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For the linear rigid rotor, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ and the electric electric dipole moment interaction is $V=-E{\hat{d}}_z$. The isotropic $\ell =0$ state has no first-order shift. Its only coupled rotational state is $\ell =1,m_{\ell }=0$, at gap ${\Delta }_{\mathrm{rot}}={\hslash }^2/I$, giving the quadratic Stark shift of a polar linear rotorUsing $1\mathrm{kV}\mathrm{c}{\mathrm{m}}^{-1}=10^5\mathrm{V}{\mathrm{m}}^{-1}$ and the supplied I and d givesThe negative sign is the expected second-order Stark effect of a nondegenerate ground state. The mixing parameter $dE/(\sqrt{3}{\Delta }_{\mathrm{rot}})\simeq 0.074$ is small, supporting the weak-field estimate. The constants h and $\hslash =h/(2\pi )$ use their defined SI values; no molecular input beyond the supplied data is needed.

The only relevant position matrix element is $z_{01}=\sqrt{\hslash /(2m{\omega }_0)}$. To first order in the radiation amplitude, the two ground–excited-state cross terms giveMultiplying by the charge $-e$ gives the induced electric dipole moment. Since $2\sqrt{\hslash /(2m{\omega }_0)}C=e{A}_0/m$, the coefficient derived in part (b) yieldsat linear order. Taking the complex conjugate inside the real part of the second term changes it to minus the same term with $e^{i\theta }$ and reversed frequency signs. ThusThe terms at ${\omega }_0$ are switch-on transients, while those at $\omega$ describe the driven response. Both are needed to make $\mathbf{P}(0)=0$. The expression is understood by continuity at $\omega ={\omega }_0$; the first quotient then tends to $t{e}^{-i{\omega }_{0}t}$, producing resonant secular growth rather than an infinite dipole at a finite time.

There is also an initial-velocity check on this harmonic-oscillator dipole response to switched radiation. The formula gives ${\dot{P}}_z(0^{+})=-2e^2{A}_0\cos (ky)/m$. Because the magnetic vector potential jumps at zero time, its time derivative contains an impulsive electric field. Equivalently, the mechanical velocity is $(p_z+e{A}_z)/m$, whose initial mean is $2e{A}_0\cos (ky)/m$. The transient therefore implements the specified switch-on correctly.

For $\omega \ne {\omega }_0$, isolate the terms proportional to $e^{-i\omega t}$ in the previous answer. Their coefficient isHence the driven electric dipole moment component isFor $t>0$, away from the instantaneous switch, the radiation electric field isThus the dynamic electric polarizability isBelow resonance the driven electric dipole moment is in phase with the electric field; above resonance its sign reverses. The same susceptibility follows from the classical Lorentz oscillator model without damping. At exact resonance the separation into bounded driven and transient terms fails: their apparent poles combine into the finite-time secular term identified in part (c). Without damping there is no bounded steady resonant response, and the first-order state expansion eventually ceases to apply.

The electron's electric dipole moment is $\mathbf{d}=-e⟨x⟩\hat{\mathbf{x}}$. It vanishes in the unperturbed ground energy eigenstate. The first correction to the wavefunction is linear in the electric field and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position expectation value of first order.

In detail, the normalized ground quantum state, to the order needed, isThe normalization correction starts at second order and does not affect the linear term below. The two cross terms giveThe absence of an $E^2$ term follows from the parity operator: $⟨x⟩$ changes sign under field reversal. Comparing with the second-order nondegenerate perturbation theory energy sum yieldsThus the induced electric dipole moment points along the electric field, with positive electric polarizability. The induced dipole moment and quadratic energy shift obey $\delta {ϵ}_0^{(2)}=-{\alpha }_{\mathrm{pol}}{E}^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the Hellmann–Feynman theorem gives $d_x=-\partial {ϵ}_0/\partial E$, with the same result.

Low rotational energy spacings can be excited in cold interstellar gas when electronic and most vibrational excitations are frozen out. A polar molecule such as carbon monosulfide has allowed electric dipole transition lines whose frequencies depend on its moment of inertia, providing a distinctive pattern for rotational spectral line assignment. Several matching spectral lines and their isotope shifts give a much stronger identification than one coincidence. The method is especially effective for polar species; a molecule without a permanent electric dipole moment lacks an ordinary pure electric-dipole rotational spectrum.

The transition strength depends on $|R_{ji}{|}^2$, so a necessary condition is that at least one Cartesian electric dipole transition moment is nonzero. An integral can be nonzero only if its integrand contains the totally symmetric representation:The harmonic vibrational ground state is totally symmetric, and a one-quantum state of normal mode $k$ transforms as ${\Gamma }^{(k)}$. For the real irreducible representations here, the fundamental is therefore symmetry-allowed precisely whenEquivalently, expand the electric dipole moment near equilibrium:The fundamental has $⟨1_k|Q_k|0_k⟩\ne 0$, so vibrational infrared activity requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving $A_{2u}$ and $E_{1u}$ as the IR-active species. Symmetry permission does not guarantee a large band intensity.

The gross requirement is an anisotropic polarizability tensor, so that changing molecular orientation changes the induced response. A permanent electric dipole moment is not required; a homonuclear diatomic can therefore have a rotational Raman scattering spectrum.

For the inelastic lines of a linear rigid rotor, the detailed rule is $\Delta J=\pm 2$, with $J\ge 0$. The $\Delta J=0$ contribution is unshifted Rayleigh scattering, not a rotational Stokes Raman scattering line. Stokes Raman scattering has $\Delta J=+2$ because the molecule gains rotational energy.

A polar diatomic molecule can absorb microwaves through its permanent electric dipole moment. In the linear rigid rotor approximation,The electric-dipole selection rule is $\Delta J=+1$ in absorption. It givesso adjacent lines are separated by $2B_0$. ThereforeAssign a line near $m\times 21.999\mathrm{GHz}$ to $J=m-1\to m$ for positive integer $m$. The first member is $0\to 1$, the next $1\to 2$, and so on; if the recorded window misses the lowest lines, their absolute frequencies fix the integer offset. At high temperature many rotational states are populated, so the strongest line need not be $0\to 1$.

This is the leading rotational pattern, not a claim that real ZnF lacks spin or hyperfine structure. Fine splittings and centrifugal distortion can perturb the ideal lines; they do not explain the three systematically displaced progressions as economically as vibrational excitation.

A nonlinear triatomic molecule has $3N-6=3$ vibrational normal modes. For water they are the symmetric O–H stretch, the H–O–H bend, and the antisymmetric O–H stretch, shown above. In the symmetric stretch both O–H lengths increase together; in the antisymmetric stretch one increases while the other decreases; the bend predominantly changes the bond angle. All three change a component of the electric dipole moment and are infrared-active. Every arrow reverses half a period later; motion of oxygen maintains the molecular centre of mass.

A diatomic molecule has $3N-5=1$ vibrational normal mode: the bond stretch. For hydrogen chloride, opposite longitudinal displacements change the electric dipole moment, so the mode has vibrational infrared activity. The following original sketches show one phase of this motion and the three water modes; arrow lengths are schematic, not mass-weighted amplitudes.

Let $Q_k$ be a vibrational normal mode coordinate with symmetry ${\Gamma }_k$. The harmonic vibrational ground state is totally symmetric; the one-quantum state transforms as $Q_k$. Integration over a symmetry-invariant domain annihilates any integrand lacking the totally symmetric representation. Thus a fundamental requiresfor some Cartesian component $\alpha$. For the one-dimensional irreducible representations used here, this means ${\Gamma }_k=\Gamma (x)$, $\Gamma (y)$ or $\Gamma (z)$. Equivalently, expanding the electric dipole moment givesthe derivative can be nonzero only if $Q_k$ and ${\mu }_{\alpha }$ have compatible symmetry. This is the symmetry criterion for vibrational infrared activity; it permits a transition without guaranteeing a large intensity.

The $1621\mathrm{c}{\mathrm{m}}^{-1}$ absorption is predominantly the carbon–carbon double-bond stretching vibration of the electron-rich vinyl ether. resonance donation from an oxygen lone pair changes both its frequency and its electric dipole moment variation during stretching. There is no carbonyl in the assigned structure, consistent with the absence of a carbon-13 NMR signal in the usual ketone or acyl chloride region.

Each site has a pi-electron population $4/3$, and each adjacent pi bond order is $2/3$. The two carbon sites supply one pi electron each and the nitrogen lone pair supplies two, giving four electrons in two occupied molecular orbitals. Orthogonality to the given Homo and LUMO fixes the remaining occupied normalized orbital, up to overall sign, asFor doubly occupied orbitals, the Hückel orbital charge density and Hückel pi bond order follow fromThe diagonal sums to four electrons. The negative $P_{13}$ is an off-diagonal density-matrix element between nonadjacent sites, not an additional directly connected chemical bond.

Relative to neutral pi contributions $(1,2,1)$, the pi-system charges are $(-e/3,+2e/3,-e/3)$: electron density has moved from nitrogen toward the two carbon sites. Thus the physical electric dipole moment vector points from the midpoint of the carbon pair toward nitrogen along the molecular symmetry bisector. The conventional chemical dipole arrow, with a crossed positive tail, points oppositely: from nitrogen toward the carbon pair. This is only the pi contribution, not a prediction of the full sigma-plus-pi molecular dipole.

An infrared fundamental must transform like a component of the electric dipole moment; a Raman spectroscopy fundamental must transform like a component of the symmetric polarizability tensor. For this centrosymmetric D4h molecular point group, the resulting classification is:

Here parallel and perpendicular refer to the principal fourfold axis. The table obeys infrared and Raman mutual exclusion.

Let $\Gamma$ be coverage in $\mathrm{mol}{\mathrm{m}}^{-2}$. Then $n_s=N_A\Gamma$ and the Helmholtz surface-dipole relation iswhere a voltage in volts has the same numerical value as the electron-energy change in electronvolts. Converting the slope from square centimeters to square meters gives $-5\times 10^5\mathrm{V}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$. ThereforeEach added molecule produces an effective outward surface dipole of about $2.2$ D, lowering the electron work function. This is the incremental adsorbate-plus-surface dipole, not necessarily the isolated molecular electric dipole moment.

Nitrogen-down adsorption places the positive hydrogen end farther from the metal, producing an outward molecular electric dipole moment and the observed decrease in work function. Its maximum normal component, however, is only $1.42$ D. Tilting alone can only reduce that component; it cannot explain an effective $2.20$ D.

The adsorbed molecule and metal therefore undergo appreciable electron density. Lone-pair donation into the metal, polarization of the substrate, image response and distortion of the adsorbate can change the effective interface dipole. The inferred increase relative to an upright isolated molecule is about $0.78$ D. For scale, transferring about $0.16$ elementary charges over $1$ Å would produce that dipole increment:This scale illustrates a plausible interfacial polarization; the work-function measurement alone does not uniquely determine a transferred charge or adsorption geometry.

A plausible explanation is partial dissociation leaving a residual surface species, such as an NH${}_x$ fragment, adsorbed nitrogen or hydrogen, after molecular ammonia has finished desorbing. A small residual population can maintain a surface electric dipole moment and a $0.2$ eV work function offset. Heating farther can remove, recombine or rearrange those species.

Test this hypothesis by recording temperature-programmed desorption signals for hydrogen and nitrogen as well as ammonia, and by measuring residual nitrogen with X-ray photoelectron spectroscopy. A surface-sensitive vibrational spectrum could distinguish NH${}_x$ groups from atomic nitrogen. Isotopically labelled ammonia would help assign desorption products. Correlating disappearance of the residual spectral signal with recovery of the clean-surface work function would be more diagnostic than the absence of an ammonia signal alone. This is a proposed explanation, not a uniquely established identification.

Use the indicated out-of-plane displacements and the equivalence of the four oxygen sites and of the two carbon sites in the isolated planar oxalate ion. In $V_1$, the two oxygen atoms on one side move oppositely to those on the other side, and the two carbon atoms also move oppositely. Their dipole changes cancel. The mode is even under molecular inversion symmetry, which independently forbids an electric-dipole transition.

In $V_2$, all four oxygen atoms move together out of the plane while both carbon atoms move in the opposite direction. The positive and negative charge distributions acquire a relative out-of-plane displacement, producing a nonzero derivative of the molecular electric dipole moment.

In $V_3$, two oxygen atoms move out and two move in, in a checkerboard arrangement, while the carbon atoms are stationary. The equivalent oxygen contributions cancel. This mode illustrates that being odd under inversion symmetry is necessary but not sufficient for vibrational infrared activity: its full symmetry does not transform like any component of an electric dipole moment.These conclusions concern the isolated-molecule displacements drawn in the source; a different crystal environment can alter selection rules through additional symmetry breaking.

Infrared spectroscopy measures absorption as the incident photon frequency is swept through molecular normal mode frequencies. In the electric-dipole approximation the interaction is $-\mathbf{p}\cdot \mathbf{E}$. Expanding the molecular electric dipole moment in a normal coordinate $Q$ givesThe fundamental one-quantum vibrational transition dipole moment is proportional to the derivative multiplying $Q$. Thusis the criterion for vibrational infrared activity; the derivative must also have a component along the incident polarization of an electromagnetic wave. A permanent molecular electric dipole moment is not necessary. Absorption occurs near $\hslash \omega =E_{v+1}-E_v$, with broadened lines reflecting finite lifetimes and the molecular environment.

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.