advanced27 min studyLesson 466 in Physics

Rovibrational selection rules

Connect dipole and polarizability matrix elements to IR and Raman activity, then calculate elementary P and R branches within stated electronic symmetry.

On this page

What you’ll understand

  • Derive the harmonic infrared vibrational rule from a dipole expansion.
  • State the restricted circumstances for P/R branches without a Q branch.
  • Distinguish infrared dipole activity from Raman polarizability activity.

An energy difference can match a photon while the corresponding transition remains weak or forbidden in a chosen interaction approximation. Spectral activity depends on a matrix element, not only on available levels. Molecular vibration changes the dipole and polarizability, while molecular rotation controls how those quantities project into the laboratory field. We will derive a simple infrared vibrational rule, attach rotational branches for a closed-shell diatomic, and explain why a homonuclear molecule can be Raman active while its electric-dipole infrared vibration is absent.

The transition operator supplies the rule

For weak light in the electric-dipole approximation, the transition amplitude involves the polarization projection of the molecular dipole between initial and final states. Separate electronic, vibrational, and rotational factors where the Born–Oppenheimer and rotation-vibration approximations allow it. Within one electronic state, expand the body-fixed dipole as a function of stretch displacement q about equilibrium. A constant dipole term gives no transition between different orthogonal vibrational states. The linear derivative term is the leading infrared contribution.

For harmonic vibrational wavefunctions, q connects levels differing by one quantum. The leading rule is therefore Δv=±1\Delta v=\pm1 when the dipole derivative is nonzero. Anharmonic wavefunctions and higher derivatives of the dipole allow overtones and modify intensities, so the harmonic rule is not an absolute prohibition in every molecular Hamiltonian. A permanent dipole is sufficient for elementary pure rotational electric-dipole activity, but vibrational activity specifically asks how that dipole changes with the normal coordinate.

Mfi=⟨f∣ϵ⋅μ∣i⟩,μ(q)=μ0+μ0′q+12μ0′′q2+⋯ ,⟨v+1∣q∣v⟩=ℏ(v+1)2μredωM_{fi}=\langle f|\boldsymbol\epsilon\cdot\boldsymbol\mu|i\rangle,\qquad \mu(q)=\mu_0+\mu'_0q+\frac12\mu''_0q^2+\cdots,\qquad \langle v+1|q|v\rangle=\sqrt{\frac{\hbar(v+1)}{2\mu_{\mathrm{red}}\omega}}
Mu without subscript red denotes electric dipole, while mu red is nuclear reduced mass. q is a physical displacement in metres. The last matrix element is for a harmonic oscillator and supports the leading fundamental rule.↔ Scroll wide equations horizontally.

Rotational branches depend on electronic symmetry

For a polar diatomic in a closed-shell 1Σ^1\Sigma electronic state, the elementary electric-dipole rovibrational transitions within that electronic state have ΔJ=±1\Delta J=\pm1. The negative change forms the P branch and the positive change the R branch. A Q branch with zero change is absent for this specified parallel Sigma-to-Sigma case. Other electronic angular momenta or polyatomic band types can admit Q branches, so the restriction must not be promoted into a universal rule for every vibration.

If upper and lower vibrational states share one approximate rotational constant, subtracting their rotor terms gives particularly simple branch positions about the vibrational band origin. P lines lie below the origin and R lines above it. There is no P(0) because angular momentum cannot become negative. Unequal rotational constants produce quadratic J-dependent shifts and possible branch heads. Nuclear-spin restrictions and population factors further determine which permitted lines appear and with what strength.

ν~R(J)=ν~0+2B~(J+1),ν~P(J)=ν~0−2B~J(J≥1)\widetilde\nu_R(J)=\widetilde\nu_0+2\widetilde B(J+1),\qquad \widetilde\nu_P(J)=\widetilde\nu_0-2\widetilde BJ\quad(J\ge1)
J labels the lower-state rotational quantum number. The formulas assume the same B tilde in both vibrational states, neglect distortion and spin, and concern the specified 1 Sigma parallel band; all quantities are in inverse centimetres.↔ Scroll wide equations horizontally.

Raman and infrared inspect different molecular properties

Raman scattering couples through the polarizability tensor, because the incident field induces a dipole. A vibrational mode can be Raman active when polarizability changes along that coordinate even if the permanent dipole remains zero. A homonuclear diatomic stretch is the standard example: inversion forbids its isolated electric-dipole vibrational transition within the same electronic state, while its polarizability can vary. For centrosymmetric molecules the familiar infrared/Raman mutual exclusion refers to the symmetry of fundamental modes within the appropriate approximations.

Raman shifts measure the energy gained or lost by the molecule, rather than a requirement that the incident photon itself match an infrared absorption frequency. Stokes light leaves the molecule more excited, while anti-Stokes light removes an initial excitation. Rotational Raman rules arise from scalar and rank-two polarizability components and differ from dipole rules. These statements identify symmetry and coupling mechanisms; actual line intensities still require populations, matrix elements, polarization geometry, and the optical frequency factors appropriate to the experiment.

An elementary infrared band

A hypothetical polar 1 Sigma diatomic has vibrational origin ν~0=1000 cm−1\widetilde\nu_0=1000\,\mathrm{cm^{-1}} and the same B~=2 cm−1\widetilde B=2\,\mathrm{cm^{-1}} in both vibrational states. Find P(1), R(0), and R(2), and identify the central gap between the nearest P and R lines.

  1. P(1) has lower J = 1 and changes to J = 0, giving 1000−2(2)(1)=996 cm−11000-2(2)(1)=996\,\mathrm{cm^{-1}}. P(0) is not an allowed label for this branch.

  2. R(0) gives 1000+2(2)(1)=1004 cm−11000+2(2)(1)=1004\,\mathrm{cm^{-1}}. R(2) gives 1000+2(2)(3)=1012 cm−11000+2(2)(3)=1012\,\mathrm{cm^{-1}}.

  3. The nearest-line gap is 1004−996=8 cm−1=4B~1004-996=8\,\mathrm{cm^{-1}}=4\widetilde B. There is no Q line at the origin in this restricted band model, even though the vibrational origin is a meaningful term difference.

The result

The lines are 996, 1004, and 1012 cm⁻¹, with an 8 cm⁻¹ central gap. The absence of a Q branch follows from the stated symmetry, not from the absence of a vibrational energy difference.

Explore the model

Interactive lab

Interactive lab · vectors

Polarization selects a projection

Use a vector as a transition-dipole projection analogy with the x axis as chosen light polarization. Reset, set Angle = 0° and Magnitude = 1, and predict x projection 1. For a fresh comparison Reset, set Angle = 90° and Magnitude = 1, and predict x projection 0.

Polarization selects a projectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis vector diagram illustrates a geometric projection only. It does not evaluate quantum transition integrals, rotate a molecule, impose selection rules, or calculate Raman intensity.

Check your understanding

What property controls leading harmonic infrared vibrational activity within one electronic state?

Try it yourself

In a harmonic model the dipole derivative is unchanged between the 0 to 1 and 1 to 2 transitions. Compare their squared vibrational transition moments before population and rotational factors. Then state whether zero permanent dipole rules out Raman activity.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The displacement matrix element grows as the square root of v plus one; intensities use its squared magnitude.

Compare with the solution

The matrix-element magnitude ratio is 2\sqrt2, so the squared vibrational transition-moment ratio is 2. This does not say the measured hot-band intensity is twice as large, because the initial excited-state population and other factors differ. A zero permanent dipole does not rule out Raman activity: a coordinate-dependent polarizability can couple the vibration to scattered light.

Take it with you

  • Transition rules arise from matrix elements of a specified interaction operator.
  • The simple no-Q P/R pattern has stated electronic and rotor assumptions.
  • IR dipole derivatives and Raman polarizability derivatives probe different symmetries and cannot be interchanged.

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