Physical Chemistry · Part 4 of 9

Exactly Solvable Systems II — Oscillator & Rotor

Quantum Chemistry, Part 4 · 9 sections · about 22,652 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

Two systems carry most of molecular spectroscopy: the oscillator behind every vibrational band, and the rotor behind every microwave line. This part solves both — the oscillator twice, once by series and once by the far more elegant ladder-operator method — and then builds the general theory of angular momentum that both depend on, including spin, which has no classical analogue at all. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 4

1The classical oscillator, the parabolic approximation, and the Schrödinger equation Free below
2The series solution, Hermite polynomials, the energy ladder and the wavefunctions
3Ladder operators, selection rules and anharmonicity
4Rotation in two dimensions: the particle on a ring
5Rotation in three dimensions: the particle on a sphere and the spherical harmonics
6The rigid rotor as a molecule: B, the spectrum, and centrifugal distortion
7The operators and their commutation relations
8The eigenvalue spectrum from the algebra alone, and the vector model
9Spin, and the addition of angular momenta

The classical oscillator, the parabolic approximation, and the Schrödinger equation

Free extract

Section D.1 of Part 4, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

Where the parabola comes from, what the reduced mass is doing there, and how the equation is put into a form with no constants left in it.

Start classically, because the classical problem is the one being replaced and its language survives in the answer. A mass m attached to a spring obeying Hooke's law feels a restoring force F = −kx proportional to the displacement x from equilibrium and directed back towards it. The constant k is the force constant, measured in N m⁻¹; a large k means a stiff spring.

F = −kx ⇒ V(x) = −∫F dx = ½kx²

Newton's second law gives m d²x/dt² = −kx, whose solution is x(t) = A sin(ωt + φ) with ω = √(k/m). The classical oscillator therefore has three properties worth holding on to, because two of them survive into quantum mechanics and one is destroyed by it:

The frequency is independent of the amplitude. ω = √(k/m) contains no A. A violin string sounds the same note whether plucked hard or softly. This survives: the quantum level spacing ℏω also does not depend on which levels you look at.
The energy is ½kA² and can take any value, since A can be anything. This does not survive. Quantum mechanically only the values (n + ½)ℏω are allowed.
The particle is confined between the turning points ±A, where all its energy is potential and it is momentarily at rest. This half-survives: the quantum particle is mostly inside the turning points but not entirely, as D.2 shows.

Why a chemist cares: the Taylor expansion of a bond

A real chemical bond does not obey Hooke's law. Its potential energy curve rises steeply when the atoms are pushed together (nuclear repulsion) and flattens out to a constant when they are pulled apart (the molecule dissociates). Nothing about that shape is parabolic. So why is the harmonic oscillator useful at all?

Because of one line of calculus. Expand the true potential V(r) in a Taylor series about the equilibrium separation r_e:

V(r) = V(r_e) + (dV/dr)_{r_e}(r − r_e) + ½(d²V/dr²)_{r_e}(r − r_e)² + …

Now dispose of the first two terms. V(r_e) is a constant, and the zero of energy is ours to choose, so set it to zero. The first derivative (dV/dr) is zero at a minimum — that is what a minimum means. So the first term that survives is the quadratic one, and comparing with V = ½kx² identifies the force constant:

k = (d²V/dr²)_{r = r_e} — the curvature of the potential at the bottom of the well

This is the whole justification, and it also tells you the limits. The parabola is the leading term of an expansion about the minimum, so it is good near the minimum and gets worse the further you go. The figure below shows exactly how far ‘near’ extends for a real molecule.

The harmonic oscillator is not a model of a bond; it is the first non-vanishing term of one. Expand any potential with a minimum at r_e in a Taylor series: the constant term is an arbitrary zero, the linear term vanishes because the slope is zero at a minimum, and the first surviving term is ½(d²V/dr²)(r−r_e)² — a parabola with k = d²V/dr² at the bottom. The green band marks where the parabola tracks the real curve to better than 2% of the well depth, computed here by scanning outwards from r_e until the discrepancy exceeds that. It comfortably contains v = 0. Already v = 1 spills past it at both ends — barely on the inner side, clearly on the outer (its turning points are 112 and 150 pm, against a band of 113–144 pm), and that lopsidedness is the anharmonicity of the real well showing through. It is why the harmonic oscillator gets the fundamental IR band nearly right and gets the overtones and the dissociation limit badly wrong. Note the two failures of the parabola that matter chemically: it rises without limit at large r (so it can never dissociate) and it is symmetric about r_e (so it predicts no thermal expansion of a bond).

Force constant, k: the second derivative of the potential energy with respect to displacement, evaluated at equilibrium. Units N m⁻¹. It measures the stiffness of a bond, not its strength: a bond's strength is its dissociation energy D_e, the depth of the whole well, while k is only the curvature at the very bottom. The two usually correlate but they are different quantities and are routinely confused in examinations.

The reduced mass, and why it is not the mass of the molecule

A diatomic molecule is two masses joined by a bond, and both of them move. The problem looks like a two-body problem, but it separates exactly into the motion of the centre of mass (free translation of the molecule as a whole, of no spectroscopic interest) and the relative motion of the two nuclei along the bond. That relative motion is a one-body problem for a fictitious particle of mass μ:

μ = m₁m₂/(m₁ + m₂) or equivalently 1/μ = 1/m₁ + 1/m₂

Two facts about μ are worth internalising because they explain a great deal of vibrational spectroscopy at a glance. First, μ is always smaller than either mass. Second, when one atom is much heavier than the other, μ tends to the mass of the lighter atom. For H–³⁵Cl, μ = 0.97959 u, only 2.8% below the mass of the hydrogen atom itself — the chlorine barely moves. That is why every X–H stretch, whatever X is, appears at high wavenumber: it is essentially a hydrogen atom vibrating against an immovable wall.

Advanced / reference layer

The separation of the two-body problem, the Schrödinger equation, its dimensionless form and the asymptotic solution.

The separation just asserted is worth doing, because the same manoeuvre reappears for the rotor in D.6 and for the hydrogen atom in Part 5. Write the two-particle Hamiltonian for motion along one axis and change to the coordinates X = (m₁x₁ + m₂x₂)/(m₁+m₂) (centre of mass) and x = x₂ − x₁ (relative displacement). The kinetic-energy operator transforms exactly, with no cross terms:

−(ℏ²/2m₁)∂²/∂x₁² − (ℏ²/2m₂)∂²/∂x₂² = −(ℏ²/2M)∂²/∂X² − (ℏ²/2μ)∂²/∂x²

with M = m₁ + m₂ the total mass. Since the potential depends only on x, the wavefunction factorises as Ψ = χ(X)ψ(x), the centre-of-mass part is a free particle, and the interesting equation is a one-dimensional oscillator of mass μ. This is not an approximation; it is exact.

−(ℏ²/2μ) d²ψ/dx² + ½kx²ψ = Eψ

Now strip the constants out. Define α = μω/ℏ with ω = √(k/μ), and use the dimensionless coordinate y = α^1/2x and the dimensionless energy ε = 2E/ℏω. Substituting and dividing through gives an equation with no parameters at all:

d²ψ/dy² + (ε − y²)ψ = 0

This is worth pausing on. Every property of the oscillator that can be stated in terms of y — the number of nodes, the shape of the density, the fraction of the probability outside the turning points — is a universal number, the same for a hydrogen molecule and a suspension bridge, because the equation that produces it contains no mass, no force constant and no ℏ. Only when you convert back to x = y/√α do the molecular parameters reappear.

The asymptotic solution, and why it must be extracted first

As y → ±∞ the term εψ becomes negligible beside y²ψ, and the equation reduces to d²ψ/dy² ≈ y²ψ. The solutions of that are approximately e^±y²/2: one grows without bound, one decays. Only the decaying one is square-integrable, so any acceptable ψ must behave as e^−y²/2 at large |y|.

The standard and essential trick is therefore to write

ψ(y) = f(y) e^−y²/2

and find the equation obeyed by the slowly varying factor f. Substituting and cancelling the exponential gives Hermite's differential equation:

f″ − 2y f′ + (ε − 1) f = 0

Note: Why the substitution is not optional. If you attack the original equation with a power series directly, the recursion relation has three terms instead of two and the analysis becomes unmanageable. More importantly, the asymptotic factor is the part that makes ψ normalisable, and it cannot be produced by a polynomial. Peeling it off first is what converts an intractable problem into a two-term recursion. The same manoeuvre — extract the asymptotic form, then solve for what is left — is used for the hydrogen radial equation in Part 5, where the extracted factor is e^−ρ/2ρ^l.

Find the force constant of H–³⁵Cl from its vibrational wavenumber, then predict the wavenumber of D–³⁵Cl Medium

Step 1 — the reduced mass. μ = m_Hm_Cl/(m_H+m_Cl) = (1.00783 × 34.96885)/(35.97668) u = 0.97959 u = 1.6267 × 10⁻²⁷ kg.

Step 2 — convert the wavenumber to an angular frequency. ω = 2πcω̃ = 2π × 2.9979 × 10⁸ m s⁻¹ × 2990.9 × 10² m⁻¹ = 5.6338 × 10¹⁴ rad s⁻¹. Note the factor of 100: wavenumbers are quoted per centimetre, and every SI calculation needs them per metre. This is the most common arithmetic slip in the whole topic.

Step 3 — invert ω = √(k/μ). k = μω² = 1.6267 × 10⁻²⁷ kg × (5.6338 × 10¹⁴ s⁻¹)² = 516.3 N m⁻¹.

Step 4 — the isotope substitution. Replacing H by D changes the mass but not the electronic structure, so k is unchanged to an excellent approximation — this is the whole basis of isotopic labelling in vibrational spectroscopy. The new reduced mass is μ(DCl) = 1.90441 u, a factor of 1.9441 larger.

Step 5 — scale the frequency. ω̃ ∝ μ^−½, so ω̃(DCl) = ω̃(HCl) × √(μ_HCl/μ_DCl) = 2990.9 × 0.7172 = 2145.1 cm⁻¹, computed here directly from k and the new μ as 2145.1 cm⁻¹.

Step 6 — the sanity check. The ratio is close to but not equal to 1/√2 = 0.7071, because the reduced mass does not quite double (the chlorine contributes a little). The computed ratio is 0.7172. A candidate who answers ‘divide by √2’ is nearly right and will lose a mark to one who uses the reduced masses properly.

Step 7 — the ³⁷Cl isotopomer, for contrast. The same calculation with ³⁷Cl gives 2988.6 cm⁻¹, a shift of only 2.3 cm⁻¹ — because the heavy atom hardly moves, changing its mass barely changes μ. Chlorine isotope splitting in an IR spectrum is small; deuteration is enormous.

⚠ Common mistakes & exam traps

Using the molecular mass instead of the reduced mass. The vibrating entity is the relative coordinate, and its inertia is μ, not m₁ + m₂. For HCl the difference is a factor of 37 in the mass and a factor of six in the predicted frequency.
Confusing the force constant with the bond energy. k is the curvature at the bottom of the well; D_e is the depth of the whole well. HF has a larger k than HI and a larger D_e, so the correlation is usually positive, but counter-examples exist and the two quantities have different units and different physical meanings.
Forgetting the factor 100 when converting cm⁻¹ to m⁻¹. This produces answers wrong by 10⁴ in k, and it is the single commonest numerical error in this topic.
Writing ω = √(k/m) with the wrong ω. ω is the angular frequency in rad s⁻¹; ν = ω/2π is the frequency in Hz; ω̃ = ν/c is the wavenumber in cm⁻¹ (with c in cm s⁻¹). Three different symbols, three different quantities, and examiners mix them deliberately.
Assuming that the parabola is a model of the bond rather than of small displacements about equilibrium. This leads to the absurd conclusion that a bond cannot be broken by stretching it, and to badly wrong overtone predictions.

Read the rest of Part 4

The remaining 8 sections of this part — The series solution, Hermite polynomials, the energy ladder and the wavefunctions, Ladder operators, selection rules and anharmonicity, Rotation in two dimensions: the particle on a ring… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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