Molecular Hamiltonian

The molecular Hamiltonian is the energy operator for all electrons and nuclei included in a molecular model. In its standard nonrelativistic, field-free form, every particle has quantum kinetic energy and every charged pair interacts through the Coulomb potential. Bonds, equilibrium geometries, rotations, vibrations, dissociation channels, and spectra are consequences of this operator; none is inserted as a separate microscopic bond term.

The laboratory operator includes free motion of the whole molecule. Separating that motion gives the internal Hamiltonian; clamping the nuclei is an additional approximation. Born–Oppenheimer scale separation explains the electronic-channel reduction, and nonadiabatic coupling develops nuclear transfer between surfaces.

Required background. The molecular overview identifies the electronic and nuclear degrees of freedom. Atomic units, translation invariance, and fermionic states supply the units, center-of-mass symmetry, and exchange convention used below.

Particles in the Coulomb model

Consider:

• $N_e$ electrons, each with mass $m_e$, charge $-e$, laboratory position $\mathbf r_i$, and spin coordinate $\sigma_i$;
• $N_N$ nuclei, with masses $M_A$, charges $Z_Ae$, positions $\mathbf R_A$, and nuclear-spin coordinates $\eta_A$;
• no external field, no quantized radiation field, and no creation or destruction of particles;
• point charges interacting instantaneously through the Coulomb potential;
• nonrelativistic kinematics.

Define the pair distances

$$\begin{aligned} r_{ij} &= \lvert\mathbf r_i-\mathbf r_j\rvert, \\ R_{AB} &= \lvert\mathbf R_A-\mathbf R_B\rvert, \\ r_{iA} &= \lvert\mathbf r_i-\mathbf R_A\rvert. \end{aligned}$$

The distinction between an isotope and an element matters. $Z_A$ fixes the nuclear charge, whereas $M_A$ and the nuclear spin depend on the isotope. In the point-nucleus model, isotopologues share the same clamped-nuclei electronic problem at the same geometry, but have different all-particle Hamiltonians. Nuclear-size and other isotope-dependent corrections modify this leading statement.

Hilbert space and statistics

Before symmetry restrictions, a coordinate-space state belongs schematically to

$$\mathcal H_{\mathrm{lab}} = L^2\!\left( \mathbb R^{3(N_e+N_N)} \right) \otimes \mathcal H_{\mathrm{spin}}.$$

The physical subspace is smaller:

• the total electron state is antisymmetric under exchange of any two electrons;
• each set of identical nuclei has the bosonic or fermionic exchange symmetry appropriate to that isotope;
• particles of different isotopes, even when they have the same nuclear charge, are distinguishable particle species.

The spin-free Coulomb operator acts as the identity on spin coordinates. Spin is nevertheless essential because spatial and spin permutation symmetries must combine into the required total exchange symmetry. The canonical constructions are developed in Fermions and Spin and Spatial Wavefunctions.

Laboratory-Frame Hamiltonian

In SI units, the full operator is

$$\begin{aligned} H_{\mathrm{lab}} ={}& T_e+T_N +V_{eN}+V_{ee}+V_{NN}, \\ T_e ={}& -\sum_{i=1}^{N_e} \frac{\hbar^2}{2m_e}\nabla_{\mathbf r_i}^2, \\ T_N ={}& -\sum_{A=1}^{N_N} \frac{\hbar^2}{2M_A}\nabla_{\mathbf R_A}^2, \\ V_{eN} ={}& -\frac{e^2}{4\pi\epsilon_0} \sum_{i,A}\frac{Z_A}{r_{iA}}, \\ V_{ee} ={}& \frac{e^2}{4\pi\epsilon_0} \sum_{i<j}\frac{1}{r_{ij}}, \\ V_{NN} ={}& \frac{e^2}{4\pi\epsilon_0} \sum_{A<B}\frac{Z_AZ_B}{R_{AB}}. \end{aligned}$$

Every coordinate in this expression is dynamical. In particular, $V_{NN}$ is an operator on nuclear coordinates, not a geometry-dependent constant, until a clamped-nuclei problem is introduced.

Atomic-unit form

Hartree atomic units set

$$\hbar=m_e=e=4\pi\epsilon_0=1.$$

When each $M_A$ is then expressed in electron-mass units,

$$H_{\mathrm{lab}} ={} -\frac12\sum_i\nabla_{\mathbf r_i}^2 -\sum_A\frac{1}{2M_A}\nabla_{\mathbf R_A}^2 - \sum_{i,A}\frac{Z_A}{r_{iA}} + \sum_{i<j}\frac{1}{r_{ij}} + \sum_{A<B}\frac{Z_AZ_B}{R_{AB}}.$$

Atomic units remove constants, not physics. The small coefficients $1/M_A$ remain and carry the electron–nucleus mass hierarchy. See Atomic Units and Scales for the AMO interpretation and Atomic Units for conversion factors.

What the five terms do

The electronic kinetic energy penalizes rapid spatial variation. Nuclear attraction favors localization, while electron repulsion and antisymmetry constrain how that localization is shared. The operator is a sum of one- and two-particle terms, but its eigenstates need not be products of orbitals, and orbital eigenvalues do not generally add to the total energy.

The pair restrictions $i<j$ and $A<B$ count each unordered pair once. An equivalent electron sum is $\frac12\sum_{i\ne j}1/r_{ij}$; combining $1/2$ with $i<j$ would undercount it. Exchange follows from the antisymmetric state, rather than from an additional force in the Hamiltonian.

Nuclear kinetic energy remains present despite its small coefficients. It produces zero-point motion, rotation, tunneling, isotope shifts, and nonadiabatic transfer. Body-fixed coordinates can reorganize it into vibrational and rotational parts, with metric factors and Coriolis couplings. Exact removal of translation alone does not justify an independent rigid rotor and harmonic oscillator.

Nuclear repulsion is coordinate dependent in the all-particle problem. It becomes a scalar only in an electronic eigenproblem at fixed geometry. Whether that scalar is included in the electronic energy or added afterward must remain consistent when computing forces and dissociation energies.

The electron–nucleus coupling makes an exact molecular wavefunction generally inseparable into independent electronic and nuclear factors. Coulomb coalescences also produce cusp conditions: appropriate kinetic and potential singularities cancel when the operator acts on the exact state. These are local constraints, not an indication that the bound-state energy diverges.

The Coulomb Operator Is Well Defined

The symbols $1/r_{ab}$ are singular, but the many-particle Coulomb Hamiltonian is not merely a formal expression. On an appropriate dense domain it defines a self-adjoint operator bounded from below. Self-adjointness supplies unitary time evolution and a real spectral problem.

For practical work, three consequences matter:

1. A basis should represent coalescence behavior well enough for the target observable.
2. Individual kinetic and potential expectation values can converge more slowly than their sum.
3. Numerical regularization must converge back to the same operator; changing a Coulomb core is a model change, not just a harmless computational trick.

The distinction between a symmetric differential expression and its self-adjoint realization is reviewed in Hermitian vs Self-Adjoint Operators.

Exact Symmetries

The field-free Coulomb Hamiltonian depends only on pair distances. Its exact symmetries organize the calculation before any molecular shape is chosen.

Translation

The total momentum

$$\mathbf P = \sum_i\mathbf p_i + \sum_A\mathbf P_A$$

commutes with the laboratory Hamiltonian:

$$[H_{\mathrm{lab}},\mathbf P]=0.$$

This is the symmetry behind exact center-of-mass separation. Translation-Invariant Hamiltonians gives the general commutator test.

Rotation

For the spin-free operator, the total orbital angular momentum

$$\mathbf L = \sum_i \mathbf r_i\times\mathbf p_i + \sum_A \mathbf R_A\times\mathbf P_A$$

is conserved:

$$[H_{\mathrm{lab}},\mathbf L]=0.$$

Spin-independent dynamics has additional independent spin symmetry. Once spin–orbit, spin–spin, or hyperfine terms are added, orbital and spin angular momenta need not be conserved separately; the appropriate total angular momentum becomes central. See Total Angular Momentum.

Parity and time reversal

Simultaneous inversion of every spatial coordinate leaves all pair distances unchanged. The field-free Hamiltonian therefore commutes with parity. With no magnetic field and no explicitly time-reversal-breaking term, it is also time-reversal invariant. These statements constrain states and transitions but do not imply that a body-fixed geometry has inversion symmetry.

Particle permutations

The operator commutes with permutations within each identical-particle species:

$$[H_{\mathrm{lab}},P_{ij}^{(e)}]=0, \qquad [H_{\mathrm{lab}},P_{AB}^{(N)}]=0$$

when nuclei $A$ and $B$ have the same charge, mass, spin species, and modeled internal structure. The physical state must occupy the correct permutation-symmetry sector.

Point groups are reduced symmetries

A point group describes operations preserving a selected nuclear geometry in a clamped-nuclei problem. It is not generally the full symmetry group of the unconstrained all-particle Hamiltonian. Exact molecular spectroscopy instead combines total rotation, parity, nuclear permutations, and inversion operations; point-group labels emerge within a chosen structural reduction.

Exact Center-of-Mass Separation

It is useful to treat all $K=N_e+N_N$ particles uniformly. Let particle $a$ have mass $m_a$, position $\mathbf x_a$, and momentum $\mathbf p_a$. Define

$$M_{\mathrm{tot}} = \sum_{a=1}^{K}m_a,$$ $$\mathbf R_{\mathrm{CM}} = \frac{1}{M_{\mathrm{tot}}} \sum_{a=1}^{K}m_a\mathbf x_a,$$

and

$$\mathbf P_{\mathrm{CM}} = \sum_{a=1}^{K}\mathbf p_a.$$

Because the Coulomb potential depends only on differences $\mathbf x_a-\mathbf x_b$, a linear canonical transformation gives

$$H_{\mathrm{lab}} = \frac{\mathbf P_{\mathrm{CM}}^2} {2M_{\mathrm{tot}}} + H_{\mathrm{int}}.$$

The first term describes free translation of the entire molecule. The second contains all internal spectra, binding, and structure.

Jacobi coordinates

One coordinate choice makes the internal kinetic energy diagonal. Define the partial masses

$$M_k=\sum_{a=1}^{k}m_a$$

and, for $k=1,\ldots,K-1$, the Jacobi vectors

$$\boldsymbol\xi_k = \mathbf x_{k+1} - \frac{1}{M_k} \sum_{a=1}^{k}m_a\mathbf x_a.$$

The associated reduced masses are

$$\mu_k = \frac{m_{k+1}M_k} {m_{k+1}+M_k}.$$

With conjugate internal momenta $\boldsymbol\pi_k$,

$$T_{\mathrm{lab}} = \frac{\mathbf P_{\mathrm{CM}}^2} {2M_{\mathrm{tot}}} + \sum_{k=1}^{K-1} \frac{\boldsymbol\pi_k^2}{2\mu_k}.$$

Jacobi coordinates diagonalize the mass metric, but a particular particle-pair distance may become a linear combination of several $\boldsymbol\xi_k$. Computational convenience therefore determines whether diagonal kinetic energy or simple pair coordinates are preferable.

Factorization of translational motion

For a total-momentum generalized eigenstate, the laboratory wavefunction can be written

$$\Psi_{\mathbf P} \left( \mathbf R_{\mathrm{CM}},\xi \right) = \exp\!\left( \frac{i\mathbf P\cdot\mathbf R_{\mathrm{CM}}}{\hbar} \right) \psi_{\mathrm{int}}(\xi),$$

with energy

$$E_{\mathrm{lab}} = \frac{P^2}{2M_{\mathrm{tot}}} + E_{\mathrm{int}}.$$

The plane wave is not square-integrable on all space; it has the usual continuum normalization. A localized molecular beam is a wavepacket in $\mathbf P$. Internal bound states can be normalizable even though the full free-molecule spectrum is continuous because of overall translation.

Reduced Mass and Mass Polarization

Jacobi coordinates are not always the most intuitive. Choose one particle, labeled $0$, as a reference and define

$$\mathbf x_i = \mathbf X_i-\mathbf X_0, \qquad i=1,\ldots,K-1.$$

After the center of mass is removed, the internal Hamiltonian takes the form

$$H_{\mathrm{int}} ={} -\sum_{i=1}^{K-1} \frac{\hbar^2}{2\mu_i} \nabla_{\mathbf x_i}^2 - \frac{\hbar^2}{m_0} \sum_{i<j} \nabla_{\mathbf x_i} \mathbin{\cdot} \nabla_{\mathbf x_j} + V(\{\mathbf x_i\}),$$

where

$$\mu_i = \frac{m_0m_i}{m_0+m_i}.$$

The cross-derivative term is called a mass-polarization, recoil, or specific-mass term, depending on context. It records the fact that motion of several particles relative to the same finite-mass reference cannot be independent.

Two warnings are important:

• reduced masses and the cross term are both finite-mass effects;
• mass polarization is coordinate-generated kinetic coupling, not an extra force.

In the formal limit $m_0\to\infty$,

$$\mu_i\to m_i, \qquad \frac{1}{m_0}\to0,$$

so the reference particle becomes immobile and the cross derivatives vanish.

Why the cross term appears

In the center-of-mass frame,

$$\mathbf p_0 = -\sum_{i=1}^{K-1}\mathbf p_i.$$

Hence the reference-particle kinetic energy contributes

$$\frac{\mathbf p_0^2}{2m_0} = \frac{1}{2m_0} \left( \sum_i\mathbf p_i \right)^2.$$

Expanding the square gives

$$\frac{\mathbf p_0^2}{2m_0} = \sum_i\frac{\mathbf p_i^2}{2m_0} + \frac{1}{m_0} \sum_{i<j} \mathbf p_i\mathbin{\cdot}\mathbf p_j.$$

In the center-of-mass frame these $\mathbf p_i$ are also the momenta conjugate to the reference-particle relative coordinates. The diagonal pieces combine with $\mathbf p_i^2/(2m_i)$ to produce $\mu_i$; the off-diagonal pieces become the cross gradients.

Two particles as the clean limit

For two particles there is one internal vector

$$\mathbf r=\mathbf x_2-\mathbf x_1$$

and no cross term. The exact separation is

$$H = \frac{\mathbf P_{\mathrm{CM}}^2}{2M} + \frac{\mathbf p^2}{2\mu} + V(r),$$

where

$$M=m_1+m_2, \qquad \mu=\frac{m_1m_2}{m_1+m_2}.$$

The reduced mass is therefore not a Born–Oppenheimer correction. It is the exact internal mass for a two-body translation-invariant problem.

Diatomic Internal Coordinates

For nuclei $A$ and $B$, the internuclear vector

$$\mathbf R=\mathbf R_B-\mathbf R_A$$

is a natural internal coordinate, with bare nuclear reduced mass

$$\mu_N = \frac{M_AM_B}{M_A+M_B}.$$

The corresponding relative kinetic piece has the form

$$T_R = -\frac{\hbar^2}{2\mu_N} \nabla_{\mathbf R}^2.$$

This does not by itself separate nuclear motion from electronic motion. Electron coordinates must be defined relative to a compatible origin, and finite-mass cross terms depend on that choice. Decomposing $\nabla_{\mathbf R}^2$ into radial and angular parts is exact, but interpreting those pieces as an isolated vibrator and rigid rotor requires a potential-surface and body-fixed reduction.

The Clamped-Nuclei Electronic Hamiltonian

The standard electronic operator at a specified nuclear geometry $R$ is, in atomic units,

$$H_e(R) ={} -\frac12\sum_i\nabla_i^2 -\sum_{i,A}\frac{Z_A}{r_{iA}} + \sum_{i<j}\frac{1}{r_{ij}} + \sum_{A<B}\frac{Z_AZ_B}{R_{AB}}.$$

It acts only on electronic variables:

$$H_e(R)\phi_\alpha(r;R) = E_\alpha(R)\phi_\alpha(r;R).$$

Here $R$ labels parameters rather than dynamical arguments of the electronic eigenfunction. The last term is therefore a geometry-dependent scalar.

An equally common convention defines

$$H_{\mathrm{el}}(R) = H_e(R)-V_{NN}(R)$$

and then forms the surface

$$U_\alpha(R) = \mathcal E_\alpha(R)+V_{NN}(R).$$

The two conventions yield the same $U_\alpha$ when used consistently.

What has changed

Passing from $H_{\mathrm{int}}$ to $H_e(R)$ is not merely removal of the center of mass. It additionally:

1. treats nuclear coordinates as parameters in an auxiliary eigenproblem;
2. omits nuclear kinetic action during that electronic diagonalization;
3. usually neglects finite-mass terms in the electronic operator at leading order;
4. restores nuclear quantum motion only in a subsequent channel or surface equation.

The exact electronic-channel expansion can still retain all surfaces and derivative couplings. The Born–Oppenheimer approximation begins when that coupled problem is truncated or its couplings are neglected. Born–Oppenheimer in Molecules follows this reduction through equilibrium structure, isotope-dependent nuclear motion, rovibrational states, and spectroscopic corrections.

Worked Hamiltonians

The hydrogen molecular ion

For $\mathrm H_2^+$ there are two protons and one electron. In atomic units, before center-of-mass removal,

$$H_{\mathrm H_2^+} ={} -\frac12\nabla_{\mathbf r}^2 -\frac{1}{2M_p}\nabla_A^2 -\frac{1}{2M_p}\nabla_B^2 - \frac{1}{r_A} - \frac{1}{r_B} + \frac{1}{R}.$$

At clamped internuclear separation $R$, place the nuclei at $\pm\mathbf R/2$. The electronic problem is

$$H_e(R) = -\frac12\nabla_{\mathbf r}^2 -\frac{1}{r_A} -\frac{1}{r_B} +\frac{1}{R}.$$

Solving this equation for many values of $R$ produces an electronic potential-energy curve. Quantized vibration and rotation are not yet included. H₂⁺ develops the two-center solution and compares minimal LCAO with accurate Born–Oppenheimer energies.

The hydrogen molecule

For $\mathrm H_2$, two electrons add one electron–electron pair:

$$H_{\mathrm H_2} ={} -\frac12 \left( \nabla_1^2+\nabla_2^2 \right) -\frac{1}{2M_p} \left( \nabla_A^2+\nabla_B^2 \right) - \sum_{i=1}^{2} \left( \frac{1}{r_{iA}}+\frac{1}{r_{iB}} \right) + \frac{1}{r_{12}} +\frac{1}{R}.$$

This four-particle operator already contains covalent binding, exchange symmetry, electron correlation, nuclear motion, and dissociation. A molecular-orbital or valence-bond wavefunction is an approximation to states of this operator or of a controlled reduction of it. Hydrogen Molecule carries the clamped-nuclei reduction into the canonical comparison of Heitler–London, restricted Hartree–Fock, and correlated dissociation.

High-precision calculations may solve the four-body nonrelativistic problem directly and then add relativistic, radiative, and finite-size corrections. That route is conceptually distinct from first constructing one potential-energy surface, though both can reach exceptional accuracy for light molecules when the necessary corrections are included consistently.

Energy Zeros and Dissociation Thresholds

Adding a constant to a Hamiltonian does not change dynamics, but molecular binding requires a common energy convention.

For a channel in which the system separates into fragments $F_1,\ldots,F_s$, define the threshold from fragment energies computed with:

• the same particle masses;
• the same relativistic or nonrelativistic model;
• the same external-field assumptions;
• compatible internal quantum numbers;
• the same energy zero.

Then a dissociation energy can be written

$$D = E_{\mathrm{threshold}} - E_{\mathrm{molecule}}.$$

A positive $D$ indicates binding relative to that channel. The “lowest threshold” must be understood within the conserved symmetry sector; a lower-energy fragmentation channel forbidden by exact quantum numbers does not automatically determine the decay width of the state under consideration.

At infinite separation, neutral fragments can have vanishing leading Coulomb interaction while retaining dispersion and multipolar tails. Ionic fragments can retain a $1/R$ Coulomb term. Therefore “set $V=0$ at infinity” is a convention that must be matched to the actual asymptotic channel.

Energy and Length Scales

Let

$$\mu=\frac{m_e}{M}$$

for a representative nuclear mass. Near a smooth, nondegenerate molecular minimum, it is useful to define

$$\kappa=\mu^{1/4}.$$

The Born–Oppenheimer hierarchy then has the schematic orders:

Sector	Typical order in electronic atomic units
Electronic structure	$E_{\mathrm{el}}\sim E_h$ and lengths $\sim a_0$
Nuclear displacement	$\Delta R/a_0\sim\kappa$ near a regular minimum
Vibrational spacing	$E_{\mathrm{vib}}/E_h\sim\kappa^2=\mu^{1/2}$
Rotational spacing	$E_{\mathrm{rot}}/E_h\sim\kappa^4=\mu$
Regular recoil terms	often proportional to one or more powers of $\mu$

These are ordering estimates, not universal numerical formulas. Force constants, equilibrium bond lengths, electronic gaps, symmetry, anharmonicity, and proximity to dissociation can alter coefficients or invalidate the regular expansion.

For a proton,

$$\frac{m_e}{M_p} \approx \frac{1}{1836},$$

so

$$\mu^{1/2}\approx2.3\times10^{-2}, \qquad \mu\approx5.4\times10^{-4}.$$

This explains why vibrational spacings are often much smaller than electronic spacings and rotational spacings smaller still. It does not prove that one electronic surface is adequate: small electronic gaps can overwhelm the mass hierarchy.

Beyond the leading Coulomb model

Point charges and $p^2/(2m)$ kinetics omit finite nuclear size, magnetic moments, relativistic interactions, and radiative effects. Spins still determine exchange symmetry, but fine, hyperfine, and spin–rotation splittings require additional terms. These corrections become essential for high precision and heavy elements.

Ionization and dissociation are continuum channels of the same fixed-particle Hamiltonian. Spontaneous emission requires a quantized radiation field; stimulated dynamics may instead use a prescribed classical field. Solvents, substrates, and collisions require additional particles or an open-system model.

External Fields

For particles with charges $q_a$, minimal coupling replaces

$$\mathbf p_a \longrightarrow \mathbf p_a-q_a\mathbf A(\mathbf x_a,t)$$

and adds scalar-potential energy $q_a\Phi$. The kinetic term becomes

$$T = \sum_a \frac{ \left[ \mathbf p_a-q_a\mathbf A(\mathbf x_a,t) \right]^2 }{2m_a}.$$

Spin couplings must be added consistently when relevant.

An electric field varying negligibly across a neutral molecule can often be expressed through an internal multipole interaction after the center of mass is handled. A magnetic field is subtler because the vector potential depends on position and ordinary canonical momentum is gauge dependent. Center-of-mass and internal motion need not separate in the field-free way; conserved pseudomomentum and the total charge become important. Magnetic Translations develops the associated symmetry.

The field-free formula

$$H_{\mathrm{lab}} = \frac{P_{\mathrm{CM}}^2}{2M_{\mathrm{tot}}} +H_{\mathrm{int}}$$

should therefore not be imported unchanged into a charged molecule in a magnetic field.

Exercises

Exercise 1: Three-particle Jacobi masses

For masses $m_1,m_2,m_3$, choose

$$\boldsymbol\xi_1 = \mathbf x_2-\mathbf x_1$$

and

$$\boldsymbol\xi_2 = \mathbf x_3 - \frac{m_1\mathbf x_1+m_2\mathbf x_2} {m_1+m_2}.$$

Derive the two Jacobi reduced masses by completing squares in the classical kinetic energy; the same linear coordinate transformation determines the quantum kinetic operator.

Solution

Write $\mathbf v_{12}=(m_1\mathbf v_1+m_2\mathbf v_2)/(m_1+m_2)$. Completing the square gives

$$\frac{m_1v_1^2+m_2v_2^2}{2} =\frac{m_1+m_2}{2}v_{12}^2 +\frac{m_1m_2}{2(m_1+m_2)} |\mathbf v_2-\mathbf v_1|^2.$$

The coefficient of $\dot{\boldsymbol\xi}_1^2/2$ is therefore

$$\mu_1 = \frac{m_1m_2}{m_1+m_2}.$$

The second measures particle $3$ relative to a cluster of mass $m_1+m_2$, so

$$\mu_2 = \frac{m_3(m_1+m_2)} {m_1+m_2+m_3}.$$

Together with $\mathbf R_{\mathrm{CM}}$, these coordinates diagonalize the free kinetic energy:

$$T = \frac{\mathbf P_{\mathrm{CM}}^2} {2(m_1+m_2+m_3)} + \frac{\boldsymbol\pi_1^2}{2\mu_1} + \frac{\boldsymbol\pi_2^2}{2\mu_2}.$$

Exercise 2: Clamp the hydrogen molecular ion

For clamped $\mathrm H_2^+$, identify which terms in the full three-particle Hamiltonian disappear from the electronic eigenproblem and which nuclear term remains as a scalar.

Solution

Both proton kinetic terms are omitted while solving the fixed-$R$ electronic problem:

$$-\frac{1}{2M_p}\nabla_A^2 -\frac{1}{2M_p}\nabla_B^2 \longrightarrow0.$$

The electron kinetic energy and both electron–proton attractions remain. The proton–proton repulsion

$$\frac{1}{R}$$

remains if the convention includes $V_{NN}$ in $H_e(R)$, but it is now a scalar parameter rather than a nuclear-coordinate operator. Nuclear quantum motion must be restored in a later equation.

Exercise 3: Point group or exact symmetry?

A bent triatomic molecule has a clamped equilibrium geometry with point group $C_{2v}$. Is $C_{2v}$ the complete exact symmetry of the freely translating and rotating all-particle Hamiltonian? Explain.

Solution

No. $C_{2v}$ classifies an electronic or rovibrational reduction relative to the selected geometry. The all-particle field-free Coulomb Hamiltonian is invariant under continuous common translations and rotations, parity, time reversal, and permutations of identical particles. A freely rotating stationary state is not fixed at one laboratory orientation. The point group remains extremely useful, but it is a reduced structural symmetry rather than the complete exact group.

Exercise 4: Isotope scaling

In the leading smooth-minimum hierarchy, estimate how vibrational and rotational spacings change when all nuclear masses double while the electronic surface is held fixed.

Solution

The schematic scalings are

$$E_{\mathrm{vib}}\propto M^{-1/2}, \qquad E_{\mathrm{rot}}\propto M^{-1}.$$

Thus

$$\frac{E_{\mathrm{vib}}(2M)} {E_{\mathrm{vib}}(M)} = \frac{1}{\sqrt2},$$

and

$$\frac{E_{\mathrm{rot}}(2M)} {E_{\mathrm{rot}}(M)} = \frac12.$$

Real isotopic shifts also contain adiabatic, nonadiabatic, geometric, and small nuclear-size effects, so this is the leading fixed-surface estimate.

Exercise 5: Consistent dissociation energy

Suppose a molecular energy includes finite nuclear mass but its separated-atom threshold was computed with infinitely heavy nuclei. Why is their difference not a controlled dissociation energy?

Solution

The two numbers are eigenvalues of different Hamiltonians. Their difference contains the desired binding energy plus an unmatched recoil contribution. One must compute molecule and fragments with the same isotope masses and physical corrections, then subtract:

$$D = E_{\mathrm{fragments}}^{\mathrm{same\ model}} - E_{\mathrm{molecule}}^{\mathrm{same\ model}}.$$

The same consistency requirement applies to relativistic, radiative, field, and finite-size corrections.

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