University Physics V · The Quantum Harmonic Oscillator · 7.1
The Oscillator Hamiltonian & Its Quadratures
Nothing in nature is exactly quadratic, yet almost everything is quadratic near the bottom. This is where that reach gets set up: which term the Taylor expansion keeps, what boundary condition the potential supplies on its own, and how three constants collapse into one dimensionless pair X and P.
Build the model
Connect the measurement to the mechanism.
The harmonic oscillator is not one system but the generic first approximation to all of them. Expand any smooth potential about a stable minimum and shift the origin there, energy zero included: the linear term dies because V′ = 0 is what a minimum means, and what survives is ½ku² with k = V″ > 0. So Ĥ = p̂²/2m + ½mω²x̂² with ω = √(V″/m), m being the reduced mass for a two-body vibration.
It acts on L²(ℝ) with no walls anywhere; the potential supplies its own boundary condition, ψ → 0 as |x| → ∞, and because V climbs without limit on both sides the spectrum is wholly discrete and non-degenerate — no continuum, no delta normalisation. Three constants ħ, m and ω admit exactly one length x₀ = √(ħ/mω), one momentum p₀ = √(mħω) and one energy ħω, conjugate by construction since x₀p₀ = ħ. Measuring x̂ and p̂ in those units gives the dimensionless quadratures X and P, still Hermitian, obeying [X, P] = i, and Ĥ = ½ħω(X² + P²) — an operator with no parameters left inside it, which is why a bond, a trapped ion and a mode of the electromagnetic field all return the same answer in different units.
The cost is the cubic term you threw away: it is what lets bonds dissociate, solids expand when heated, phonons scatter and overtones appear, and none of it survives a strictly quadratic Hamiltonian.
- Simple definition
- The quantum harmonic oscillator is the Hamiltonian p̂²/2m + ½mω²x̂² — the leading surviving term when any potential is expanded about a stable minimum — which in its own natural units becomes ½ħω(X² + P²) with [X, P] = i.
- Example
- For H³⁵Cl, k = V″ = 516 N m⁻¹ and μ = 0.980 u give ω = 5.63 × 10¹⁴ rad s⁻¹, so ħω = 0.371 eV and x₀ = √(ħ/μω) = 10.7 pm — 8% of the 127 pm bond, which is why the quadratic term rules.
The minimum kills the linear term and stability makes V″ > 0. Everything past u² is what you agree to drop.
V′ = 0 at the minimum and V(xₘᵢₙ) is shifted to zero; V″ in J m⁻²
H³⁵Cl: k = 516 N m⁻¹ and μ = 0.980 u give ω = 5.63 × 10¹⁴ rad s⁻¹, i.e. 2991 cm⁻¹.
m is the reduced mass μ = m₁m₂/(m₁+m₂) for a diatomic; ω in rad s⁻¹
Fixes the width of every state before a single eigenvector has been found.
x₀ in m, p₀ in kg m s⁻¹, energy scale ħω in J; no other combination has these dimensions
Replaces [x̂, p̂] = iħ: the ħ moves out of the algebra and into the units.
both dimensionless and both still Hermitian, so both remain observables
One operator for every oscillator, and the only form a grid solver can condition.
ξ = u/x₀ and ε = E/ħω, so every coefficient in the eigenvalue problem is 1 or ½
|λ₃| ≪ 1 is the licence to drop it. H³⁵Cl has λ₃ ≈ −0.094, small but not zero.
the cubic energy in quanta, evaluated at the ground state's own width
Every stable minimum is quadratic first
Take any smooth potential with a stable minimum at xₘᵢₙ and expand: V(x) = V(xₘᵢₙ) + V′u + ½V″u² + ⅙V‴u³ + …, with u = x − xₘᵢₙ. Two terms go immediately. V′(xₘᵢₙ) = 0 is what “minimum” means, and V(xₘᵢₙ) is only a choice of energy origin, so set it to zero. What is left at leading order is ½ku² with k = V″(xₘᵢₙ), and stability is exactly the statement k > 0. Write k = mω² and you have Ĥ = p̂²/2m + ½mω²x̂² with ω = √(V″/m). For a diatomic the two-body problem separates into a free centre of mass plus one relative coordinate, so m is the reduced mass μ = m₁m₂/(m₁+m₂): H³⁵Cl has k = 516 N m⁻¹ and μ = 0.980 u, hence ω = 5.63 × 10¹⁴ rad s⁻¹. The construction fails in one place only. Where V″(xₘᵢₙ) = 0 — a quartic minimum, a lattice sitting at a structural transition — there is no ω to build, the leading term is quartic, and none of the machinery below ever starts.
The operator, its domain, and the boundary condition
Ĥ acts on L²(ℝ): x̂ multiplies by x, p̂ = −iħ d/dx, and Ĥ is essentially self-adjoint on the Schwartz functions, so it has one self-adjoint extension and a real spectrum. There are no walls anywhere — the boundary condition is inherited, not imposed. Square-integrability together with V → ∞ as |x| → ∞ forces ψ → 0 at both ends, and faster than any power of x. That one fact separates this problem from every well of the previous unit: with no asymptotic floor for a continuum to sit above, the spectrum is entirely discrete, and there is no δ-normalised sector and no scattering state to match. The one-dimensional Wronskian argument makes each level non-degenerate too, so a single quantum number labels the whole spectrum. Two symmetries come free: V is even about the minimum, so [Ĥ, Π̂] = 0 for the parity operator Π̂ψ(x) = ψ(−x) and every eigenvector may be chosen with definite parity; and once the scales below are used the Fourier transform maps Ĥ to itself, which is why x and p play interchangeable roles in everything that follows.
Three constants, and only one length among them
ħ, m and ω are the whole content of Ĥ, and exactly one length can be built from them. [ħ] = M L² T⁻¹, [m] = M and [ω] = T⁻¹, so ħ/(mω) has dimension L² and x₀ = √(ħ/mω) is forced. The same counting gives one momentum p₀ = √(mħω) and one energy ħω, and the pair is conjugate by construction: x₀p₀ = √((ħ/mω)(mħω)) = ħ. Nothing has been solved and the width of every state is already fixed. The numbers show why the classical world looks classical. For H³⁵Cl, x₀ = 10.7 pm against a 127 pm bond — 8%, so the molecule lives inside the quadratic region. For one ⁴⁰Ca⁺ ion trapped at ω/2π = 1 MHz, x₀ = 15.9 nm and ħω/kB = 48 μK, which is why ground-state cooling needs lasers rather than a fridge. For a 1 g mass on a 1 Hz spring, x₀ = 1.3 × 10⁻¹⁶ m, a tenth of a proton radius, and a millijoule of amplitude is 1.5 × 10³⁰ quanta.
Non-dimensionalise: the quadratures and [X, P] = i
Measure x̂ and p̂ in their own units. X = x̂/x₀ = √(mω/ħ) x̂ and P = p̂/p₀ = p̂/√(mħω) are dimensionless and still Hermitian, so both remain observables. Their commutator inherits the scaling: [X, P] = [x̂, p̂]/(x₀p₀) = iħ/ħ = i. The ħ has not vanished; it has moved into the units. Substituting back, p̂²/2m = ½ħωP² and ½mω²x̂² = ½ħωX², so Ĥ = ½ħω(X² + P²) — one energy multiplying an operator with no free parameters at all. That is the content of this topic: a bond, a trapped ion and a mode of the electromagnetic field are the same operator wearing different values of ħω. The name comes from the phase space. The Heisenberg equations dX/dt = (i/ħ)[Ĥ, X] = ωP and dP/dt = −ωX are a rigid rotation at rate ω, so in (X, P) coordinates the classical orbit is a circle rather than an ellipse and X and P are the cosine and sine amplitudes of the motion — a quarter cycle apart, in quadrature. The contour at energy E is a circle of radius √(2E/ħω).
What the commutator forces, and what a solver needs
Two payoffs arrive before any eigenvector is found. First a bound: for any state ⟨X²⟩ ≥ (ΔX)² and likewise for P, so ⟨Ĥ⟩ = ½ħω(⟨X²⟩ + ⟨P²⟩) ≥ ħω ΔX ΔP ≥ ħω/2, by AM–GM and then Robertson applied to [X, P] = i. The oscillator cannot sit at the bottom of its own well, and the floor is set by the commutator, not by anything in the potential. Second, the numerics. In ξ = u/x₀ with ε = E/ħω the eigenvalue problem reads −½ψ″(ξ) + ½ξ²ψ(ξ) = εψ(ξ): every coefficient is 1 or ½. Hand a finite-difference second derivative on ξ ∈ [−8, 8] with a few hundred points to numpy.linalg.eigh and the lowest eigenvalues come back 0.500, 1.500, 2.500 to four figures. Feed the same routine SI quantities and its matrix entries span thirty-odd orders of magnitude and the small eigenvalues drown. Non-dimensionalising here is not cosmetic; it is what conditions the problem.
The cubic term you agreed to throw away
Restore the next term. In scaled units it is ħωλ₃ξ³ with λ₃ = V‴x₀³/(6ħω) — the cubic energy counted in quanta, at the ground state's own width. H³⁵Cl gives λ₃ ≈ −0.094: small enough to perturb with, far too large to forget. Parity makes ⟨n|X³|n⟩ = 0, so the shift is second order and looks mild, and the measured anharmonicity ωₑ xₑ/ωₑ = 1.8% agrees. What is not mild is what a strictly quadratic Ĥ forbids outright. Its ladder is infinite and evenly spaced, so a bond never breaks; HCl in fact holds about 28 levels and then dissociates. Its well is symmetric, so ⟨u⟩ = 0 in every state at every temperature — a harmonic crystal does not expand when heated, and thermal expansion is a purely cubic effect. Its normal modes are exact eigenvectors, so phonons never scatter off one another and an insulator's thermal conductivity would be infinite. With a linear dipole it allows only Δn = ±1, so the overtone bands every infrared spectrum shows could not exist.
Change one variable at a time
Make the relationship visible.
Slide a⋅x₀ from 0.02 to 0.40: the solid curve peels outward while stiffening inward, and that asymmetry — the cubic term — is why heated solids expand. At 0.40 the well is only 3.1 ħω deep, so push E past it and the filled outer marker vanishes. The harmonic one never does.
CUBIC STRENGTH λ₃-0.090
WELL DEPTH D15.4 ħω
HARMONIC TURNING PT2.24 x₀
TRUE OUTER TURNING PT2.86 x₀
Live interpretationCUBIC STRENGTH λ₃: −0.090. WELL DEPTH D: 15.4 ħω. HARMONIC TURNING PT: 2.24 x₀. TRUE OUTER TURNING PT: 2.86 x₀
Catch the common trap
Explain before calculating.
Replacing ¹H by ²H in H³⁵Cl leaves the electronic potential — and so the force constant k = V″(rₑ) — untouched, but raises the reduced mass from 0.980 u to 1.904 u. What happens to ω, to ħω, and to the length scale x₀ = √(ħ/mω)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe stretching mode of ¹²C¹⁶O has force constant k = 1902 N m⁻¹. Treating it as a harmonic oscillator in the relative coordinate, find the reduced mass, ω, the energy quantum ħω in eV, and the natural scales x₀ and p₀.
- The two-body problem separates into a free centre of mass plus one relative coordinate, so the reduced mass is what enters: μ = (12.000 × 15.995)/(27.995) u = 6.856 u = 1.1385 × 10⁻²⁶ kg. Neither atomic mass alone appears anywhere.
- ω = √(k/μ) = √(1902 / 1.1385 × 10⁻²⁶) = 4.087 × 10¹⁴ rad s⁻¹. Dividing by 2πc puts it where a spectroscopist would write it: 2170 cm⁻¹.
- ħω = 1.0546 × 10⁻³⁴ × 4.087 × 10¹⁴ = 4.310 × 10⁻²⁰ J = 0.269 eV.
- μω = 4.653 × 10⁻¹² kg s⁻¹, so x₀ = √(ħ/μω) = √(2.266 × 10⁻²³) = 4.76 × 10⁻¹² m = 4.76 pm, and p₀ = ħ/x₀ = 2.22 × 10⁻²³ kg m s⁻¹. The two are conjugate by construction: x₀p₀ = ħ exactly.
- Sanity, using numbers the model itself supplied: x₀ is 4.2% of the 112.8 pm bond, and ħω is 10.4 times kBT at 300 K. The mode sits in its ground state at room temperature and never leaves the quadratic part of the well — which is precisely what licenses treating it as quadratic.
Answerμ = 1.139 × 10⁻²⁶ kg, ω = 4.09 × 10¹⁴ rad s⁻¹ (2170 cm⁻¹), ħω = 0.269 eV, x₀ = 4.76 pm and p₀ = 2.22 × 10⁻²³ kg m s⁻¹.
MediumTwo argon atoms interact through V(r) = 4ε[(σ/r)¹² − (σ/r)⁶] with ε/kB = 119.8 K and σ = 3.405 Å. Expand about the minimum to get the force constant, then find ω, ħω and x₀ for the Ar₂ dimer, and say what fraction of the well depth zero-point motion consumes.
- V′ = 0 gives (σ/r)⁶ = ½, so rₘ = 2¹⁄⁶σ = 382 pm. Put the origin there with u = r − rₘ; the linear term is gone by construction and the constant −ε is only an energy offset.
- V″(r) = 4ε(156σ¹²r⁻¹⁴ − 42σ⁶r⁻⁸), and at rₘ this collapses to 72ε/(2¹⁄³σ²) = 57.15 ε/σ². With ε = 119.8 × 1.3806 × 10⁻²³ = 1.654 × 10⁻²¹ J and σ = 3.405 × 10⁻¹⁰ m, k = 0.815 N m⁻¹ — three orders of magnitude softer than the covalent bond above.
- For a homonuclear dimer μ = m/2 = 39.948 × 1.6605 × 10⁻²⁷/2 = 3.317 × 10⁻²⁶ kg, so ω = √(0.815 / 3.317 × 10⁻²⁶) = 4.96 × 10¹² rad s⁻¹, i.e. 26.3 cm⁻¹.
- ħω = 5.23 × 10⁻²² J = 3.26 meV, or 37.9 K in temperature units; x₀ = √(ħ/μω) = 25.3 pm, which is 6.6% of rₘ.
- Zero-point energy is ħω/2 = 0.158 ε: the dimer is born with 16% of its binding already spent, and 2ε/ħω = 6.3 says only a handful of levels fit. Repeat the arithmetic for He₂, lighter and far more weakly bound, and ħω/2 comes out at 2.1 ε — the harmonic model predicts no bound state at all, which is why helium will not solidify under its own vapour pressure.
Answerk = 0.815 N m⁻¹, ω = 4.96 × 10¹² rad s⁻¹ (26.3 cm⁻¹), ħω = 3.26 meV, x₀ = 25.3 pm. Zero-point energy is 0.158 ε, so 16% of the well depth is gone before any excitation.
HardH³⁵Cl is measured at ωₑ = 2990.9 cm⁻¹ and ωₑ xₑ = 52.82 cm⁻¹. Model the bond as a Morse well V(u) = Dₑ(1 − e(−au))², whose expansion is Dₑ(a²u² − a³u³ + …). Find the dimensionless cubic strength λ₃ = V‴x₀³/(6ħω), and say what a strictly harmonic Ĥ gets wrong.
- Match the quadratic term: V″(0) = 2Dₑ a² = μω², so ω = a√(2Dₑ/μ). The Morse spectrum Eₙ = ħω(n+½) − (ħω)²(n+½)²/(4Dₑ) identifies xₑ = ħω/(4Dₑ), so Dₑ = ωₑ/(4xₑ) = 2990.9/(4 × 0.017660) = 4.234 × 10⁴ cm⁻¹ = 5.25 eV.
- The only dimensionless number the well and the quantum scale can form is a x₀. From 2Dₑ a² = μω² and x₀² = ħ/(μω) it follows that (a x₀)² = ħω/(2Dₑ) = 2990.9/(2 × 42 340) = 0.03532, so a x₀ = 0.188: the quantum width is a fifth of the length over which the potential changes shape.
- Now the cubic. V‴(0) = −6Dₑ a³, so λ₃ = −6Dₑ a³x₀³/(6ħω) = −(Dₑ/ħω)(a x₀)³ = −14.16 × 0.188³ = −0.094. For a Morse well this is always just −a x₀/2.
- Parity gives ⟨n|X³|n⟩ = 0, so the cubic term shifts nothing at first order; it enters at second order and, with the quartic, reproduces the measured −ωₑ xₑ(n+½)². The harmonic model puts the fundamental at ωₑ = 2991 cm⁻¹; the anharmonic one at ωₑ − 2ωₑ xₑ = 2885 cm⁻¹, and the band is observed at 2886 cm⁻¹ — a 3.6% error from a term worth a tenth.
- The qualitative losses are worse than 3.6%. A harmonic ladder is infinite and evenly spaced, so the bond never breaks, where the Morse fit gives ωₑ/(2ωₑ xₑ) ≈ 28 levels and then dissociation. A harmonic well is symmetric, so ⟨u⟩ = 0 in every state and the bond never lengthens on heating; and with a linear dipole only Δn = ±1 survives, so the observed overtone near 5665 cm⁻¹ would be missing.
Answerλ₃ = −(Dₑ/ħω)(a x₀)³ = −0.094, with a x₀ = 0.188 and Dₑ = 5.25 eV. The harmonic Ĥ misplaces the fundamental by 3.6% (2991 against 2886 cm⁻¹) and loses dissociation, thermal expansion and overtones outright.