University Physics IV · The Quantum Harmonic Oscillator · 8.3
Asymptotic Form, Hermite Polynomials & Quantisation
No walls, no nodes imposed by hand — so where does the ladder come from? This lesson earns the oscillator spectrum the honest way: factor out the Gaussian the equation demands at infinity, hand what remains to a power series, and let normalisability kill every energy except Eₙ = (n + ½)ħω.
Build the model
Connect the measurement to the mechanism.
Scaled to its natural units the oscillator equation reads ψ″ = (ξ² − K)ψ, with everything about the particular oscillator absorbed into K = 2E/ħω. Far out, the equation forces ψ toward e(±ξ²/2); only the decaying branch can be a state, so write ψ = h(ξ)e(−ξ²/2) and ask what h must be. A power series answers with a two-step recursion, aⱼ₊₂ = aⱼ(2j + 1 − K)/((j + 1)(j + 2)) — and here is the trap: for generic K the recursion never ends, its tail matches the series of e(+ξ²) term for term, and the "solution" rebuilds the diverging exponential already thrown away.
Every K solves the differential equation; almost no K yields a normalisable state. Rescue comes only when a numerator vanishes: K = 2n + 1, that is Eₙ = (n + ½)ħω, with the still-running opposite-parity chain set to zero by hand. Quantisation here is a boundary condition at infinity playing exactly the role the walls played for the box, and because the recursion steps by two, every eigenfunction comes out even or odd.
The price of the method is the assumption it leans on: the potential must stay exactly quadratic out to infinity, which no real bond does.
- Simple definition
- The oscillator's bound states are Hermite polynomials times a Gaussian, and they exist only at Eₙ = (n + ½)ħω because only there does the power series multiplying the Gaussian terminate rather than outgrow it.
- Example
- At K = 2E/ħω = 5 the even chain stops at once: a₂ = −2, a₄ = 0, so h = 1 − 2ξ² ∝ H₂ and E₂ = 2.5 ħω — for a CO-like bond with ħω = 0.266 eV, that is 0.665 eV.
Every oscillator becomes the same equation; quantisation becomes a statement about the single number K.
ξ = x/x₀ with x₀ = √(ħ/mω) in metres; K = 2E/ħω is a pure number
The Gaussian carries the decay at infinity, so h only has to avoid growing like e(+ξ²).
an exact change of unknown; e(+ξ²/2) is discarded as unnormalisable
Stepping by two is why even and odd never mix — parity is settled before any energy is.
h = Σ aⱼξʲ; the seeds a₀ and a₁ start an even and an odd chain
The whole spectrum — equal spacing ħω and the ½ħω floor — from a boundary condition at infinity.
an unterminated chain grows as e(+ξ²) and destroys normalisability
Hₙ has n real zeros and parity (−1)ⁿ: read the quantum number straight off a plotted state.
convention: leading coefficient 2ⁿ; Hₙ₊₁ = 2ξHₙ − 2nHₙ₋₁
The closed form the rest of the unit quotes for every expectation value and matrix element.
orthonormal, ∫ψₘψₙ dx = δₘₙ; ξ = x/x₀ is dimensionless
One number survives the scaling
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 over 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.
Keep only the tail that dies
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.
Two chains that never mix
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.
The unterminated series is the enemy in disguise
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.
Termination quantises — and one chain must be killed by hand
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.
Name the polynomials, then audit one
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.
Change one variable at a time
Make the relationship visible.
Start at K = 1, the bare Gaussian, then step K by 0.05: the tails peel off the axis at once and only lie down again at K = 5. Flip parity to odd and the curve is pinned through zero at the origin, terminating at K = 3 and 7 instead.
ENERGY E0.50 ħω
n = (K − 1)/20.00
TAIL ψ(ξ = 4)0.00
DETUNING |K − (2n+1)|0.00
Live interpretationENERGY E: 0.50 ħω. n = (K − 1)/2: 0.00. TAIL ψ(ξ = 4): 0.00. DETUNING |K − (2n+1)|: 0.00
Catch the common trap
Explain before calculating.
Solving ψ″ = (ξ² − K)ψ with the substitution ψ = h(ξ)e(−ξ²/2) and a power series for h, what actually forces the discrete spectrum Eₙ = (n + ½)ħω?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySet K = 7 and run the odd chain from a₁ = 1. Show the series terminates, build the resulting polynomial, and confirm it is proportional to H₃ = 8ξ³ − 12ξ. What energy does this state carry?
- An odd state starts from a₁ = 1 with a₀ = 0. The recursion is aⱼ₊₂ = aⱼ(2j + 1 − K)/((j + 1)(j + 2)) with K = 7.
- j = 1: a₃ = a₁(3 − 7)/(2 × 3) = −4/6 = −2/3.
- j = 3: a₅ = a₃(7 − 7)/(4 × 5) = 0 — the numerator hits zero, and every later coefficient inherits it. The chain terminates.
- So h = ξ − (2/3)ξ³. Multiply by −12: −12h = 8ξ³ − 12ξ = H₃(ξ), so h ∝ H₃; the overall constant is fixed later by normalisation.
- Termination at j = n = 3 means K = 2n + 1 = 7, so E₃ = (3 + ½)ħω = 3.5 ħω.
Answerh ∝ H₃ = 8ξ³ − 12ξ: an odd, three-node state with K = 7 and E₃ = 3.5 ħω.
MediumA student proposes E = 1.2 ħω for an even bound state of the oscillator. Run the even chain far enough to show what goes wrong, compare its tail with the series for e(+ξ²), and locate the allowed even energies either side.
- K = 2E/ħω = 2 × 1.2 = 2.4. An even trial starts from a₀ = 1 with a₁ = 0.
- Run the recursion: a₂ = (1 − 2.4)/2 = −0.700; a₄ = a₂(5 − 2.4)/12 = −0.152; a₆ = a₄(9 − 2.4)/30 = −0.0334; a₈ = a₆(13 − 2.4)/56 = −0.00632. No numerator (2j + 1 − 2.4) can ever vanish: 2j + 1 runs through the odd integers and 2.4 is not one.
- Watch the ratio aⱼ₊₂/aⱼ = (2j + 1 − 2.4)/((j + 1)(j + 2)): at j = 6 it is 10.6/56 = 0.189; by j = 40 it is 78.6/1722 = 0.0456, closing on 2/(j + 2) = 0.0476.
- The banished exponential e(+ξ²) = Σ ξ(2m)/m! steps its coefficients by exactly 1/(m + 1) = 2/(j + 2) — the same tail. So h grows like e(+ξ²), and ψ = h⋅e(−ξ²/2) grows like e(+ξ²/2).
- A wavefunction growing as e(+ξ²/2) has a divergent norm, so E = 1.2 ħω supports no state. The nearest even terminations are K = 1 and K = 5: E₀ = 0.5 ħω and E₂ = 2.5 ħω.
AnswerNo — the chain never terminates, ψ grows as e(+ξ²/2), and the norm diverges. E = 1.2 ħω falls in the gap between the even eigenvalues E₀ = 0.5 ħω and E₂ = 2.5 ħω.
HardH³⁵Cl vibrates with ħω = 0.358 eV; the reduced mass is μ = 1.63 × 10⁻²⁷ kg. From the terminated n = 1 chain, construct the normalised eigenfunction, locate the peaks of its probability density in picometres, and give E₁ in electronvolts.
- n = 1 terminates the odd chain at once: a₃ = a₁(3 − 3)/6 = 0, so h = a₁ξ ∝ H₁ = 2ξ and ψ₁ = C ξ e(−ξ²/2), with ξ = x/x₀ and x₀ = √(ħ/μω).
- Normalise: ∫|ψ₁|² dx = C²x₀ ∫ ξ²e(−ξ²) dξ = C²x₀(√π/2) = 1, so C = √2/(π¹⁄⁴√x₀) — exactly the closed form ψ₁(x) = (μω/πħ)¹⁄⁴ √2 ξ e(−ξ²/2).
- The density |ψ₁|² ∝ ξ²e(−ξ²) has derivative 2ξ(1 − ξ²)e(−ξ²), zero at ξ = 0 (the node) and ξ = ±1: the peaks sit one natural length either side of centre, x = ±x₀.
- Numbers: ω = 0.358 × 1.602 × 10⁻¹⁹ J ÷ 1.055 × 10⁻³⁴ J s = 5.44 × 10¹⁴ s⁻¹, so x₀ = √(1.055 × 10⁻³⁴ ÷ (1.63 × 10⁻²⁷ × 5.44 × 10¹⁴)) = 1.09 × 10⁻¹¹ m = 10.9 pm.
- E₁ = (1 + ½)ħω = 1.5 × 0.358 = 0.537 eV. The peaks sit at ±10.9 pm — about 9% of the 127 pm bond length, small enough that truncating the true potential at its quadratic term is still defensible on this rung.
Answerψ₁(x) = (μω/πħ)¹⁄⁴√2 ξ e(−ξ²/2) with ξ = x/x₀; the density peaks at x = ±x₀ = ±10.9 pm with a node at centre, and E₁ = 0.537 eV.