The oscillator equation and its scales

Start from −(ħ²/2m)d²ψ/dx² + ½mω²x²ψ = Eψ and inventory what it contains: ħ, m and ω, and nothing else. Their signatures are [ħ] = M L² T⁻¹, [m] = M and [ω] = T⁻¹ — three quantities built from three independent base dimensions, so the group count n − k gives 3 − 3 = 0. There is no dimensionless number to be formed from the constants of this problem. That single fact drives the whole lesson. It means the constants can make exactly one length, one momentum, one energy and one time, each unique up to a pure number, and it means every eigenvalue must be ħω times a number the equation supplies rather than the potential. Compare the neighbours. The infinite well imports its length from outside — the wall separation L — and its energies go as ħ²/2mL². The finite well carries two scales, depth V₀ and half-width a, so it does own a dimensionless group, z₀ = (a/ħ)√(2mV₀), and its bound-state count, roughly 2z₀/π, depends on it. The oscillator has no such dial to turn.

Two routes give the same x₀. Dimensionally, ħ/mω has signature (M L² T⁻¹)/(M T⁻¹) = L², so x₀ = √(ħ/mω) is the only length available and p₀ = ħ/x₀ = √(mħω) the only momentum; note x₀p₀ = ħ, which is twice the uncertainty floor ħ/2: these are scales, not the uncertainties themselves. Physically, the two terms of the Hamiltonian want opposite things. Squeeze the state into a width a and the uncertainty bound forces Δp ≈ ħ/2a, so the kinetic cost is ħ²/8ma² while the potential cost is ½mω²a². Add them, differentiate and set to zero: −ħ²/4ma³ + mω²a = 0 gives a² = ħ/2mω, that is a = x₀/√2, and the energy there is ħω/4 + ħω/4 = ħω/2. Both numbers are exact for this potential — the true ground state really does have Δx = x₀/√2 — because a quadratic potential's ground state is a minimum-uncertainty Gaussian. So x₀ is not an arbitrary yardstick: it fixes the width, x₀/√2, at which confining and spreading cost the same, and ħω fixes what that compromise costs, ħω/2.

Put x = x₀ξ. The chain rule gives d²ψ/dx² = (1/x₀²)d²ψ/dξ², so the kinetic coefficient becomes ħ²/2mx₀² = (ħ²/2m)(mω/ħ) = ħω/2, and the potential coefficient becomes ½mω²x₀² = ½mω²(ħ/mω) = ħω/2. Both have collapsed to the same constant — which is precisely what x₀ was chosen to do. The equation now reads −(ħω/2)ψ″ + (ħω/2)ξ²ψ = Eψ, and dividing through by ħω/2 leaves d²ψ/dξ² = (ξ² − ε)ψ with ε = 2E/ħω. Not one of ħ, m or ω survives. Two bookkeeping points follow. The factor 2 in ε is chosen so the reduced equation has no fractions; the price is that the eventual spectrum reads ε = 2n + 1, the odd integers, not n + ½. And normalisation moves with the variable: since dx = x₀ dξ, writing ψ(x) = x₀^(−1/2)u(ξ) with ∫|u|²dξ = 1 is what keeps ∫|ψ|²dx = 1 and restores ψ's m^(−1/2) units on the way back.

The single parameter has a picture attached. Classically the particle stops where all its energy is potential, ½mω²x² = E, so xₜₚ = √(2E/mω²); divide by x₀ = √(ħ/mω) and the masses and frequencies cancel to leave ξₜₚ = √(2E/ħω) = √ε. So ε is not merely "the energy in some units" — it is the square of the classical turning point measured in natural lengths, and the allowed region is −√ε ≤ ξ ≤ √ε. That is what organises every solution. Inside the region ξ² − ε is negative, so ψ″ and ψ carry opposite signs and the curve bends back toward the axis: oscillation, and nodes. Outside it the sign flips, ψ bends away from the axis, and the only behaviour that stays finite is decay. The turning point is where the character changes, and the reduced equation locates it with one number. On the spectrum ε = 2n + 1, so ξₜₚ = √(2n+1): 1 at n = 0, 3 at n = 4, 4.58 at n = 10. Energy climbs linearly in n while reach climbs only as its square root, which is E = ½mω²A² read backwards.

Take the parabola literally and V → ∞ in both directions. Then for any energy, however large, the classically allowed region is the finite interval |ξ| ≤ √ε, and beyond it ψ must decay. That is a boundary condition at both ends of an infinite line. A second-order equation has two independent solutions at each energy; demanding decay on the right kills one combination and demanding decay on the left kills another, so a solution satisfying both exists only at special values of ε. Hence a discrete spectrum, all the way up, with no threshold anywhere. Contrast the neighbours: a finite well of depth V₀ has a finite number of bound states and a continuum above V₀, while a step or a barrier has a continuum only. The oscillator has no continuum at all and infinitely many levels, because there is no energy at which the particle can escape. That is the model's cost written plainly — an exactly quadratic, infinitely deep potential cannot dissociate, cannot ionise, and cannot run out of rungs. Real bonds do all three.

The scales answer "how quantum is this?" on their own. An electron bound harmonically at ω = 2.0×10¹⁵ rad s⁻¹ has x₀ = 0.241 nm and ħω = 1.32 eV — a length like a chemical bond and a spacing in the near infrared, at 942 nm. The CO stretch at 2170 cm⁻¹, so ω = 4.09×10¹⁴ rad s⁻¹ with reduced mass 6.86 u, has x₀ = 4.76 pm and ħω = 0.269 eV; the natural length is 4.2% of the 112.8 pm bond, which is exactly why a quadratic fit survives the low levels. A 1.0 g bob on a spring at ω = 10 rad s⁻¹ has x₀ = 1.03×10⁻¹⁶ m, about an eighth of a proton's charge radius, and ħω = 6.6×10⁻¹⁵ eV. Since the classical amplitude is A = x₀√ε with ε = 2n + 1, the quantum number is n ≈ ½(A/x₀)², so a 1.0 mm swing gives n ≈ 4.7×10²⁵. The model's limits arrive by the same arithmetic: CO's 11.09 eV dissociation energy is only 41 harmonic quanta, while the harmonic ladder itself never stops.