The oscillator Hamiltonian and its quadratures

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 a non-degenerate minimum 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.

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.

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/k_B = 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, about a sixth of a proton radius, and a millijoule of oscillation energy is 1.5 × 10³⁰ quanta.

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).

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 as 0.5, 1.5 and 2.5 to better than 0.1%. Feed the same routine SI quantities and it returns the same spectrum times ħω, since floating point has no preferred scale; what SI costs is everything around the solve, because the box, the grid spacing and every absolute tolerance must be re-chosen in metres and joules for each new oscillator. Non-dimensionalising here is not cosmetic; it turns every oscillator into one problem, solved once.

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; a Morse fit to HCl 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, at leading order, a 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.