The ladder and zero-point energy

Eₙ = (n + ½)ħω contains exactly one physical input, ω = √(k/m); the rest is arithmetic. Two habits of notation cause most of the errors. First, n counts quanta, not levels: n = 0 is the ground state, so what is loosely called the second excited state is n = 2, at 2.5ħω. Second, the sum runs to infinity — a parabola never stops rising, so every state is bound, the spectrum stays discrete all the way up, and nothing in the model can ever come apart. Put numbers on it with carbon monoxide. Its harmonic wavenumber, fitted from the infrared spectrum, is 2170 cm⁻¹ (the band centre itself sits lower, at 2143 cm⁻¹, for the reason the last section gives), and 1 cm⁻¹ = 1.2398×10⁻⁴ eV, so ħω = 0.269 eV. The reduced mass is μ = (12 × 16)/28 u = 6.86 u = 1.138×10⁻²⁶ kg, and ω = 2πc × 2170 cm⁻¹ = 4.09×10¹⁴ rad s⁻¹, so the force constant is k = μω² = 1.90×10³ N m⁻¹. The levels then sit at 0.135, 0.404, 0.673, 0.942 eV and onward, forever, above the bottom of the well.

Compare the two standard bound problems. In an infinite well the walls do not move: raising the energy only shortens the wavelength inside a fixed length L, so Eₙ = n²h²/8mL² and the gaps widen as (2n + 1)E₁ — for an electron in a 0.50 nm well, 4.51 eV, then 7.52, then 10.53. In a parabola the walls move with the particle. The classical turning point of level n is Aₙ = √((2n + 1)ħ/mω), which grows as √E, and that widening is exactly what cancels the shortening wavelength. The sharpest way to see it is to count states by phase-space area, since the number of levels below E is that area divided by h. The oscillator's orbit is an ellipse with semi-axes √(2E/mω²) and √(2mE), so its area is 2πE/ω — linear in E, hence one level per ħω, evenly. The box's orbit is a rectangle of area 2L√(2mE), which grows only as √E, so its levels thin out. The WKB quantisation rule makes it exact rather than suggestive: ∮p dx = (n + ½)h, which is Bohr–Sommerfeld's nh plus the half that the two soft turning points contribute, returns E = (n + ½)ħω for the oscillator, half-quantum and all.

Nothing in the argument for ½ħω needs the differential equation. In a symmetric well a stationary state has ⟨x⟩ = 0 and ⟨p⟩ = 0, so ⟨x²⟩ = (Δx)², ⟨p²⟩ = (Δp)², and ⟨E⟩ = (Δp)²/2m + ½mω²(Δx)². Squeezing the particle towards the bottom cuts the second term and inflates the first, because ΔxΔp ≥ ħ/2. Take the bound as an equality, put Δp = ħ/2Δx, and minimise E(u) = ħ²/8μ + ½mω²u over u = (Δx)²: dE/du = −ħ²/8mu² + ½mω² = 0 gives u = ħ/2mω. Substituting back, the kinetic term is ħω/4, the potential term is ħω/4, and Eₘᵢₙ = ½ħω. Two things follow. The bound is saturated — the true ground state has ⟨x²⟩ = ħ/2mω and ΔxΔp = ħ/2 exactly — so this variational estimate is the answer, not merely a floor under it. And the width is small without being zero: for CO, Δx = 3.37 pm against a bond length of 113 pm, a 3% wobble that no refrigerator removes.

The usual objection is that ½ħω is a constant, and constants in energy are conventions. That would hold if it were the same constant everywhere, and it is not: it is ½ħ√(k/m), so anything that changes the stiffness or the mass moves it. Isotopic substitution is the cleanest case, because the Born–Oppenheimer electronic curve is fixed by nuclear charges, not nuclear masses. H₂ and D₂ therefore share one curve and one well depth, Dₑ = 4.747 eV, while μ(D₂) = 2μ(H₂) drops ωₑ from 4401 cm⁻¹ to 4401/√2 = 3112 cm⁻¹. Their zero-point energies are 0.273 eV and 0.193 eV, so the energies actually needed to break the bonds, D₀ = Dₑ − ½ħω, are 4.474 eV and 4.554 eV. Deuterium is bound by 0.080 eV more than hydrogen purely because it sits lower in the same well; the measured difference is 0.078 eV. That gap of a few tens of meV is what drives kinetic isotope effects in chemistry, and the isotope separations built on them.

The second piece of evidence is a substance that will not freeze. Two helium atoms attract with a pair potential only about 10 K deep — ε/k_B ≈ 10.2 K, or 0.9 meV — with a Lennard-Jones σ of 2.56 Å. Localising an atom inside a cage that size costs a zero-point energy of order h²/mσ², and for ⁴He that scale is roughly seven times ε. The square root of that ratio is the de Boer parameter Λ = h/σ√(mε), which is 2.68 for ⁴He and 3.08 for the lighter ³He. Crystallising means localising, and here localising costs more than the binding returns, so ⁴He stays liquid at saturated vapour pressure all the way to absolute zero and needs roughly 25 atm of applied pressure to solidify at all. The comparison that settles it is neon: the same kind of atom with the same kind of potential, but Λ = 0.58, so its zero-point energy is a third of its well depth and it freezes at 24.6 K like anything else. Zero-point energy is not a small correction; in helium it decides the phase.

All of it rests on the potential being exactly quadratic, and no bond is. Real vibrational levels are fitted by G(v) = ωₑ(v + ½) − ωₑ xₑ(v + ½)², whose gaps shrink linearly: ΔG = ωₑ − 2ωₑ xₑ(v + 1). For HCl, ωₑ = 2991 cm⁻¹ and ωₑ xₑ = 52.8 cm⁻¹, so the fundamental is 2991 − 106 = 2885 cm⁻¹, while the first overtone lands at 5665 cm⁻¹ instead of twice the fundamental, 5771 cm⁻¹ — short by 106 cm⁻¹, or 1.8%, and that shortfall is the direct measurement of the anharmonicity. Extrapolating the shrinking gaps to zero, a Birge–Sponer plot, puts the last bound level near v = 27; the true count is smaller, because the Morse form over-extrapolates, but either way it is a few dozen and then a continuum, where the parabola promised infinitely many. So equal spacing is a claim about the bottom of a well: good for the lowest levels of a stiff bond, worse above them. The zero-point energy, being a property of the ground state, survives the failure far better than the equal spacing does.