Asymptotic form, Hermite polynomials and quantisation

Topic 8.2 built the yardsticks: the length x₀ = √(ħ/mω) and the energy ħω. Measure position in x₀ — set ξ = x/x₀ — and the time-independent equation collapses to d²ψ/dξ² = (ξ² − K)ψ with K = 2E/ħω. Mass, frequency and ħ have all been absorbed. For H³⁵Cl the yardstick is x₀ ≈ 10.9 pm; for a trapped ion in a megahertz trap it is roughly a thousand times larger — yet both face the identical dimensionless equation. That is the power of the reduction: the question "what are the energies of this oscillator?" becomes "for which pure numbers K does this one equation possess a solution that dies at infinity?" Answer it once, and every harmonic oscillator in nature is solved.

Far from the origin ξ² dwarfs K and the equation is approximately ψ″ = ξ²ψ. Try ψ = e^(±ξ²/2): differentiating gives ψ″ = (ξ² ± 1)e^(±ξ²/2), which is ξ²ψ up to a correction that fades as ξ grows — so both exponentials are legitimate asymptotic behaviours, and a general solution carries a mixture of them. A normalisable state can carry none of the growing one. So make the survivor explicit: write ψ(ξ) = h(ξ)e^(−ξ²/2) exactly — a change of unknown, not an approximation — and substitute back. The Schrödinger equation becomes Hermite's equation, h″ − 2ξh′ + (K − 1)h = 0. Nothing is quantised yet. The move has simply reassigned the labour: the Gaussian handles the decay, and h's only obligation is to grow slowly enough not to wreck it.

Feed Hermite's equation the series h = Σ aⱼξʲ and equate coefficients of each power: aⱼ₊₂ = aⱼ(2j + 1 − K)/((j + 1)(j + 2)). Two seeds, a₀ and a₁, generate everything — exactly the two free constants a second-order equation owes you — and because the recursion steps by two, the even-j chain and the odd-j chain never communicate. Any solution splits into an even function plus an odd one, so the eventual eigenfunctions will have definite parity: a direct consequence of stepping by two, itself a consequence of the potential being even in ξ. Watch a chain die in real time at K = 5: a₂ = (1 − 5)/2 = −2, then a₄ = a₂(5 − 5)/12 = 0, and every later even coefficient inherits the zero. The chain has snapped shut at h = 1 − 2ξ², which is H₂ up to the factor −2.

Suppose no numerator ever vanishes. For large j the ratio aⱼ₊₂/aⱼ = (2j + 1 − K)/((j + 1)(j + 2)) → 2/j. Now compare the function banished at the start: e^(+ξ²) = Σ ξ^(2m)/m! steps its coefficients by 1/(m + 1) = 2/(j + 2) — the same tail. So an unterminated h grows like e^(+ξ²), and ψ = h⋅e^(−ξ²/2) grows like e^(+ξ²/2): the series has quietly rebuilt the diverging exponential the factorisation threw away. This is the crux of the topic. The differential equation itself is perfectly happy at every K — solutions exist for E = 0.3 ħω, for 1.2 ħω, for π ħω — but all of them blow up at infinity. Solving the equation was never the hard part; being a state is.

The only escape is a numerator hitting zero: 2n + 1 − K = 0 for some integer n, so K = 2n + 1 and Eₙ = (n + ½)ħω. Equal rungs ħω apart, and a lowest rung of ½ħω that nobody inserted — the zero-point energy drops out of the termination itself. But termination stops one chain only. If n is even, the odd chain still runs forever and would drag its e^(+ξ²) tail along, so its seed must be set to zero by hand: a₁ = 0. If n is odd, a₀ = 0 instead. Each allowed energy therefore owns exactly one state — no degeneracy in one dimension — a polynomial of a single parity times the Gaussian. Count what the argument used: no wall, no node-counting, no operator algebra. Only the demand that ∫|ψ|² dx be finite.

With the convention that the leading coefficient is 2ⁿ, the terminated chains are the Hermite polynomials: H₀ = 1, H₁ = 2ξ, H₂ = 4ξ² − 2, H₃ = 8ξ³ − 12ξ, linked by Hₙ₊₁ = 2ξHₙ − 2nHₙ₋₁. Hₙ has n real zeros, so ψₙ has n nodes, and parity (−1)ⁿ — the quantum number is readable off a graph of the state. Never trust a quoted polynomial: audit H₂ against Hermite's equation at K = 5. h″ = 8; −2ξh′ = −16ξ²; (K − 1)h = 4(4ξ² − 2) = 16ξ² − 8. Sum: 8 − 16ξ² + 16ξ² − 8 = 0. It passes. Gaussian integrals then fix the normalisation, ψₙ(x) = (mω/πħ)¹⁄⁴(2ⁿn!)^(−1/2) Hₙ(ξ) e^(−ξ²/2) — the closed form every later topic in this unit quotes.