Selection rules

A transition needs something to grab hold of. Light is an oscillating electric field. For a molecule to absorb or emit a photon, that field must be able to push and pull on some oscillating or rotating charge distribution in the molecule — its dipole moment. This is captured formally by the transition dipole moment integral:

where: $\psi_i$ and $\psi_f$ are the initial- and final-state wavefunctions, $\hat{\mu}$ is the dipole moment operator, and $d\tau$ is integration over all space. A transition is allowed if $\mu_{fi} \neq 0$ and forbidden if $\mu_{fi} = 0$. You are not expected to evaluate this integral symbolically in CY11001 — only to use its physical consequence: no dipole to couple to means no absorption, no matter how well the photon's energy matches $\Delta E$.

Two separate conditions must both be met. The first is a gateway condition — does the transition have a dipole to couple to at all? The second is a "which levels" condition — given that gateway is open, which specific jumps in the quantum number are allowed?

• Rotational gateway: the molecule needs a permanent dipole moment. A polar molecule like $\text{HCl}$ presents an oscillating dipole to a stationary observer as it tumbles; a homonuclear diatomic like $\text{N}_2$ or $\text{O}_2$ has zero dipole at every bond length and is microwave-inactive.
• Vibrational gateway: the dipole must change during the vibration, $d\mu/dr \neq 0$. $\text{HCl}$'s dipole changes as the bond stretches, so it is IR-active; $\text{N}_2$'s dipole stays zero throughout the stretch by symmetry, so it is IR-inactive.
• Rotational levels: for the rigid rotor, $\Delta J = \pm 1$ (absorption: $\Delta J = +1$).
• Vibrational levels: for the simple harmonic oscillator, $\Delta v = \pm 1$ (absorption: $\Delta v = +1$); real anharmonic molecules show weak overtone bands with $\Delta v = \pm2, \pm3,\ldots$, but these are not the primary quantitative content here.
• Electronic transitions: governed qualitatively by orbital overlap — $\sigma \to \sigma^*$, $n \to \sigma^*$, $\pi \to \pi^*$ and $n \to \pi^*$ jumps differ in how strongly allowed they are, with good spatial overlap giving large absorption intensity ($\varepsilon$) and poor overlap giving weak bands.

The figure below separates the two conditions cleanly: a molecule can fail the gateway test (no dipole at all, like $\text{N}_2$), pass the gateway but fail the level test (right dipole, wrong $\Delta J$ or $\Delta v$), or pass both and show a real spectral line.

The physical picture behind both conditions is the same coupling requirement, applied at two different levels of detail, and no group theory is needed to see it.

Why the gateway condition is about a dipole, specifically. A photon's electric field oscillates in time. For the molecule to absorb energy from it, the molecule must have some charge separation that can be pushed by that field — a dipole moment — and that dipole must itself be changing (rotating, for absorption of rotational energy; stretching in magnitude, for vibrational energy). A perfectly symmetric charge distribution, like $\text{N}_2$'s, has nothing for the field to grab: the positive and negative charge centres coincide at every instant, at every bond length, so there is no dipole to drive.

Why the gateway condition survives motion that "looks" active. $\text{N}_2$ genuinely rotates and vibrates — it has real rotational and vibrational kinetic energy. But motion alone does not create a coupling; only a changing dipole does. This is the trap students fall into most: mechanical activity is not the same thing as spectroscopic activity.

Why the level rule restricts $\Delta J$ and $\Delta v$ to $\pm 1$ (qualitatively). A photon carries a fixed, small quantum of angular momentum and a fixed quantum of energy per absorption event. In the simplest rigid-rotor and harmonic-oscillator pictures used in this course, one photon absorbed drives one single step up the ladder of rotational or vibrational levels — not a two-rung jump. This is why $J = 2 \to J = 3$ is a clean single-photon event but $J = 2 \to J = 4$ is not, even when a photon with exactly the right total energy exists: selection rules constrain which single-step transitions the coupling mechanism can drive, on top of (not instead of) the requirement that photon energy matches $\Delta E$.

Problem: A rigid-rotor diatomic shows a strong microwave absorption line for $J = 2 \to J = 3$. A student proposes that $J = 2 \to J = 4$ should also appear, since a suitable photon energy exists for that gap too. Also explain, using selection-rule reasoning, why $\text{CO}$ shows both rotational and vibrational spectra while $\text{N}_2$ shows neither.

Step 1. Check the rotational level rule for the proposed transitions. The rigid-rotor rule is $\Delta J = \pm 1$, and absorption requires $J$ to increase by exactly $1$. For $J = 2 \to J = 3$, $\Delta J = +1$: this satisfies the rule, so the transition is allowed, matching the observed line.

Step 2. Check $J = 2 \to J = 4$ the same way. Here $\Delta J = +2 \neq \pm 1$, so the level rule is violated: the transition is forbidden, regardless of whether a photon of the right energy exists. Energy matching and the selection rule are independent conditions, and both must hold; this transition fails the second one.

Step 3. Apply the gateway condition to $\text{CO}$. Carbon and oxygen differ in electronegativity, so $\text{CO}$ carries a small permanent dipole ($\mu \approx 0.11\ \text{D}$). As it rotates, this dipole sweeps around in space (rotational gateway open), and as the bond stretches, the dipole magnitude changes, $d\mu/dr \neq 0$ (vibrational gateway open). Both spectroscopies are therefore accessible, and the observed lines obey $\Delta J = \pm 1$ and $\Delta v = +1$.

Step 4. Apply the gateway condition to $\text{N}_2$. By symmetry, the two identical nitrogen atoms give zero dipole moment at every bond length, so $\mu = 0$ always and $d\mu/dr = 0$ always. Both gateways are shut, independent of the level rule. Answer: $J = 2 \to J = 4$ is forbidden by the $\Delta J = \pm 1$ rule even though energy conservation alone would permit it; $\text{CO}$ is both microwave- and IR-active because it has a permanent dipole that also changes on stretching, while $\text{N}_2$ is invisible to both because its dipole is exactly zero at all bond lengths.

• Treating "the molecule is rotating/vibrating" as the same as "the transition is spectroscopically active" — $\text{N}_2$ does both mechanically but shows neither spectrum, because it fails the dipole gateway.
• Applying the vibrational gateway to the equilibrium dipole instead of its change. $\text{CO}_2$ has zero permanent dipole yet is IR-active in its bending and asymmetric-stretch modes, because those specific motions do change the dipole.
• Assuming a photon with the exact right energy is automatically absorbed. The resonance condition $\Delta E = h\nu$ and the selection rule are separate hurdles; a transition must clear both.
• Reaching for group-theory or character-table arguments. CY11001 only requires the physical dipole-moment reasoning and the resulting $\Delta J$, $\Delta v$ rules — not a symmetry-based derivation.

Ask the tutor: ask me to check whether a specific molecule (like $\text{HCl}$, $\text{Cl}_2$, $\text{H}_2\text{O}$ or $\text{SO}_2$) is microwave-active, IR-active, or both, and I'll walk through the gateway and level tests with you.