University Physics IV · The Quantum Harmonic Oscillator · 8.2
The Oscillator Equation & Its Scales
Before solving anything, ask what ħ, m and ω can build. They make exactly one length and one energy, and dividing by them leaves a single dimensionless parameter — one equation that covers a trapped electron, a CO bond, and a gram on a spring.
Build the model
Connect the measurement to the mechanism.
Write the time-independent equation for V = ½mω²x² and inventory what is in it: ħ, m and ω, three constants built from three base dimensions, so by the counting of dimensional analysis there are n − k = 0 dimensionless groups to be formed. That single fact drives everything. With no dimensionless parameter available, the constants make exactly one length, x₀ = √(ħ/mω), one energy, ħω, and one momentum, ħ/x₀ = √(mħω), each unique up to a pure number — so every answer the problem can produce must be one of those times a number the equation itself supplies, not one the potential supplies.
Substituting ξ = x/x₀ collapses both coefficients to ħω/2 and leaves d²ψ/dξ² = (ξ² − ε)ψ with ε = 2E/ħω: one equation, one parameter, solved once for every oscillator that will ever be posed, from a 4.76 pm CO bond to a gram on a spring whose natural length is 10⁻¹⁶ m. What it costs is honesty about the potential. Taking the parabola as exact and infinitely deep makes the classically allowed region −√ε ≤ ξ ≤ √ε finite at every energy, however large, so ψ must decay on both sides for every E: every state is bound, the spectrum is discrete all the way up, and there is no continuum, no threshold and no dissociation anywhere in the model.
Real bonds have all three.
- Simple definition
- Non-dimensionalising the oscillator means measuring x in units of the natural length x₀ = √(ħ/mω) and E in units of ħω, which reduces the Schrödinger equation to d²ψ/dξ² = (ξ² − ε)ψ with the single parameter ε = 2E/ħω.
- Example
- For the CO bond, ω = 4.09×10¹⁴ rad s⁻¹ and reduced mass 6.86 u give x₀ = 4.76 pm and ħω = 0.269 eV, so the n = 2 level sits at ε = 5 with turning points at x₀√5 = 10.6 pm — under a tenth of the 112.8 pm bond.
Only ħ, m and ω appear, and no length or energy is imposed from outside — unlike the well, which is handed its width L.
m in kg, ω in rad s⁻¹, E in J; ψ carries m(−1/2) in one dimension
Nothing dimensionless can be formed, so every eigenvalue must be ħω times a pure number fixed by the equation alone.
n = 3 constants (ħ, m, ω) built from k = 3 base dimensions M, L, T
Electron at ω = 2.0×10¹⁵ rad s⁻¹: x₀ = 0.241 nm, p₀ = 4.38×10⁻²⁵ kg m s⁻¹, ħω = 1.32 eV.
x₀ in m, p₀ in kg m s⁻¹; x₀p₀ = ħ exactly, so the natural pair already sits at the uncertainty floor
One equation for every oscillator: solve it once in ξ, then convert back with x₀ and ħω.
ξ and ε dimensionless; both coefficients of the original equation became ħω/2
ε is the turning point squared: it marks where ψ stops oscillating and starts decaying, with no other input.
on the spectrum ε = 2n + 1, so ξₜₚ = √(2n+1), which is 1 at n = 0 and 3 at n = 4
Forget it and every expectation value comes out wrong by a power of x₀.
dx = x₀ dξ, so the x₀(−1/2) restores ψ's m(−1/2) units when you convert back
Count what is in the equation before solving it
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.
Where the natural length comes from
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, so the natural pair already sits at the uncertainty floor. 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 is the width at which confining and spreading cost the same, and ħω is what that compromise costs.
Substituting: dividing the constants out
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.
ε is the classical turning point, squared
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.
Why every state is bound and the spectrum stays discrete
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.
Reading real systems off the two scales
The scales answer "how quantum is this?" on their own. An electron in an optical trap 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 visible in an optical spectrum. 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.
Change one variable at a time
Make the relationship visible.
Hold n = 0 and drag m and ω: the parabola reshapes and the ticks move, but the dot always lands on the tick — ε = 1 means the ground-state turning point is exactly one natural length. Then climb n and watch the dot walk out as √(2n+1) while the ticks stay put.
NATURAL LENGTH x₀4.76 pm
ENERGY QUANTUM ħω0.263 eV
REDUCED ENERGY ε1
TURNING POINT x₀√ε4.76 pm
Live interpretationNATURAL LENGTH x₀: 4.76 pm. ENERGY QUANTUM ħω: 0.263 eV. REDUCED ENERGY ε: 1. TURNING POINT x₀√ε: 4.76 pm
Catch the common trap
Explain before calculating.
An electron and a proton sit in traps with the same ω = 1.0×10¹⁴ rad s⁻¹; the electron's x₀ is 1.08 nm. What are the proton's scales?
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 of angular frequency ω = 2.0×10¹⁵ rad s⁻¹. Find the natural length x₀, the natural momentum p₀ and the energy quantum ħω, then locate the classical turning point of the ground state. Take ħ = 1.0546×10⁻³⁴ J s and mₑ = 9.109×10⁻³¹ kg.
- Natural length: x₀ = √(ħ/mω) = √(1.0546×10⁻³⁴ ÷ (9.109×10⁻³¹ × 2.0×10¹⁵)) = √(5.789×10⁻²⁰ m²) = 2.406×10⁻¹⁰ m = 0.241 nm.
- Energy quantum: ħω = 1.0546×10⁻³⁴ × 2.0×10¹⁵ = 2.109×10⁻¹⁹ J = 1.32 eV, so consecutive rungs sit 1.32 eV apart.
- Natural momentum: p₀ = ħ/x₀ = 1.0546×10⁻³⁴ ÷ 2.406×10⁻¹⁰ = 4.38×10⁻²⁵ kg m s⁻¹, which equals √(mħω) and satisfies x₀p₀ = ħ exactly.
- Ground state: E₀ = ½ħω = 0.658 eV, so ε = 2E₀/ħω = 1 and ξₜₚ = √ε = 1. The classical turning point sits at exactly one natural length, xₜₚ = x₀ = 0.241 nm.
Answerx₀ = 0.241 nm, ħω = 1.32 eV, p₀ = 4.38×10⁻²⁵ kg m s⁻¹; at n = 0, ε = 1 and the turning point is x₀ itself, 0.241 nm.
MediumStarting from −(ħ²/2m)d²ψ/dx² + ½mω²x²ψ = Eψ, substitute x = x₀ξ with x₀ = √(ħ/mω) and reduce the equation to the form d²ψ/dξ² = (ξ² − ε)ψ. Identify ε, say what √ε measures, then evaluate both for a particle with E = 0.50 eV in a trap where ħω = 0.20 eV.
- Substitute x = x₀ξ. By the chain rule d/dx = (1/x₀)d/dξ, so d²ψ/dx² = (1/x₀²)d²ψ/dξ².
- Kinetic coefficient: ħ²/(2mx₀²) = (ħ²/2m)(mω/ħ) = ħω/2, using x₀² = ħ/mω.
- Potential coefficient: ½mω²x₀² = ½mω²(ħ/mω) = ħω/2. Both terms have collapsed to the same constant — that is exactly what choosing x₀ bought.
- The equation reads −(ħω/2)d²ψ/dξ² + (ħω/2)ξ²ψ = Eψ. Divide by ħω/2: d²ψ/dξ² = (ξ² − ε)ψ with ε = 2E/ħω. No trace of ħ, m or ω survives.
- Meaning of √ε: setting ½mω²x² = E gives xₜₚ/x₀ = √(2E/ħω) = √ε, so ε is the classical turning point squared, in natural lengths.
- Numbers: ε = 2 × 0.50 eV ÷ 0.20 eV = 5.0, so ξₜₚ = √5 = 2.24 — the classical excursion is 2.24 natural lengths. And 5 is odd: 5 = 2n + 1 with n = 2, so E = 2.5ħω is exactly the third rung.
Answerd²ψ/dξ² = (ξ² − ε)ψ with ξ = x/x₀ and ε = 2E/ħω, which equals ξₜₚ². For E = 0.50 eV with ħω = 0.20 eV, ε = 5.0 and ξₜₚ = 2.24 — the n = 2 level.
HardA 1.0 g bob on a spring with ω = 10 rad s⁻¹ oscillates with amplitude 1.0 mm. Find its natural length x₀ and energy quantum ħω, then the reduced energy ε and the quantum number n. Finally work out the fractional change in amplitude caused by one quantum, and say what that shows about the classical limit.
- Natural length: x₀ = √(ħ/mω) = √(1.0546×10⁻³⁴ ÷ (1.0×10⁻³ × 10)) = √(1.0546×10⁻³²) = 1.03×10⁻¹⁶ m — about an eighth of a proton's charge radius.
- Energy quantum: ħω = 1.0546×10⁻³⁴ × 10 = 1.05×10⁻³³ J = 6.58×10⁻¹⁵ eV.
- Reduced energy from the amplitude: the classical turning point is A and xₜₚ = x₀√ε, so ε = (A/x₀)² = (1.0×10⁻³ ÷ 1.0269×10⁻¹⁶)² = (9.74×10¹²)² = 9.48×10²⁵.
- Quantum number: ε = 2n + 1, so n = (ε − 1)/2 = 4.74×10²⁵. Cross-check classically: E = ½mω²A² = ½ × 1.0×10⁻³ × 100 × 1.0×10⁻⁶ = 5.0×10⁻⁸ J, and E/ħω = 5.0×10⁻⁸ ÷ 1.05×10⁻³³ = 4.74×10²⁵ = n + ½.
- One rung: ΔE/E = 1/(n + ½) = 2.11×10⁻²⁶. Since E ∝ A², ΔA/A = ½ΔE/E = 1.05×10⁻²⁶, so one quantum shifts the amplitude by 1.05×10⁻²⁹ m — fourteen orders of magnitude below a proton radius.
- Nothing about the equation has changed; it is still d²ψ/dξ² = (ξ² − ε)ψ. Only ε has gone from 1 to 10²⁶. The classical limit is a large value of one dimensionless number, not a different physics.
Answerx₀ = 1.03×10⁻¹⁶ m, ħω = 6.58×10⁻¹⁵ eV, ε = 9.48×10²⁵ and n = 4.74×10²⁵; one quantum shifts the amplitude by only 1.05×10⁻²⁹ m.