University Physics IV · The Quantum Harmonic Oscillator · 8.1
The Classical Oscillator & Harmonic Approximation
Before any wavefunction, the quadratic well has to be earned. Expand a real potential about its minimum, see what the leading term buys — a frequency from curvature alone, turning points read off the energy line, a density that piles up at the edges — and see exactly what the discarded cubic term takes with it.
Build the model
Connect the measurement to the mechanism.
Nothing in nature is a spring, yet almost everything behaves like one near equilibrium. Expand a smooth potential about a stable minimum x₀: the constant is only a datum, the linear term vanishes because U′(x₀) = 0, and the first surviving term is ½U″(x₀)ξ². That single number — the curvature — is the whole harmonic model.
It buys a linear restoring force, an angular frequency ω = √(U″(x₀)/m) that no amplitude can shift, turning points at ±√(2E/k) read straight off the energy line, and a probability density 1/(π√(A²−x²)) that is smallest at the centre and piles up at the edges because that is where the particle is slowest. The cost is everything beyond ξ². The cubic term carries the asymmetry that lets a bond stretch more easily than it compresses, that makes solids expand when heated, and that finally lets a molecule dissociate — none of which a parabola can do.
So the harmonic approximation is not the claim that a well is quadratic; it is the claim that ξ stays small enough for |U‴₀|ξ/(3U″₀) to remain well below one. For HCl at its zero-point amplitude that ratio is already about 0.2, which is exactly how much anharmonicity Unit 8 will have to apologise for later.
- Simple definition
- The harmonic approximation replaces a potential near a stable minimum by the leading quadratic term of its Taylor series, ½U″(x₀)ξ², so the system becomes a mass on a spring whose stiffness is the curvature of the curve.
- Example
- For HCl the bond curvature at the minimum is U″ = 480 N m⁻¹, so with μ = 1.627 × 10⁻²⁷ kg the model gives ω = 5.43 × 10¹⁴ rad s⁻¹, or 2.88 × 10³ cm⁻¹ — within 0.1% of the observed 2886 cm⁻¹ fundamental.
The constant is only a datum and the linear term is identically zero, so the quadratic term is the entire leading model.
ξ = x − x₀ in m; U′(x₀) = 0 defines the minimum; U″₀ has units J m⁻² = N m⁻¹
One second derivative fixes the frequency; no other feature of the curve enters at leading order.
k in N m⁻¹, m in kg, ω in rad s⁻¹; the point is a minimum only if U″(x₀) > 0
Energy fixes the amplitude, and the period ignores it — the isochronism the even quantum ladder inherits.
E in J, A in m, T in s; the turning points are where the energy line cuts U(x)
Time per unit length goes as 1/|v|, so the particle is least likely at the centre and piles up at the turning points.
P in m⁻¹, normalised over −A ≤ x ≤ A; its minimum value is P(0) = 1/(πA)
Turns the density into something a stopwatch can check: 41% of every period is spent in the outer fifth of the range.
Dimensionless, x₁ ≤ A. Putting x₁ = 0.8A gives 0.590, so 0.410 lies outside.
The test that decides whether the harmonic model may be used, and it contains no system size — only derivatives.
a in m⁻¹, Dₑ in J. The force-level ratio |U‴₀|ξ/(2U″₀) is 1.5 times larger.
Why every smooth minimum begins as a parabola
Take any smooth potential U(x) with a stable equilibrium at x₀ and write ξ = x − x₀. Taylor's theorem gives U = U(x₀) + U′(x₀)ξ + ½U″(x₀)ξ² + ⅙U‴(x₀)ξ³ + … . The constant only says where you count energy from, and the linear term is identically zero because equilibrium is defined by U′(x₀) = 0. So the first term that does any physics is ½U″(x₀)ξ², and if the curvature is positive the force F = −dU/dξ = −U″(x₀)ξ is linear and restoring. Check the units: U″ is joules per metre squared, which is newtons per metre — a stiffness. Nothing about springs was assumed anywhere. Take U(x) = 5.00x⁴ − 20.0x² + 12.0 J: U′ = 20.0x(x² − 2) puts minima at x = ±1.414 m, and U″ = 60.0x² − 40.0 gives 80.0 N m⁻¹ there, so a 0.400 kg particle oscillates at √(80.0/0.400) = 14.1 rad s⁻¹ even though the curve is nowhere a pure parabola.
Energy sets the amplitude; it does not touch the period
Once U is ½kξ², the equation mξ̈ = −kξ has the solution ξ = A cos(ωt + φ) with ω = √(k/m), and the total energy is E = ½kA² at every instant. Read that backwards. The turning points sit at ξ = ±A = ±√(2E/k), exactly where a horizontal energy line cuts the parabola, and the particle can never be found beyond them because the kinetic energy E − U(ξ) would have to be negative. Raise E and the turning points move out as √E. What does not move is the period, T = 2π√(m/k), which contains no A at all: a 0.250 kg glider on a 40.0 N m⁻¹ spring takes 0.497 s per cycle whether the amplitude is 1 mm or 16 cm. That isochronism belongs to the quadratic potential alone, and it is the classical shadow of the evenly spaced ladder Eₙ = (n + ½)ħω this unit is heading for.
Dwell time, not passage count, sets the density
Ask where a photograph taken at a random instant would catch the particle. The chance of finding it in a strip dx is the fraction of the period spent there: two passes per cycle, each lasting dx/|v|, so P(x) = 2/(T|v|). With |v| = ω√(A² − x²) and T = 2π/ω the ω cancels and P(x) = 1/(π√(A² − x²)). It is smallest at the centre, P(0) = 1/(πA), and it diverges at ±A — integrably, since the integral from −A to A is exactly 1. Integrating gives the working form t(|x| < x₁)/T = (2/π)arcsin(x₁/A). For A = 0.158 m, P(0) = 2.01 m⁻¹, and the outer fifth of the range, |x| > 0.8A, holds the particle for 1 − (2/π)arcsin(0.800) = 41.0% of every period. The particle is fast in the middle, so it is rarely caught there.
What the discarded cubic term was doing
A parabola is symmetric, and that symmetry is a lie about every real bond. Keep the next term, U = ½kξ² + ⅙U‴₀ξ³ with U‴₀ negative for a bond, and three things change at once. The well becomes easier to stretch than to compress, so the time-averaged position drifts outward as the energy rises — that drift is thermal expansion, and a strictly harmonic solid does not expand when heated at all. The level spacing stops being constant, so real vibrational ladders converge as the quantum number climbs instead of stepping by a fixed ħω. And the curve finally flattens to a dissociation limit, which a parabola never does: a harmonic bond cannot be broken by any finite energy. The harmonic model is therefore excellent for the frequency, serviceable for the zero-point energy, and worthless for dissociation.
How small is small: the number that decides
The honest criterion compares terms, not lengths. The kept term is ½U″₀ξ² and the dropped one is ⅙U‴₀ξ³, so their ratio is |U‴₀|ξ/(3U″₀); the matching force-level ratio |U‴₀|ξ/(2U″₀) is 1.5 times larger. A Morse curve makes the arithmetic painless. Expanding U = Dₑ[1 − e(−aξ)]² gives U = Dₑ(a²ξ² − a³ξ³ + …), so k = 2Dₑ a² and U‴₀ = −6Dₑ a³, and the potential-level ratio collapses to just aξ. For H³⁵Cl with k = 480 N m⁻¹ and Dₑ = 4.62 eV, a = √(k/2Dₑ) = 1.80 × 10¹⁰ m⁻¹, and the classical turning point of a vibration carrying ½ħω is A = 10.9 pm. So aA = 0.197: the amplitude is only 8.6% of the 127.5 pm bond, yet the anharmonic term is already a fifth of the harmonic one.
What the classical picture hands to the quantum one
Everything else in Unit 8 is built on the potential this topic produces, U = ½mω²x², whose only inputs are m and ω. Two scales follow at once and both are classical in origin. The energy ħω is the step the ladder will take, and the length √(ħ/mω) is the width of the ground state — and that length is exactly the classical turning point of a particle carrying ½ħω, since A = √(2E/k) with E = ½ħω and k = mω² collapses to √(ħ/mω). For HCl it is 10.9 pm by either route. The classical density P(x) is the second bequest: it is the target that |ψₙ|² must average to as n grows, which is how the correspondence principle gets tested here. What the classical picture gets wrong is equally sharp — it permits any E ≥ 0, it forbids |x| > A absolutely, and it lets the particle sit still at the bottom. The quantum solution overturns all three.
Change one variable at a time
Make the relationship visible.
Drag the probe out towards ±A: the speed readout collapses while the density readout climbs, and at 0.6A the particle already spends 59% of each period further out still. Raising E pushes the turning points apart; raising k pulls them in and shortens the period.
AMPLITUDE A0.632 m
SPEED AT PROBE0.876 m/s
DENSITY AT PROBE0.629 /m
TIME BEYOND PROBE0.590 of T
Live interpretationAMPLITUDE A: 0.632 m. SPEED AT PROBE: 0.876 m/s. DENSITY AT PROBE: 0.629 /m. TIME BEYOND PROBE: 0.590 of T
Catch the common trap
Explain before calculating.
A classical particle oscillates in U = ½kx² with fixed total energy and turning points at ±A. A photograph is taken at a random instant. Where is the particle most likely to be found, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle of mass 0.400 kg moves in the potential U(x) = 5.00x⁴ − 20.0x² + 12.0 J, with x in metres. Locate the equilibria and classify them, find the small-oscillation angular frequency and period about a stable one, and give the turning points if the particle is released from rest 0.050 m from that minimum.
- U′(x) = 20.0x³ − 40.0x = 20.0x(x² − 2), so the equilibria are at x = 0 and x = ±√2 = ±1.414 m.
- U″(x) = 60.0x² − 40.0. At x = 0 it is −40.0 N m⁻¹, an unstable maximum; at x = ±1.414 m it is 60.0(2) − 40.0 = +80.0 N m⁻¹, so both outer points are stable minima.
- The harmonic approximation sets k = U″₀ = 80.0 N m⁻¹, so ω = √(80.0/0.400) = √200 = 14.1 rad s⁻¹ and T = 2π/ω = 0.444 s.
- Released from rest at ξ = 0.050 m the amplitude is A = 0.050 m, so the oscillation energy is ½kA² = ½(80.0)(0.050)² = 0.100 J and the turning points sit at ξ = ±0.050 m — that is x = 1.364 m and x = 1.464 m about the right-hand minimum.
AnswerStable minima at x = ±1.414 m with k = 80.0 N m⁻¹; ω = 14.1 rad s⁻¹ and T = 0.444 s; E = 0.100 J with turning points 0.050 m either side, at x = 1.364 m and 1.464 m.
MediumA 0.250 kg glider on a spring of stiffness 40.0 N m⁻¹ carries total energy 0.500 J. Find its amplitude and period, the classical probability density at the centre of the motion, and the fraction of each period it spends in the outer fifth of its range, |x| > 0.8A.
- Amplitude from the energy line: E = ½kA², so A = √(2E/k) = √(2 × 0.500/40.0) = √0.0250 = 0.158 m.
- ω = √(k/m) = √(40.0/0.250) = √160 = 12.65 rad s⁻¹, so T = 2π/ω = 0.497 s. Note that A never enters — the period is the same at any energy.
- Classical density P(x) = 1/(π√(A² − x²)). At the centre P(0) = 1/(πA) = 1/(π × 0.1581) = 2.01 m⁻¹, which is its smallest value anywhere in the range.
- Time inside |x| < x₁ is (2/π)arcsin(x₁/A) of a period. With x₁ = 0.8A: (2/π)arcsin(0.800) = (2/π)(0.9273) = 0.590, so the outer fifth takes 1 − 0.590 = 0.410 of every period.
AnswerA = 0.158 m, T = 0.497 s (independent of energy), P(0) = 2.01 m⁻¹, and 41.0% of each period is spent in the outer 20% of the range.
HardH³⁵Cl has bond force constant k = 480 N m⁻¹, reduced mass μ = 1.627 × 10⁻²⁷ kg and dissociation depth Dₑ = 4.62 eV, and its bond potential is modelled by the Morse curve U(ξ) = Dₑ[1 − e(−aξ)]². Find the harmonic wavenumber, the Morse range parameter a, and how large the discarded cubic term is at the classical turning point of a vibration carrying the zero-point energy ½ħω.
- ω = √(k/μ) = √(480/1.627 × 10⁻²⁷) = √(2.950 × 10²⁹) = 5.43 × 10¹⁴ rad s⁻¹, so ν̃ = ω/(2πc) = 5.43 × 10¹⁴ ÷ (2π × 2.998 × 10¹⁰ cm s⁻¹) = 2.88 × 10³ cm⁻¹, within 0.1% of the observed 2886 cm⁻¹ fundamental.
- Expanding the Morse form gives U = Dₑ(a²ξ² − a³ξ³ + …), so k = 2Dₑ a² and U‴₀ = −6Dₑ a³. With Dₑ = 4.62 eV = 7.40 × 10⁻¹⁹ J, a = √(k/2Dₑ) = √(480/1.480 × 10⁻¹⁸) = 1.80 × 10¹⁰ m⁻¹.
- Zero-point energy: E₀ = ½ħω = ½(1.055 × 10⁻³⁴)(5.43 × 10¹⁴) = 2.86 × 10⁻²⁰ J = 0.179 eV.
- Its classical turning point: A = √(2E₀/k) = √(5.73 × 10⁻²⁰ ÷ 480) = 1.09 × 10⁻¹¹ m = 10.9 pm — only 8.6% of the 127.5 pm bond length.
- Cubic against quadratic in U: |U‴₀|A/(3k) = 6Dₑ a³A ÷ (6Dₑ a²) = aA = (1.80 × 10¹⁰)(1.09 × 10⁻¹¹) = 0.197. At force level the ratio is 1.5aA = 0.295.
Answerν̃ ≈ 2.88 × 10³ cm⁻¹, a = 1.80 × 10¹⁰ m⁻¹, and aA = 0.197 — the cubic term is already about a fifth of the quadratic one at the zero-point turning point, which is why HCl's vibrational spacings converge.