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The diatomic molecule containing one hydrogen atom and one iodine atom. Its measured spectroscopic wavenumbers include anharmonic corrections beyond the two-term Morse oscillator model.^[ref]

By Klaus P. Huber and Gerhard H. Herzberg. NIST Chemistry WebBook compilation of diatomic constants.

Cited row: ground $X{}^1{\Sigma }^{+}$ state of hydrogen iodide, ${\omega }_e=2309.014\mathrm{c}{\mathrm{m}}^{-1}$ and ${\omega }_e{x}_e=39.6435\mathrm{c}{\mathrm{m}}^{-1}$. Higher anharmonic coefficients are also tabulated, so the two-constant Morse oscillator extrapolation should not be confused with a measurement of the actual dissociation threshold.

The two extra progressions are rotational transitions within thermally populated $v=1$ and $v=2$ vibrational levels. Isotopic purity rules out attributing them to different isotope masses. Their rotational constants areThe Boltzmann distribution makes each higher vibrational manifold less populated, so its rotational spectrum is weaker. A Morse oscillator explores larger separations as it is excited: the bond's vibrationally averaged moment of inertia increases and the rotational spacing decreases. More precisely, $B_v$ involves an average of the inverse square separation; it is not simply the inverse square of the mean separation.

The observed decrement is constant to the given precision:This fits the leading vibration-rotation interaction law $B_v=B_e-{\alpha }_e(v+\frac{1}{2})$.

The strong band is $0\to 1$, while the weak band is the first vibrational overtone $0\to 2$. Let $w={\tilde{\omega }}_e$ and $q={\tilde{\omega }}_e{x}_e$. Subtracting the Morse oscillator levels givesTwice the fundamental is $881.2\mathrm{c}{\mathrm{m}}^{-1}$, slightly above the overtone because the higher vibrational gap is smaller. Eliminate $w$:soThe implied dimensionless anharmonicity constant is $x_e\simeq 0.004164$. The overtone's weakness follows from its being forbidden in the leading harmonic/linear-dipole model, rather than a lack of population in its lower state.

For one mole, $Nk=R$, so the three-level Morse oscillator result isFor the full quantum harmonic oscillator with the given wavenumber,The specified anharmonic calculation is about $1.7\%$ higher. Anharmonicity lowers the first excitation spacings and increases excited-state occupation. Both results remain small because $100\mathrm{K}$ is well below the vibrational temperature. The comparison also mixes a three-level anharmonic truncation with an infinite harmonic sum; omitted higher Morse levels would make a further small positive correction at this temperature.

Subtracting the $v=0$ energy in the Morse oscillator givesThe first three relative wavenumbers are $0$, $213$, and $424{\mathrm{cm}}^{-1}$. Thus, with ${\theta }_1=213/0.695\mathrm{K}$ and ${\theta }_2=424/0.695\mathrm{K}$,This is a three-level partition function, as stipulated, rather than an unrestricted sum of the anharmonic formula to arbitrarily large $v$.

Write the upper Morse oscillator term values in wavenumber units asThe spacing is $G(v+1)-G(v)={\tilde{\omega }}_e-2a(v+1)$. A Birge-Sponer plot therefore falls linearly in this model and can be extrapolated to zero spacing. If its horizontal coordinate is $x=v+1/2$, the plotted line is $y=({\tilde{\omega }}_e-a)-2ax$. Its area from $x=0$ to its zero givesEquivalently a plot against $v+1$ has intercept ${\tilde{\omega }}_e$ and triangle area ${\tilde{D}}_e={\tilde{\omega }}_e^2/(4a)$; subtracting $G(0)={\tilde{\omega }}_e/2-a/4$ recovers ${\tilde{D}}_0$. These conventions distinguish dissociation from the potential minimum and from the lowest vibrational level.

Measured consecutive members fix a local slope and intercept, so unobserved members can be supplied by extrapolation under an explicit spacing law. The upper-state limit on the absorption scale is ${\tilde{\nu }}_{\lim }={\tilde{\nu }}_{00}+{\tilde{D}}_0$. A local high-v fit can likewise estimate the remaining energy above an observed member without knowing its absolute v index. Near dissociation a real potential need not remain Morse-like, so an extrapolated triangle has model uncertainty; missing lines must not be silently assumed consecutive.

Let $u=r-r_e$ measure displacement from the equilibrium bond length and put the minimum of potential energy at zero. The harmonic model and the Morse potential areThe harmonic approximation is a symmetric parabola and has no dissociation limit. The Morse potential has a steep compressed side and a shallower extended side, approaching $D_e$ as $r\to \infty$. Matching the curvatures at the minimum gives $k=2D_e{\alpha }^2$. With $\tilde{\omega }$ denoting a spectroscopic wavenumber rather than an angular frequency, the energy levels areHere $x_e$ is the dimensionless spectroscopic anharmonicity constant. The harmonic spacing is constant, while the Morse oscillator spacing is $E_{v+1}-E_v=hc\tilde{\omega }[1-2x_e(v+1)]$ and decreases as the levels approach dissociation. Only finitely many Morse oscillator levels are bound. The exactly solvable potential was introduced to model this anharmonic behaviour.^[ref]

For the harmonic model put $q=\sqrt{\mu \omega /\hslash }u$, with $\omega =2\pi c\tilde{\omega }$, and $N={\pi }^{-1/4}$. The first three normalized reduced-coordinate wavefunctions areTheir numbers of nodes are zero, one and two, and their parities are even, odd and even. The $v=2$ nodes are at $q=\pm 1/\sqrt{2}$. The diagrams below use analytic curves and arbitrary illustrative scales, with both potentials given the same local curvature.

Write $s=v+1/2$ and complete the square in the Morse oscillator energy:The finite bound-level sequence approaches the dissociation limit at the vertex of this extrapolated parabola. Thus the bond potential-well depth isThe continuous vertex is a convenient way to identify the limit; it is not an additional normalizable bound level at the threshold.

The ground-state dissociation energy starts at the lowest vibrational level, rather than the minimum of the potential. The zero-point energy of the Morse oscillator isSubtracting it from the bond potential-well depth givesThis distinction matters whenever a dissociation energy is inferred from a vibrational spectrum.

The molecule with the larger inferred dissociation energy is assigned the stronger hydrogen–halogen bond: $\mathrm{A}=\mathrm{HBr}$ and $\mathrm{B}=\mathrm{HI}$. Moving from bromine to the larger iodine atom generally lengthens the bond and reduces effective overlap with the compact hydrogen $1s$ orbital, so hydrogen iodide is more weakly bound. Independently, the tabulated ground-state harmonic constants near $2649\mathrm{c}{\mathrm{m}}^{-1}$ for hydrogen bromide and $2309\mathrm{c}{\mathrm{m}}^{-1}$ for hydrogen iodide agree with the inferred assignment.^[ref][ref] The ordering is the identification criterion; the low-level Morse oscillator extrapolations above should not be presented as measured dissociation energies.

Let $f$ and $o$ denote the measured fundamental and first-overtone wavenumbers, and put $y=\tilde{\omega }{x}_e$. Differences of the Morse oscillator levels giveConsequently $x_e=y/\tilde{\omega }$, $D_e/(hc)={\tilde{\omega }}^2/(4y)$ and $D_0/(hc)=D_e/(hc)-(\tilde{\omega }/2-y/4)$. Substituting the two pairs of measured bands gives the following derived quantities:The conversion is $1\mathrm{c}{\mathrm{m}}^{-1}=0.0119626566\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, after multiplication by $N_{A}hc$ and consistent length units. Therefore the requested ground-state dissociation energies are $D_0(\mathrm{A})=448.4\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and $D_0(\mathrm{B})=388.5\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The corresponding potential-well depths are $D_e(\mathrm{A})=464.1$ and $D_e(\mathrm{B})=402.2\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are extrapolations under the stipulated Morse oscillator model; two low-lying bands do not establish the accuracy of the real potential near dissociation.

True for the finite bound-level Morse model with nonzero adjacent control couplings. The Morse oscillator energies have the formOver the physical bound-level range these adjacent gaps are positive, and $B\ne 0$ makes them pairwise distinct. Label the retained levels consecutively. With $H_1={\sum }_j{d}_j(E_{j,j+1}+E_{j+1,j})$, all $d_j\ne 0$, the squared adjoint action of $i{H}_0$ has eigenvalue $-(E_{j+1}-E_j{)}^2$ on each adjacent skew-Hermitian coupling. A polynomial chosen by Lagrange interpolation can therefore isolate any one adjacent coupling from $i{H}_1$.

Commuting an isolated coupling once with the drift produces its independent antisymmetric quadrature. The bracket between the two quadratures yields a traceless diagonal generator; brackets along the connected chain produce every nonadjacent off-diagonal generator. Thus connected nondegenerate transition chain generates special unitary control proves $\mathfrak{su}(N)\subseteq \mathfrak{g}$, which gives density-state controllability. If the physical drift has nonzero trace, subtracting its traceless part also produces the identity direction, giving $\mathfrak{u}(N)$ and full operator controllability. Shifting the energy origin may remove that extra phase direction but does not change state controllability. The assertion concerns the finite-level model in the question, not the entire bound-plus-dissociation Hilbert space of the untruncated Morse potential.

In the quadratic vibrational term expression $E_v/(hc)={\tilde{\omega }}_e(v+1/2)-{\tilde{\omega }}_e{x}_e(v+1/2{)}^2$, the constant $x_e$ is dimensionless, while ${\tilde{\omega }}_e{x}_e$ has units of spectroscopic wavenumber. A positive value makes adjacent level spacings decrease. For an exact Morse oscillator, $D_e/(hc)={\tilde{\omega }}_e/(4x_e)$. Using only two low-lying transitions to extrapolate this identity to a real molecule assumes the higher-order anharmonic terms are negligible all the way to dissociation.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.