hump whose peak, at $J_{\max}\approx\sqrt{k_BT/2hcB}$ , walks outward as $\sqrt{T}$ — a spectroscopic thermometer.
Key idea: one vibrational transition fragments into P ( $\Delta J=-1$ ) and R ( $\Delta J=+1$ ) branches separated by a $\sim2B$ gap at the usually-missing band origin, with a Q branch ( $\Delta J=0$ ) appearing only when the molecule carries electronic angular momentum along its axis.

The core insight

A single vibrational transition fragments into P ( $\Delta J=-1$ ) and R ( $\Delta J=+1$ ) branches separated by a $\sim2B$ gap at the (usually missing) band origin, with the Q branch ( $\Delta J=0$ ) appearing only when electronic angular momentum permits it.

Where this lives today

Every time a satellite measures greenhouse gases, it is reading vibration-rotation branches. The OCO-2 spacecraft and ground-based Fourier-transform infrared (FTIR) spectrometers resolve the P and R lines of CO $_2$ , CH $_4$ , CO, and H $_2$ O; the branch positions identify the molecule and the line intensities, fit against the Boltzmann envelope, return column density and temperature. The reference line lists that make this possible — HITRAN and GEISA — tabulate exactly the quantities in this article: band origins, $B_v$ values, $\alpha_e$ , and line strengths for millions of individual P, Q, and R transitions. The same physics runs the other way in the laboratory. Tunable diode-laser absorption spectroscopy (TDLAS) parks a laser on a single rotational line and reads the gas temperature from the intensity ratio of two lines — equation (6) used two points at a time — giving microsecond, non-intrusive thermometry inside engines and shock tubes where no thermocouple survives. Astronomers use the same fingerprints in emission: the CO band at $4.7\,\mathrm{\mu m}$ traces warm molecular gas in protoplanetary disks, and the presence or absence of a Q branch still tells an observer at a glance whether the emitting molecule carries electronic angular momentum.

The physical picture

Think of each vibrational level as a shelf and the rotational levels as a ladder of rungs rising from it, the rungs spreading apart as $J(J+1)$ . An infrared photon lifts the molecule to the next shelf up, but the rotational selection rule forbids it from landing on the rung it left: it must step up one rung (R) or down one rung (P). Because a room-temperature gas populates many rungs at once, the single vibrational jump is replayed at dozens of slightly different photon energies — one line per starting rung — and the band is born. The empty centre is the diagnostic feature. Staying on the same rung ( $\Delta J=0$ ) while changing shelf is exactly the forbidden move for a molecule with no electronic angular momentum about its axis, so no photon sits at the bare vibrational frequency $\tilde\nu_0$ : the first R line perches at $\tilde\nu_0+2B$ , the first P line at $\tilde\nu_0-2B$ , framing a $4B$ -wide hole. Give the molecule a whirl about its own bond axis — a $^2\Pi$ state like NO, or the bending mode of a linear polyatomic — and the constraint relaxes: a pile-up of $\Delta J=0$ lines fills the gap as a sharp, bright Q branch.

Figure

Term diagram for a vibration-rotation band: two vibrational shelves v equals 0 and v equals 1, each carrying a rotational ladder spaced as J times J plus 1. R transitions with ΔJ equal plus 1 (blue) sit above the origin, P transitions with ΔJ equal minus 1 (amber) below it, and the ΔJ equal 0 move (red, crossed) is forbidden. The vibrational quantum shown is the HCl origin ν₀ ≈ 2886 cm⁻¹

The heights of the lines tell a second story. A rung of quantum number $J$ has $2J+1$ magnetic sublevels — a rising target count — but sits higher in energy, so the Boltzmann factor $e^{-hcBJ(J+1)/k_BT}$ thins its population. The product peaks at an intermediate $J$ , giving each branch a rounded hump whose position $J_{\max}\approx\sqrt{k_BT/2hcB}$ is a thermometer: heat the gas and both humps march outward. And on close inspection the fence is not quite symmetric — the R lines creep together while the P lines drift apart, the gentle signature of a bond that is a little longer, and so a little floppier, in the upper vibrational state.

Figure

The resulting HCl fundamental band as a stick spectrum: P branch (amber) below the origin, R branch (blue) above it, adjacent lines spaced by 2B ≈ 21 cm⁻¹, and no line at the origin ν₀ ≈ 2886 cm⁻¹ (Q forbidden). Line heights trace the thermal envelope (2J plus 1) times exp of minus hcBJ(J plus 1) over kT, peaking near J ≈ 3 at 300 K

The core equations

Under Born-Oppenheimer the total internal energy is a simple sum of a vibrational and a rotational term, each already quantized by its own article. The key point is that both parts add before any photon is absorbed, so a transition energy is a difference of two such sums. In the wavenumber units ( $\mathrm{cm^{-1}}$ ) spectroscopists use, the term value is:

E_{v,J}=\left(v+\tfrac12\right)hc\,\tilde\nu_e+hc\,B_vJ(J+1) \qquad (1)

where: $v$ — vibrational quantum number $0,1,2,\dots$ ; $J$ — rotational quantum number $0,1,2,\dots$ ; $\tilde\nu_e$ — harmonic vibrational wavenumber at equilibrium; $B_v$ — rotational constant in level $v$ ; $h,c$ — Planck constant and speed of light. Physically: the coarse first term sets which shelf, the fine second term which rung, and the band is the set of photons that jump one shelf while stepping one rung. The rotational constant is not the same on every shelf. Because a real bond is anharmonic, its average length grows with $v$ , the moment of inertia $I=\mu r^2$ swells, and $B\propto1/I$ shrinks — linearly in $v$ to leading order:

B_v=B_e-\alpha_e\left(v+\tfrac12\right) \qquad (2)

where: $B_e$ — constant at the equilibrium bond length; $\alpha_e$ — rotation-vibration coupling constant, $\alpha_e\ll B_e$ (for HCl, $B_e\approx10.6$ but $\alpha_e\approx0.31\,\mathrm{cm^{-1}}$ ). Physically $\alpha_e$ is the entire reason the branches are asymmetric: set it to zero and $B_1=B_0$ and the fence is a perfect mirror. A transition wavenumber is the upper term value minus the lower. Putting $J'=J+1$ over $J''=J$ and collecting powers of $J$ gives the R branch:

\tilde\nu_R(J)=\tilde\nu_0+2B_1+(3B_1-B_0)J+(B_1-B_0)J^2,\qquad J=0,1,2,\dots \qquad (3)

where: $\tilde\nu_0$ — band origin (the forbidden $\Delta J=0$ energy, $\approx\tilde\nu_e$ minus a small anharmonic shift); $B_0,B_1$ — constants of the lower ( $v$ ) and upper ( $v+1$ ) shelves. The first R line, $\tilde\nu_R(0)=\tilde\nu_0+2B_1$ , is the innermost tooth on the high side. The mirror move, $J'=J-1$ over $J''=J$ , gives the P branch, which exists only for $J\ge1$ :

\tilde\nu_P(J)=\tilde\nu_0-(B_1+B_0)J+(B_1-B_0)J^2,\qquad J=1,2,\dots \qquad (4)

where: $J$ — lower-state rotational quantum number. The first P line sits at $\tilde\nu_P(1)=\tilde\nu_0-2B_0$ , so the innermost P and R teeth straddle the origin with a gap $2B_0+2B_1\approx4B$ . Both branches are one parabola in disguise. Introduce a signed index $m$ , positive in R and negative in P, and the two formulae collapse into the Fortrat parabola that generates the whole band:

\tilde\nu(m)=\tilde\nu_0+(B_1+B_0)m+(B_1-B_0)m^2,\qquad m=+(J+1)\ \text{for R},\ \ m=-J\ \text{for P} \qquad (5)

where: $m$ runs $\dots,-2,-1,+1,+2,\dots$ and skips $m=0$ , and that skipped value is precisely the empty origin. The linear term spaces the lines by roughly $2B$ ; the small negative quadratic term (since $B_1<B_0$ ) is what bends R inward and flings P outward. Line heights are set not by the transition but by how many molecules start in the absorbing rung. Degeneracy pushes up, temperature pulls down:

I(J)\propto(2J+1)\,e^{-hcB_0J(J+1)/k_BT} \qquad (6)

where: $2J+1$ — rotational degeneracy (the count of $M_J$ sublevels); $k_B$ — Boltzmann constant; $T$ — absolute gas temperature. The competition peaks at an intermediate $J$ , which is why each branch is humped rather than flat, and why the envelope is a thermometer. When $B_1<B_0$ the Fortrat parabola also turns over at $\tilde\nu_{\text{head}}\approx\tilde\nu_0-(B_1+B_0)^2/4(B_1-B_0)$ , the band head where the R lines stop advancing — reached within a few lines in electronic bands but hidden at impossibly high $J$ in infrared fundamentals.

Derivation AP and R line positions from term-value differences

The goal is to manufacture the branch formulae and the unifying Fortrat parabola directly from the additive term value of equation (1), showing that the rigid limit is a perfect comb of spacing $2B$ with a $4B$ hole, and that the only source of the asymmetry is the tiny inequality $B_1\ne B_0$ . We subtract term values once for each selection rule, then fold both into a single signed-index expression whose skipped value is the missing origin.

Step 1. Write the term value with anharmonicity and centrifugal distortion switched off, $T(v,J)=\tilde\nu_e(v+\tfrac12)+B_vJ(J+1)$ , and let $\tilde\nu_0$ absorb the whole vibrational difference between the two shelves. Why: that spacing is a fixed offset common to every line, so parking it in $\tilde\nu_0$ leaves only the rotational bookkeeping that creates the branches.

Step 2. For the R branch impose $\Delta J=+1$ : upper rung $J'=J+1$ over lower $J''=J$ , so $\tilde\nu_R=\tilde\nu_0+B_1(J+1)(J+2)-B_0J(J+1)=\tilde\nu_0+2B_1+(3B_1-B_0)J+(B_1-B_0)J^2$ . Why: it is a bare subtraction of two rotational energies, so every R line is fixed once $B_0$ , $B_1$ , and the origin are known.

Step 3. For the P branch impose $\Delta J=-1$ : now $J'=J-1$ over $J''=J$ , giving $\tilde\nu_P=\tilde\nu_0+B_1(J-1)J-B_0J(J+1)=\tilde\nu_0-(B_1+B_0)J+(B_1-B_0)J^2$ , valid for $J\ge1$ . Why: spinning down is forbidden from $J=0$ , which is why the P branch starts a full gap below the origin.

Step 4. Unify with a signed index, $m=+(J+1)$ in R and $m=-J$ in P: both collapse to $\tilde\nu(m)=\tilde\nu_0+(B_1+B_0)m+(B_1-B_0)m^2$ , and $m=0$ is never used. Why: the single parabola is the compact truth of the band, and the forbidden $\Delta J=0$ transition is literally the missing integer at the vertex.

Step 5. Take the rigid limit $B_1=B_0=B$ : the quadratic term vanishes and $\tilde\nu=\tilde\nu_0+2Bm$ , a uniform comb of spacing $2B$ with a $4B$ hole across the skipped origin. Why: recovering the evenly spaced comb confirms the algebra and shows the asymmetry is a pure $B_1\ne B_0$ effect, nothing more. *As a check: the HCl fundamental gives R(0) = 2906.2 cm⁻¹ and P(1) = 2865.0 cm⁻¹ from ν₀ = 2885.9, B₀ = 10.44, B₁ = 10.13 cm⁻¹ — a gap of 41.1 cm⁻¹ that equals 4B to three figures, confirming the branch formulae.*

Go deeper · Derivation B · 1 minThe most-populated rotational level and the intensity envelope

Derivation BThe most-populated rotational level and the intensity envelope

The goal is to explain why every branch is humped, not flat, and why the humps walk outward when the gas is heated, by finding the rung $J_{\max}$ that the most molecules occupy. We treat the population as a continuous product of a rising degeneracy and a falling Boltzmann factor, maximize it, and read off the $\sqrt{T}$ scaling that turns a resolved band into a thermometer.

Step 1. Write the absorbing-level population as $N_J\propto(2J+1)\,e^{-hcB_0J(J+1)/k_BT}$ , the product of $2J+1$ degenerate sublevels and the Boltzmann weight of $E_J=hcB_0J(J+1)$ . Why: absorption strength scales with how many molecules wait in the lower rung, so the intensity envelope is a direct image of this thermal distribution.

Step 2. Because many rungs are populated at room temperature (for HCl $k_BT/hc\approx208\,\mathrm{cm^{-1}}$ versus $B_0\approx10\,\mathrm{cm^{-1}}$ ), treat $J$ as continuous and set $\mathrm{d}N_J/\mathrm{d}J=0$ . Why: with dozens of rungs in play the distribution is smooth enough that a derivative locates the peak to better than one unit of $J$ .

Step 3. Differentiate the product; the exponential divides out and, using $\mathrm{d}[J(J+1)]/\mathrm{d}J=2J+1$ , the condition becomes $2-\dfrac{hcB_0}{k_BT}(2J+1)^2=0$ . Why: the two competing trends — a factor rising like $2J+1$ and an exponent falling like $J(J+1)$ — balance exactly at the peak, which this equation pins down.

Step 4. Solve for the peak rung, $J_{\max}=\sqrt{\dfrac{k_BT}{2hcB_0}}-\dfrac12$ , which grows as $\sqrt{T}$ . Why: raising the temperature fourfold doubles $J_{\max}$ and so doubles the separation between the two humps, giving a calibrated, non-contact gas thermometer. *As a check: CO with $B_0=1.931\,\mathrm{cm^{-1}}$ gives $J_{\max}=6.85\approx7$ at 300 K and $14.2\approx14$ at 1200 K — the fourfold increase in $T$ exactly doubles the peak rung, consistent with the observed shift of its infrared fundamental.*

Limiting cases

Case	Result	Meaning
Rigid rotor ( $B_1=B_0=B$ )	$\tilde\nu_R(J)=\tilde\nu_0+2B(J+1),\ \ \tilde\nu_P(J)=\tilde\nu_0-2BJ$	Both branches an exact comb of spacing $2B$ , with a symmetric $4B$ hole across the empty origin.
Real rotor ( $B_1<B_0$ )	$\tilde\nu=\tilde\nu_0+(B_1+B_0)m+(B_1-B_0)m^2$	The negative quadratic term makes R converge (toward a distant head) and P diverge — one parabola spans both.
Q branch allowed ( $\Delta J=0$ , $\Lambda\ne0$ )	$\tilde\nu_Q(J)=\tilde\nu_0+(B_1-B_0)J(J+1)$	Lines cluster near $\tilde\nu_0$ because the spacing rides only on the tiny $B_1-B_0$ , making a sharp intense spike at the band centre.
Hot band / high $T$	$J_{\max}\approx\sqrt{\dfrac{k_BT}{2hcB}}-\tfrac12$	The peak rung, and hence the envelope width, grows as $\sqrt{T}$ — a direct spectroscopic thermometer.
Centrifugal stretch (large $J$ )	add $-D\,J^2(J+1)^2$ to the term value	Fast rotation lengthens the bond, pulling high- $J$ lines below their rigid positions.

Common trapThe band centre is empty, not bright — do not expect a line at the origin

The single most common error is to draw a strong line at the band origin $\tilde\nu_0$ , imagining the "pure vibrational" transition. For an ordinary closed-shell ( $^1\Sigma$ ) diatomic the $\Delta J=0$ Q branch is forbidden, so the origin is a gap: the innermost lines sit at $\tilde\nu_0\pm2B$ and the centre is dark. A Q branch appears only when the molecule carries electronic angular momentum along its axis ( $\Lambda\ne0$ , as in NO), or in a degenerate bending mode of a linear polyatomic. The second trap is to treat $B$ as one number for the whole band: it is $B_1\ne B_0$ that makes the branches asymmetric, so collapsing them to a single $B$ erases the band head and the coupling constant $\alpha_e$ entirely.

Worked examples

1 · Bond length of HCl from its infrared band (confidence builder)

The resolved lines of the HCl fundamental are spaced by an average of about $2B\approx20.9\,\mathrm{cm^{-1}}$ , so $B_0\approx10.44\,\mathrm{cm^{-1}}$ . Find the moment of inertia and the equilibrium bond length. Use the reduced mass of $^1$ H $^{35}$ Cl.

Step 1. Convert the spacing to a moment of inertia through $I=h/(8\pi^2cB_0)$ . With $h=6.626\times10^{-34}\,\mathrm{J\,s}$ , $c=2.998\times10^{10}\,\mathrm{cm/s}$ (the cgs speed, because $B$ is in $\mathrm{cm^{-1}}$ ), and $B_0=10.44\,\mathrm{cm^{-1}}$ : $I=6.626\times10^{-34}/(78.96\times2.998\times10^{10}\times10.44)=\mathbf{2.68\times10^{-47}\ kg\,m^2}$ .

Step 2. Compute the reduced mass $\mu=m_\mathrm{H}m_\mathrm{Cl}/(m_\mathrm{H}+m_\mathrm{Cl})$ . With $m_\mathrm{H}=1.008$ u and $m_\mathrm{Cl}=34.97$ u, $\mu=(1.008\times34.97)/(35.98)=0.980$ u $=0.980\times1.6605\times10^{-27}=\mathbf{1.627\times10^{-27}\ kg}$ .

Step 3. Invert $I=\mu r^2$ : $r=\sqrt{I/\mu}=\sqrt{2.68\times10^{-47}/1.627\times10^{-27}}=\mathbf{1.28\times10^{-10}\ m}$ . So the HCl bond is about 128 pm, within a percent of the accepted 127.5 pm — read straight off the spacing of an infrared picket fence, no microscope required.

2 · The intensity envelope as a thermometer (CO)

Carbon monoxide has $B_0\approx1.931\,\mathrm{cm^{-1}}$ . Find the most-populated rotational level $J_{\max}$ at 300 K and in a 1200 K flame, and say what happens to the band.

Step 1. Evaluate the thermal energy in wavenumbers, $k_BT/hc=(k_B/hc)\,T=0.6950\,\mathrm{cm^{-1}\,K^{-1}}\times T$ . At 300 K that is $208.5\,\mathrm{cm^{-1}}$ ; at 1200 K it is $834.0\,\mathrm{cm^{-1}}$ .

Step 2. Apply $J_{\max}=\sqrt{k_BT/2hcB_0}-\tfrac12$ . At 300 K: $\sqrt{208.5/(2\times1.931)}-0.5=\sqrt{54.0}-0.5=7.35-0.5=\mathbf{6.85\approx7}$ . At 1200 K: $\sqrt{834.0/3.862}-0.5=\sqrt{216}-0.5=14.7-0.5=\mathbf{14.2\approx14}$ .

Step 3. Because $J_{\max}\propto\sqrt{T}$ , quadrupling the temperature exactly doubles the peak rung (7 → 14), so the two envelope maxima pull twice as far apart. Measuring that separation on a single resolved band returns the gas temperature with no probe touching it — the working principle of combustion and plasma thermometry. J_max ≈ 7 at 300 K and ≈ 14 at 1200 K, so the envelope width doubles.

3 · Watching the branches go asymmetric (HCl line positions)

For the HCl fundamental take $\tilde\nu_0=2885.9\,\mathrm{cm^{-1}}$ , $B_0=10.44\,\mathrm{cm^{-1}}$ , $B_1=10.13\,\mathrm{cm^{-1}}$ ( $\alpha_e=0.31$ ). Locate the first two R lines and first two P lines, and show that R converges while P diverges.

Step 1. R branch, equation (3). For $J=0$ : $2885.9+2B_1=\mathbf{2906.2\ cm^{-1}}$ . For $J=1$ : $2885.9+20.26+(30.39-10.44)-0.31=\mathbf{2925.8\ cm^{-1}}$ . Their spacing $19.6\,\mathrm{cm^{-1}}$ is already less than $2B_0=20.9$ — the R lines are drawing together.

Step 2. P branch, equation (4). For $J=1$ : $2885.9-20.57-0.31=\mathbf{2865.0\ cm^{-1}}$ . For $J=2$ : $2885.9-41.14-1.24=\mathbf{2843.5\ cm^{-1}}$ . Their spacing $21.5\,\mathrm{cm^{-1}}$ is more than $2B_0$ — the P lines are fanning apart.

Step 3. The innermost lines R(0) and P(1) straddle the empty origin with a gap $2906.2-2865.0=41.2\,\mathrm{cm^{-1}}$ , equal to $2B_0+2B_1=41.1\,\mathrm{cm^{-1}}$ — the $\sim4B$ hole. The converging-R, diverging-P pattern is the direct fingerprint that $B_1<B_0$ , i.e. the bond is longer in $v=1$ . R(0)=2906.2, R(1)=2925.8, P(1)=2865.0, P(2)=2843.5 cm⁻¹; R converges, P diverges, gap ≈ 4B.

Physics of the portrait

The vibration-rotation band is a place where quantum theory was caught in the act: its two branches and missing centre were predicted from the old quantum theory, resolved into individual lines just before wave mechanics arrived, and codified into a whole industry of molecular constants by one towering figure.

Origins · 1 minGerhard Herzberg and how this was discovered

Gerhard Herzberg

1904-1999 · infrared and molecular spectroscopy of diatomics

Gerhard Herzberg turned the reading of molecular spectra from an art into a discipline. Trained as a physicist in Germany, he fled the Nazi regime in 1935 and settled at the University of Saskatchewan and then the National Research Council of Canada in Ottawa, where he built one of the world's great spectroscopy laboratories. His genius was systematic: given a photographic plate crowded with the P, Q, and R lines of a band, he showed how to extract the vibrational frequency $\tilde\nu_e$ , the rotational constants $B_e$ and $\alpha_e$ , the centrifugal term $D$ , the bond length, the dissociation energy, and the electronic term symbol — the full molecular fingerprint. His three-volume Molecular Spectra and Molecular Structure, begun in 1939, remains the reference every spectroscopist still reaches for; the analysis in this article is essentially his. He detected new molecules and free radicals in the laboratory and in interstellar space, and received the 1971 Nobel Prize in Chemistry for this work on the structure and geometry of molecules — a physicist's prize awarded by the chemists, fittingly for the man who taught both fields to read the picket fence.

History

1881

— Abney and Festing photograph the near-infrared absorption of organic liquids, the first molecular vibration spectra

1905

— Coblentz systematically measures the infrared spectra of hundreds of compounds, mapping the vibration region

1912

— Bjerrum predicts that quantized rotation splits each infrared vibration band into separated branches

1919

— Imes resolves the individual rotational lines of HCl and HBr and finds the missing line at the band centre

1926

— wave mechanics fixes the rotational selection rule ΔJ = ±1 for a Σ state, explaining the empty origin

1950

— Herzberg's Molecular Spectra and Molecular Structure I codifies the full P, Q, R branch analysis

1971

— Herzberg receives the Nobel Prize in Chemistry for molecular spectroscopy

Going further

Texts. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules — the definitive reference for branch analysis · Demtröder, Laser Spectroscopy, Vol. 1, chapter on molecular spectra · Bernath, Spectra of Atoms and Molecules, 3rd ed. · Banwell and McCash, Fundamentals of Molecular Spectroscopy · Foot, Atomic Physics, molecular chapter. Key numerical anchors for this article: HCl $B_0\approx10.44\,\mathrm{cm^{-1}}$ gives a bond length of 128 pm; HCl line positions R(0)=2906.2 and P(1)=2865.0 cm⁻¹ follow from $\tilde\nu_0=2885.9$ , $B_0=10.44$ , $B_1=10.13\,\mathrm{cm^{-1}}$ ; CO $B_0=1.931\,\mathrm{cm^{-1}}$ gives $J_{\max}\approx7$ at 300 K and $\approx14$ at 1200 K. The R and P branch formulae follow from term-value subtraction with $J'=J\pm1$ ; $J_{\max}=\sqrt{k_BT/2hcB_0}-\tfrac12$ from maximizing $(2J+1)e^{-hcB_0J(J+1)/k_BT}$ ; and the band-head vertex is the turning point of the Fortrat parabola $\tilde\nu(m)=\tilde\nu_0+(B_1+B_0)m+(B_1-B_0)m^2$ .

Builds on: The Rigid Rotor Model · The Harmonic Oscillator Model of Vibration · Selection Rules for Molecular Transitions · The Born-Oppenheimer Approximation. Leads to: Anharmonicity and the Morse Potential · Electronic Transitions in Molecules · The Franck-Condon Principle · Raman Scattering and the Rotational and Vibrational Raman Effect.

Ask the tutor: Want to see how the same P, Q, R machinery reshapes into a band head and Franck-Condon envelope once the molecule makes an electronic jump instead of a vibrational one?

Why does one vibrational transition produce so many lines instead of one?

Because the molecule changes two things at once. Absorbing an infrared photon flips $v\to v+1$ , but the dipole rule forces $J$ to change too, by one unit. A room-temperature gas is spread over dozens of rotational rungs, and each starting rung gives a slightly different total photon energy — the vibrational quantum plus or minus a rotational step. The result is one line per populated rung, spaced by roughly $2B$ . Freeze rotation out (everything in $J=0$ near absolute zero) and you would see essentially a single line, the first R line at $\tilde\nu_0+2B$ .

What exactly determines whether a Q branch appears?

The Q branch is the set of $\Delta J=0$ transitions, and whether it exists depends on the molecule's electronic angular momentum about the internuclear axis, the quantum number $\Lambda$ . For a closed-shell $^1\Sigma$ diatomic $\Lambda=0$ , the move is forbidden and the band centre is a gap. If $\Lambda\ne0$ — like NO in its $^2\Pi$ ground state — or in a degenerate bending vibration of a linear polyatomic, $\Delta J=0$ becomes allowed and a Q branch appears. It is sharp and intense because its lines cluster near $\tilde\nu_0$ : their positions scale only with the tiny $B_1-B_0$ , so they pile up almost on top of each other.

What is the difference between B_e, B_0, and B_1, and why does it matter?

$B_e$ is the rotational constant the molecule would have sitting motionless at the bottom of its potential well, at the equilibrium bond length — a fictional reference. $B_0$ and $B_1$ are the real constants averaged over the vibrational motion in $v=0$ and $v=1$ . Because an anharmonic bond spends more time stretched than compressed, its average length grows with $v$ , the moment of inertia grows, and $B$ shrinks: $B_1<B_0<B_e$ , tied together by $B_v=B_e-\alpha_e(v+\tfrac12)$ . This is the entire origin of the branch asymmetry: if $B_0=B_1$ the P and R branches would be perfect mirror images, and the small inequality is what makes R converge, P diverge, and band heads possible.

How does reading a band give me the bond length?

The average spacing between adjacent lines is approximately $2B$ , and $B=h/(8\pi^2cI)$ ties the constant to the moment of inertia $I=\mu r^2$ . So you measure the spacing, halve it to get $B$ , invert to get $I$ , divide by the reduced mass $\mu$ (known from atomic masses), and take a square root to get $r$ — worked example 1 does exactly this for HCl and lands on 128 pm. Resolving the small branch asymmetry gives $B_0$ and $B_1$ separately, hence $\alpha_e$ and how the geometry changes with vibration — a remarkable amount of structural detail from a single infrared trace.

Why do the line intensities rise and then fall across a branch?

Two effects fight. The degeneracy $2J+1$ counts the magnetic sublevels of a rung, so higher rungs offer more molecules to excite — intensity rising with $J$ . But higher rungs cost more energy, so the Boltzmann factor $e^{-hcBJ(J+1)/k_BT}$ thins their population — intensity falling with $J$ . Near the origin the degeneracy is small; far out the exponential has crushed everything; the product peaks in between at $J_{\max}=\sqrt{k_BT/2hcB}-\tfrac12$ , giving each branch its hump. Because $J_{\max}$ grows as $\sqrt{T}$ , that hump's position is a thermometer.

Is a band head the same thing as the band origin?

No — they are almost opposites. The band origin $\tilde\nu_0$ is the (usually empty) centre of the band, the forbidden $\Delta J=0$ energy. A band head is a sharp edge out in a branch, where the Fortrat parabola passes its vertex and the lines stop advancing and turn back on themselves. Setting $\mathrm{d}\tilde\nu/\mathrm{d}m=0$ locates it at $\tilde\nu_{\text{head}}\approx\tilde\nu_0-(B_1+B_0)^2/4(B_1-B_0)$ . For an infrared fundamental the constants differ so little ( $\alpha_e\sim0.3\,\mathrm{cm^{-1}}$ ) that the head lies at astronomically high $J$ and is never reached, so you see two open branches. In an electronic transition the bond length changes a lot, $B_1-B_0$ is a sizeable fraction of $B$ , and the head appears within a dozen lines as the striking degraded edge of molecular band spectra.