University Physics IV · The Quantum Harmonic Oscillator · 8.6

Eigenfunctions, Nodes & the Forbidden Region

You have the spectrum twice over — series and ladder. Now learn to read the states themselves: count nodes to get n, use parity and orthogonality to kill integrals before computing them, and take seriously the 16% of the ground state that lives where classical mechanics says nothing can. And stop expecting an eigenstate to move.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The first five topics of this unit fixed the spectrum twice over; this one looks at the states themselves, and almost everything worth knowing is visible in a plot. Each ψₙ is a single Gaussian envelope e^{(−x²/2x₀²)}, with x₀ = √(ħ/mω), reshaped by a Hermite polynomial that can add wiggles but can never beat the envelope's decay. The polynomial's degree shows up as n interior zeros, so the plot labels itself: count nodes and you have the energy.

Alternating even and odd symmetry follows from the symmetric potential, orthogonality from the hermiticity of Ĥ, and together they turn the ladder into a basis in which any state is a list of amplitudes. The price of eigenstate-hood is a threefold strangeness. The density stretches past the classical turning points ±√(2n+1)x₀ — 15.7% of ground-state position measurements land where a classical oscillator of that energy cannot go.

The ground state piles up at the centre, precisely where a classical oscillator spends the least time. And nothing moves: the time dependence is one overall phase, so |Ψₙ|² is frozen and an oscillator eigenstate does not oscillate. Classical motion has not been lost — it is hiding in superpositions of rungs, which is exactly where the next topic goes looking for it.

Simple definition: The oscillator eigenfunctions are Hermite-polynomial-times-Gaussian states with n nodes and parity (−1)ⁿ, orthogonal across the ladder, frozen in time, and extending measurably past the classical turning points ±√(2n+1)x₀.
Example: For the ground state the turning points sit at x = ±x₀ = ±√(ħ/mω), and P = erfc(1) = 0.157: about one position measurement in six finds the particle where a classical oscillator with energy ½ħω can never be.

Normalised eigenfunctionsψₙ(x) = (mω/πħ)^¼ (2ⁿn!)^(−½) Hₙ(x/x₀) e^{(−x²/2x₀²)}

Every state is one Gaussian envelope reshaped by a polynomial: the envelope guarantees normalisability, and Hₙ can only add structure inside it.

x₀ = √(ħ/mω) is the oscillator length in m; Hₙ is the degree-n Hermite polynomial, dimensionless.

Nodes and parityψₙ has exactly n nodesψₙ(−x) = (−1)ⁿ ψₙ(x)

Count the zeros on a plot and you have read off n and Eₙ with no equation; parity kills ⟨x⟩ in every eigenstate and half of all dipole integrals.

A node is an interior zero with a sign change; parity (−1)ⁿ follows because Hₙ has only even or only odd powers.

Orthonormality∫ ψₘ(x) ψₙ(x) dx = δₘₙ

Makes the ladder a basis: cₙ = ∫ψₙΨ dx extracts each amplitude, |cₙ|² is the probability of measuring Eₙ, and expectation values carry no cross terms.

δₘₙ = 1 if m = n, else 0; ψ carries m^(−½), so the integral is dimensionless. The ψₙ are real, so no conjugate is needed.

Classical turning points½mω²xₙ² = Eₙ ⟹ xₙ = ±√(2n+1) x₀

Beyond them V − E > 0 and ψ'' has the same sign as ψ: oscillation gives way to decay, and the inflection points of ψₙ sit exactly on the turning points.

Where V(x) reaches Eₙ = (n+½)ħω and a classical oscillator of that energy stops; xₙ in m, growing as √(2n+1).

Ground-state leakagePₒᵤₜ = (2/√π) ∫₁^∞ e^(−ξ²) dξ = erfc(1) ≈ 0.157

One position measurement in six finds the ground-state particle where classical mechanics forbids it — for any m and ω, since the fraction is scale-free in x₀.

Both tails counted; ξ = x/x₀. For n = 1 the fraction is 0.112, for n = 2 it is 0.095 — falling toward the classical zero.

Stationary densityΨₙ(x, t) = ψₙ(x) e^{(−iEₙt/ħ)} ⟹ |Ψₙ|² = |ψₙ(x)|²

No eigenstate moves. Anything that oscillates — ⟨x⟩ swinging at ω — needs at least two rungs, which is where the coherent states of topic 8.8 come in.

Eₙ/ħ is a phase rate in rad s⁻¹; it cancels in the modulus squared, and with it in every probability.

One Gaussian, reshaped n times

Every eigenfunction is built from the same two parts: ψₙ(x) = (mω/πħ)^¼ (2ⁿn!)^(−½) Hₙ(ξ) e^(−ξ²/2), with ξ = x/x₀ and x₀ = √(ħ/mω) the only length in the problem. The Gaussian envelope is non-negotiable — it is the large-distance behaviour that survives normalisation, and it decays faster than any polynomial grows, so the Hermite factor can reshape the state but never unbind it. H₀ = 1, H₁ = 2ξ, H₂ = 4ξ² − 2, H₃ = 8ξ³ − 12ξ: each new degree adds one zero and one sign change inside the envelope. Get a feel for x₀ from carbon monoxide: reduced mass μ = 1.14 × 10⁻²⁶ kg and ω = 4.04 × 10¹⁴ rad s⁻¹ give x₀ = √(ħ/μω) = 4.8 pm, about 4% of the 113 pm bond length. The ground-state vibration is a small, stiff tremble about the minimum — which is precisely why the harmonic approximation of topic 8.1 works at all.

Count the nodes and you have read the energy

ψₙ has exactly n interior zeros — a theorem for one-dimensional bound states, not an accident of this potential. The physics behind the count is curvature: ψ''/ψ = −2m(E − V)/ħ², so a higher rung bends more sharply, and more bending between the same decaying tails forces more crossings. That makes the node count a free measurement: hand someone a plotted oscillator state with three zeros and they know n = 3 and E = 3.5 ħω without solving anything. Parity comes along too. Because V(−x) = V(x), each Hₙ carries only even or only odd powers, so ψₙ(−x) = (−1)ⁿψₙ(x), alternating up the ladder. That sign is a workhorse: ⟨x⟩ = 0 in every eigenstate because x|ψₙ|² is odd, and ⟨m|x̂|n⟩ vanishes whenever m and n share a parity — half of all dipole integrals die before pen touches paper, and the ladder algebra of topic 8.5 then cuts the survivors down to Δn = ±1.

Orthogonality makes the ladder a basis

Eigenstates of a Hermitian operator with different eigenvalues are orthogonal, so ∫ψₘψₙ dx = δₘₙ across the whole ladder: Gaussians times Hermite polynomials of different degree integrate exactly to zero against each other. That promotes the ψₙ from a list of solutions to a coordinate system — any normalisable Ψ is Σ cₙψₙ with cₙ = ∫ψₙΨ dx, each amplitude extracted by one overlap integral with no interference from the others. Measurement statistics then become bookkeeping: |cₙ|² is the probability of finding Eₙ, and expectation values pick up no cross terms. For Ψ = (ψ₀ + ψ₁)/√2, ⟨E⟩ = ½(½ħω) + ½(3/2)ħω = ħω exactly — and no single measurement ever returns it, because ħω sits between rungs. Orthogonality also audits your algebra for free: a superposition whose |cₙ|² do not sum to 1 was mis-normalised, no further inspection required.

At the turning points, the wavefunction changes character

A classical oscillator with energy Eₙ stops where the potential absorbs all of it: ½mω²x² = (n+½)ħω gives xₙ = ±√(2n+1) x₀ — at ±x₀ for the ground state, at ±√3 x₀ ≈ ±1.73 x₀ for n = 1. The eigenfunction registers the same boundary. Inside, E > V makes ψ''/ψ < 0, and the state curves back toward the axis: oscillation, nodes, structure. Outside, E < V flips the sign of ψ''/ψ and the state can only curve away from the axis and decay, with no further zeros; the turning points are precisely the inflection points of ψₙ. The far tail dies as e^(−ξ²/2) — faster than the plain e^(−κx) tails of unit 7's flat barriers, because this barrier keeps steepening forever. So the penetration is real but short: measured in x₀, everything interesting about the forbidden-region density happens within about one oscillator length of the classical edge.

Sixteen percent where no classical particle can be

For the ground state the forbidden fraction is one integral: P = 2⋅(1/√π)∫₁^∞ e^(−ξ²) dξ = erfc(1) = 0.157, or 0.079 per tail. Nothing in it depends on m or ω — measured in x₀, every ground state and its turning points look identical, so an electron in a quantum dot and a proton in a nuclear well leak the same 16%. Climbing the ladder shrinks the fraction: 0.112 for n = 1, 0.095 for n = 2, falling toward the classical zero. And there is no energy paradox in the tail. Kinetic energy is ⟨p̂²⟩/2m, a property of the whole state; E − V(x) at a point is a classical decomposition the theory does not offer. Try to certify the particle beyond x₀ and the measurement confining it to a tail of width about x₀ forces a momentum spread Δp ≳ ħ/2x₀, injecting energy of the order of the deficit it was meant to expose. The forbidden-region density is measurable — the same physics as the evanescent tail behind the STM — but 'the particle sat there with negative kinetic energy' is not a statement quantum mechanics makes.

Stationary means stationary

Attach the time dependence and each eigenstate becomes Ψₙ(x, t) = ψₙ(x)e^{(−iEₙt/ħ)}. The exponential is one global phase: it cancels in |Ψₙ|², in every probability, and in the expectation of every time-independent operator. The n = 5 state is not a particle sloshing five times harder than n = 1 — nothing in it moves at all, which is why these are called stationary states and why an atom parked in one does not radiate. The disagreement with classical mechanics is sharpest at the bottom of the ladder: a classical oscillator's time-averaged density, 1/(π√(xₘₐₓ² − x²)), piles up at the turning points where it moves slowest and is smallest at the centre, while |ψ₀|² peaks at the centre and falls monotonically — the two pictures are opposites, and no single eigenstate reconciles them. Motion reappears only when rungs are mixed: in (ψ₀e^(−iωt/2) + ψ₁e^(−3iωt/2))/√2 the cross term beats at exactly (E₁ − E₀)/ħ = ω, ⟨x⟩ swings sinusoidally, and the classical frequency re-emerges as a difference of phase rates — the door topic 8.8 walks through.

Change one variable at a time

Make the relationship visible.

Interactive model

Quantum number n0

Probe position ξ1.4 x₀

Step n from 0 to 3 and count zeros — one new node per rung, parity alternating. Then park the probe just outside a turning point: V − E turns positive there and the curve decays instead of oscillating. Note the n = 0 peak at ξ = 0, exactly where a classical oscillator spends the least time.

ENERGY Eₙ0.5 ħω

TURNING POINT ξₜₚ1.00 x₀

NODES0

V − E AT PROBE0.48 ħω

Live interpretationENERGY Eₙ: 0.5 ħω. TURNING POINT ξₜₚ: 1.00 x₀. NODES: 0. V − E AT PROBE: 0.48 ħω

Catch the common trap

Explain before calculating.

Misconception check

The word oscillator misleads: students picture the n = 3 state as a particle swinging back and forth at frequency ω, faster or wider than in n = 1. No eigenstate does anything of the kind. The full time dependence of Ψₙ is a single overall phase e^{(−iEₙt/ħ)}, which cancels in |Ψₙ|², so the probability density of every eigenstate is rigorously frozen — and by parity ⟨x⟩ = 0 at every instant, so there is no swing hiding in the averages either. The frequency ω never appears in an eigenstate as motion; it appears as a level spacing. Motion needs at least two rungs: superpose ψₙ and ψₙ₊₁ and the interference term beats at exactly (Eₙ₊₁ − Eₙ)/ħ = ω, which is where the classical frequency has been hiding all along. Saying 'the particle in ψ₂ vibrates at ω' is not a harmless simplification — it attributes to the state a time dependence it provably does not have.

An oscillator sits in its ground state, energy ½ħω, with classical turning points at ±x₀. What is the probability that a position measurement finds the particle beyond them, and how does that fraction behave for higher eigenstates?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is held in a harmonic trap with level spacing ħω = 3.0 eV, roughly a small quantum dot. Find the oscillator length x₀, then analyse ψ₁: its node, its parity, and the most probable positions of the electron.

Convert the spacing to an angular frequency: ω = (3.0 × 1.602 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 4.56 × 10¹⁵ rad s⁻¹.
x₀ = √(ħ/mω) = √(1.055 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 4.56 × 10¹⁵)) = √(2.54 × 10⁻²⁰ m²) = 1.59 × 10⁻¹⁰ m ≈ 0.16 nm.
ψ₁ ∝ ξ e^(−ξ²/2) with ξ = x/x₀: one interior zero at x = 0 with a sign change — one node, confirming n = 1 — and ψ₁(−x) = −ψ₁(x), odd parity, which gives ⟨x⟩ = 0 with no integral.
Most probable positions maximise |ψ₁|² ∝ ξ²e^(−ξ²): the derivative 2ξ(1 − ξ²)e^(−ξ²) vanishes at ξ = ±1, so the peaks sit at x = ±x₀ ≈ ±0.16 nm — well inside the classical turning points at ±√3 x₀ ≈ ±0.28 nm.

Answerx₀ = 0.159 nm. ψ₁ has one node at x = 0, odd parity, and most probable positions x = ±x₀ ≈ ±0.16 nm, inside the turning points at ±0.28 nm.

MediumA molecular vibration with ħω = 0.20 eV is prepared in the state Ψ = (ψ₀ + 2ψ₂)/√5. What are the possible outcomes of an energy measurement and their probabilities, what is ⟨E⟩, and does ⟨x⟩ ever differ from zero?

Orthonormality does the normalisation check: ⟨Ψ|Ψ⟩ = (1² + 2²)/5 = 1, with the cross term ∫ψ₀ψ₂ dx = 0 contributing nothing.
An energy measurement can only return rungs present in the expansion: E₀ = ½ħω = 0.10 eV with probability |c₀|² = 1/5 = 0.20, and E₂ = (5/2)ħω = 0.50 eV with |c₂|² = 4/5 = 0.80.
⟨E⟩ = 0.20 × 0.10 + 0.80 × 0.50 = 0.020 + 0.400 = 0.42 eV — a weighted average that is never itself an outcome, since 0.42 eV = 2.1 ħω sits between rungs.
For ⟨x⟩, every needed matrix element is an odd integral: ψ₀ and ψ₂ are both even and x̂ is odd, so ⟨0|x̂|0⟩ = ⟨2|x̂|2⟩ = ⟨0|x̂|2⟩ = 0. The oscillating cross phase e^{(−i(E₂−E₀)}t/ħ) multiplies a zero, so ⟨x⟩ = 0 at every instant.
The state is still not stationary: ⟨0|x̂²|2⟩ ≠ 0 because x̂² is even and connects Δn = 2, so the width ⟨x²⟩ breathes at (E₂ − E₀)/ħ = 2ω. Mixing rungs restores time dependence — just not, with these parities, in ⟨x⟩.

AnswerOutcomes 0.10 eV (P = 0.20) and 0.50 eV (P = 0.80); ⟨E⟩ = 0.42 eV; ⟨x⟩ = 0 for all time by parity, though the density's width oscillates at 2ω.

HardShow that the n = 1 turning points sit at ±√3 x₀ and find the probability that a position measurement lands beyond them. Use ∫ₐ^∞ ξ²e^(−ξ²) dξ = (a/2)e^(−a²) + (√π/4)erfc(a), with e⁻³ = 0.0498 and erfc(√3) = 0.0143. Compare with the ground state's 0.157.

Turning points: ½mω²x² = E₁ = (3/2)ħω gives x² = 3ħ/mω = 3x₀², so xₜₚ = ±√3 x₀ ≈ ±1.73 x₀ — further out than the ground state's ±x₀, because the rung is higher.
Normalised density in ξ = x/x₀: |ψ₁|² dx = (2/√π) ξ²e^(−ξ²) dξ. Check the normalisation: (2/√π)∫₋∞^∞ ξ²e^(−ξ²) dξ = (2/√π)(√π/2) = 1.
By symmetry take twice the right-hand tail: P₁ = 2 · (2/√π) ∫_(√3)^∞ ξ²e^(−ξ²) dξ = (4/√π)[(√3/2)e⁻³ + (√π/4)erfc(√3)].
Evaluate the two pieces: (4/√π)(√3/2)e⁻³ = (2√3/√π)(0.0498) = 1.954 × 0.0498 = 0.0973, and (4/√π)(√π/4)erfc(√3) = erfc(√3) = 0.0143.
P₁ = 0.0973 + 0.0143 = 0.112. That is below the ground state's erfc(1) = 0.157: the higher rung leaks a smaller fraction, and the sequence 0.157, 0.112, 0.095, … falls toward the classical zero as n grows.

Answerxₜₚ = ±√3 x₀; P₁ = (2√3/√π)e⁻³ + erfc(√3) = 0.112, down from 0.157 for n = 0 — each rung up is proportionally more classical.