University Physics IV · The Quantum Harmonic Oscillator · 8.4
The Ladder & Zero-Point Energy
Two numbers describe a vibrating bond: which rung it sits on, and how tall the rungs are. This lesson takes the ladder apart — why every rung is the same height when a box's rungs widen without limit, why the bottom rung floats half a quantum above the floor, and what would be measured differently if it did not.
Build the model
Connect the measurement to the mechanism.
The spectrum of a quadratic well is an arithmetic progression: Eₙ = (n + ½)ħω, with n counting quanta from zero and one frequency fixing everything. Two features of it are not obvious, and both have causes. The gaps are all equal because the parabola's turning points spread as √E, so the box holding the particle widens exactly fast enough to cancel the shortening wavelength; the phase-space area enclosed at energy E is 2πE/ω, linear in E, and the semiclassical count E/ħω is not an approximation here but exact.
An infinite well's walls do not move, its enclosed area grows only as √E, and its gaps therefore widen as (2n + 1)E₁. The offset ½ is not bookkeeping either. Set ⟨x⟩ = ⟨p⟩ = 0, put Δp = ħ/2Δx into ⟨E⟩ = (Δp)²/2m + ½mω²(Δx)², and the minimum over Δx is exactly ½ħω, reached at the width Δx = √(ħ/2mω) that the true ground state turns out to have.
A particle in a parabola cannot rest at the bottom, because resting there would fix position and momentum together. What all of this costs is the word strictly: both results are properties of x², and no real bond is x². Its gaps close as the levels climb, and it dissociates after a few dozen of them, where the parabola promises infinitely many equal ones.
- Simple definition
- The harmonic ladder is the spectrum Eₙ = (n + ½)ħω for n = 0, 1, 2, …: infinitely many levels separated by the same ħω, the lowest of them — the zero-point energy — sitting half a quantum above the bottom of the well rather than on it.
- Example
- CO absorbs at 2170 cm⁻¹, so ħω = 0.269 eV and its levels sit at 0.135, 0.404, 0.673 and 0.942 eV above the bottom of the well; cooled to absolute zero the bond still holds that first 0.135 eV, and every step up the ladder absorbs at the same 4.61 μm.
One frequency fixes the entire spectrum: there is no second parameter, and no top level.
ħ = 1.0546×10⁻³⁴ J s; k in N m⁻¹, m the reduced mass in kg; n counts quanta, not levels
A harmonic band is one line, not a comb — which is how anharmonicity announces itself.
CO: ħω = 0.269 eV = 2170 cm⁻¹, the same step from n = 0 as from n = 20
The lowest state sits half a rung above the floor, and its energy splits evenly between the two terms.
CO: E₀ = 0.135 eV = 2.16×10⁻²⁰ J, held at every temperature
Recovers ½ħω with no differential equation, and names the width that achieves it.
uses ΔxΔp ≥ ħ/2 with ⟨x⟩ = ⟨p⟩ = 0; Δx in m, E in J
Fixed walls make the gaps widen; the parabola's turning points move out as √E and hold them level.
electron in L = 0.50 nm: E₁ = 1.50 eV, then gaps of 4.51, 7.52, 10.53 eV
Turns the half-quantum into a measurement: D₀(D₂) − D₀(H₂) = 0.080 eV predicted, 0.078 eV observed.
H₂ and D₂ share one Dₑ = 4.747 eV; ωₑ = 4401 and 3112 cm⁻¹
Reading the whole spectrum off one number
Eₙ = (n + ½)ħω contains exactly one physical input, ω = √(k/m); the rest is arithmetic. Two habits of notation cause most of the errors. First, n counts quanta, not levels: n = 0 is the ground state, so what is loosely called the second excited state is n = 2, at 2.5ħω. Second, the sum runs to infinity — a parabola never stops rising, so every state is bound, the spectrum stays discrete all the way up, and nothing in the model can ever come apart. Put numbers on it with carbon monoxide. Its infrared band sits at 2170 cm⁻¹, and 1 cm⁻¹ = 1.2398×10⁻⁴ eV, so ħω = 0.269 eV. The reduced mass is μ = (12 × 16)/28 u = 6.86 u = 1.138×10⁻²⁶ kg, and ω = 2πc × 2170 cm⁻¹ = 4.09×10¹⁴ rad s⁻¹, so the force constant is k = μω² = 1.90×10³ N m⁻¹. The levels then sit at 0.135, 0.404, 0.673, 0.942 eV and onward, forever, above the bottom of the well.
Why the rungs stay level while a box's spread apart
Compare the two standard bound problems. In an infinite well the walls do not move: raising the energy only shortens the wavelength inside a fixed length L, so Eₙ = n²h²/8mL² and the gaps widen as (2n + 1)E₁ — for an electron in a 0.50 nm well, 4.51 eV, then 7.52, then 10.53. In a parabola the walls move with the particle. The classical turning point of level n is Aₙ = √((2n + 1)ħ/mω), which grows as √E, and that widening is exactly what cancels the shortening wavelength. The sharpest way to see it is to count states by phase-space area, since the number of levels below E is that area divided by h. The oscillator's orbit is an ellipse with semi-axes √(2E/mω²) and √(2mE), so its area is 2πE/ω — linear in E, hence one level per ħω, evenly. The box's orbit is a rectangle of area 2L√(2mE), which grows only as √E, so its levels thin out. Bohr–Sommerfeld makes it exact rather than suggestive: ∮p dx = (n + ½)h returns E = (n + ½)ħω for the oscillator, half-quantum and all.
Where the half-quantum comes from
Nothing in the argument for ½ħω needs the differential equation. In a symmetric well a stationary state has ⟨x⟩ = 0 and ⟨p⟩ = 0, so ⟨x²⟩ = (Δx)², ⟨p²⟩ = (Δp)², and ⟨E⟩ = (Δp)²/2m + ½mω²(Δx)². Squeezing the particle towards the bottom cuts the second term and inflates the first, because ΔxΔp ≥ ħ/2. Take the bound as an equality, put Δp = ħ/2Δx, and minimise E(u) = ħ²/8μ + ½mω²u over u = (Δx)²: dE/du = −ħ²/8mu² + ½mω² = 0 gives u = ħ/2mω. Substituting back, the kinetic term is ħω/4, the potential term is ħω/4, and Eₘᵢₙ = ½ħω. Two things follow. The bound is saturated — the true ground state has ⟨x²⟩ = ħ/2mω and ΔxΔp = ħ/2 exactly — so this variational estimate is the answer, not merely a floor under it. And the width is small without being zero: for CO, Δx = 3.37 pm against a bond length of 113 pm, a 3% wobble that no refrigerator removes.
The offset cannot be defined away
The usual objection is that ½ħω is a constant, and constants in energy are conventions. That would hold if it were the same constant everywhere, and it is not: it is ½ħ√(k/m), so anything that changes the stiffness or the mass moves it. Isotopic substitution is the cleanest case, because the Born–Oppenheimer electronic curve is fixed by nuclear charges, not nuclear masses. H₂ and D₂ therefore share one curve and one well depth, Dₑ = 4.747 eV, while μ(D₂) = 2μ(H₂) drops ωₑ from 4401 cm⁻¹ to 4401/√2 = 3112 cm⁻¹. Their zero-point energies are 0.273 eV and 0.193 eV, so the energies actually needed to break the bonds, D₀ = Dₑ − ½ħω, are 4.474 eV and 4.554 eV. Deuterium is bound by 0.080 eV more than hydrogen purely because it sits lower in the same well; the measured difference is 0.078 eV. That gap of a few tens of meV is what drives kinetic isotope effects in chemistry, and the isotope separations built on them.
Helium: a well too shallow for its own zero-point energy
The second piece of evidence is a substance that will not freeze. Two helium atoms attract with a pair potential only about 10 K deep — ε/kB ≈ 10.2 K, or 0.9 meV — with a Lennard-Jones σ of 2.56 Å. Localising an atom inside a cage that size costs a zero-point energy of order h²/mσ², and for ⁴He that scale is roughly seven times ε. The ratio is normally quoted as the de Boer parameter Λ = h/σ√(mε), which is 2.68 for ⁴He and 3.08 for the lighter ³He. Crystallising means localising, and here localising costs more than the binding returns, so ⁴He stays liquid at saturated vapour pressure all the way to absolute zero and needs roughly 25 atm of applied pressure to solidify at all. The comparison that settles it is neon: the same kind of atom with the same kind of potential, but Λ = 0.58, so its zero-point energy is a third of its well depth and it freezes at 24.6 K like anything else. Zero-point energy is not a small correction; in helium it decides the phase.
Where the even ladder stops being even
All of it rests on the potential being exactly quadratic, and no bond is. Real vibrational levels are fitted by G(v) = ωₑ(v + ½) − ωₑ xₑ(v + ½)², whose gaps shrink linearly: ΔG = ωₑ − 2ωₑ xₑ(v + 1). For HCl, ωₑ = 2991 cm⁻¹ and ωₑ xₑ = 52.8 cm⁻¹, so the fundamental is 2991 − 106 = 2885 cm⁻¹, while the first overtone lands at 5665 cm⁻¹ instead of twice the fundamental, 5771 cm⁻¹ — short by 106 cm⁻¹, or 1.8%, and that shortfall is the direct measurement of the anharmonicity. Extrapolating the shrinking gaps to zero, a Birge–Sponer plot, puts the last bound level near v = 27; the true count is smaller, because the Morse form over-extrapolates, but either way it is a few dozen and then a continuum, where the parabola promised infinitely many. So equal spacing is a claim about the bottom of a well: good for the lowest levels of a stiff bond, worse above them. The zero-point energy, being a property of the ground state, survives the failure far better than the equal spacing does.
Change one variable at a time
Make the relationship visible.
Leave the anharmonicity at zero and sweep the marked level: the arrow never changes length, and that is what evenly spaced means. Raise ħω and the shaded zero-point band grows with it, always half a rung. Then raise the anharmonicity and watch the upper rungs close in.
ZERO-POINT ENERGY E₀0.135 eV
MARKED LEVEL Eₙ0.135 eV
GAP Eₙ₊₁ − Eₙ0.270 eV
GAP AS A FRACTION OF ħω1.000
Live interpretationZERO-POINT ENERGY E₀: 0.135 eV. MARKED LEVEL Eₙ: 0.135 eV. GAP Eₙ₊₁ − Eₙ: 0.270 eV. GAP AS A FRACTION OF ħω: 1.000
Catch the common trap
Explain before calculating.
Infrared spectroscopy puts carbon monoxide's vibrational quantum at ħω = 0.269 eV. Treating the bond as a harmonic oscillator, what is the energy of the n = 2 level above the bottom of the well, and what is the least vibrational energy the molecule can have?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCarbon monoxide absorbs strongly in the infrared at a wavenumber of 2170 cm⁻¹. Treating the bond as a harmonic oscillator, find ħω in eV, the zero-point energy, the energy of the n = 3 level, the gap between n = 2 and n = 3, and the wavelength of a photon that drives one step up the ladder. Use 1 cm⁻¹ = 1.2398×10⁻⁴ eV and hc = 1240 eV nm.
- Convert the band position: ħω = 2170 cm⁻¹ × 1.2398×10⁻⁴ eV per cm⁻¹ = 0.2690 eV.
- Zero-point energy: E₀ = ½ħω = 0.5 × 0.2690 = 0.1345 eV. That belongs to the ground state n = 0, and it is held at every temperature.
- Third excited state: E₃ = (3 + ½)ħω = 3.5 × 0.2690 = 0.9416 eV above the bottom of the well.
- Gap: E₃ − E₂ = ħω = 0.2690 eV. Because the gap carries no n, the same answer serves for E₁ − E₀ or for the step from n = 20 to n = 21.
- Photon: λ = hc/ΔE = 1240 eV nm ÷ 0.2690 eV = 4610 nm = 4.61 μm — which is just 1/(2170 cm⁻¹) rewritten, as it had to be.
Answerħω = 0.269 eV; E₀ = 0.135 eV; E₃ = 0.942 eV; E₃ − E₂ = 0.269 eV; λ = 4.61 μm.
MediumA single ⁸⁸Sr⁺ ion of mass 1.46×10⁻²⁵ kg is held in a trap with ω = 2π × 1.00 MHz. Without solving the Schrödinger equation, use ⟨x⟩ = ⟨p⟩ = 0 and ΔxΔp ≥ ħ/2 to find the lowest mean energy the ion can have, the width Δx at which it occurs, and how that energy divides between kinetic and potential parts. Then quote the result as a temperature E/kB.
- With ⟨x⟩ = ⟨p⟩ = 0, ⟨x²⟩ = (Δx)² and ⟨p²⟩ = (Δp)², so ⟨E⟩ = (Δp)²/2m + ½mω²(Δx)². Take the bound as an equality, Δp = ħ/2Δx, and set u = (Δx)²: E(u) = ħ²/8μ + ½mω²u.
- Minimise: dE/du = −ħ²/8mu² + ½mω² = 0, so u² = ħ²/4m²ω² and u = ħ/2mω. Hence Δx = √(ħ/2mω).
- Substitute back: the kinetic term is ħ²/(8m · ħ/2mω) = ħω/4 and the potential term is ½mω² · ħ/2mω = ħω/4, so Eₘᵢₙ = ½ħω, split equally between them.
- Numbers: ω = 2π × 1.00×10⁶ = 6.283×10⁶ rad s⁻¹, so ħω = 1.0546×10⁻³⁴ × 6.283×10⁶ = 6.63×10⁻²⁸ J and Eₘᵢₙ = 3.31×10⁻²⁸ J = 2.07 neV.
- Width: Δx = √(1.0546×10⁻³⁴ ÷ (2 × 1.46×10⁻²⁵ × 6.283×10⁶)) = √(5.75×10⁻¹⁷ m²) = 7.58×10⁻⁹ m, about 7.6 nm.
- As a temperature: Eₘᵢₙ/kB = 3.31×10⁻²⁸ ÷ 1.381×10⁻²³ = 2.40×10⁻⁵ K = 24 μK. Cooling below that strips quanta, never the half, which is why ground-state cooling is quoted as a mean occupation rather than as zero energy.
AnswerEₘᵢₙ = ½ħω = 3.31×10⁻²⁸ J = 2.07 neV, at Δx = 7.6 nm, divided as ħω/4 kinetic and ħω/4 potential; as a temperature, 24 μK.
HardH₂ and D₂ move on the same Born–Oppenheimer electronic curve, of depth Dₑ = 4.747 eV measured from the bottom of the well. H₂ has ωₑ = 4401 cm⁻¹, and the reduced mass of D₂ is twice that of H₂. Predict ωₑ for D₂, the two zero-point energies and the two dissociation energies D₀ measured from the ground vibrational level, compare with the observed D₀(H₂) = 4.478 eV and D₀(D₂) = 4.556 eV, and say what the residual means. Use 1 cm⁻¹ = 1.2398×10⁻⁴ eV.
- The electronic curve is fixed by the charges, not the nuclear masses, so k and Dₑ are identical for the two molecules and only μ changes. With ω = √(k/μ) and μ(D₂) = 2μ(H₂), ωₑ(D₂) = 4401/√2 = 3112 cm⁻¹ — the observed value is 3116 cm⁻¹.
- Zero-point energies: ½ωₑ = 2200.5 cm⁻¹ for H₂ and 1556.0 cm⁻¹ for D₂, that is 0.2728 eV and 0.1929 eV.
- Dissociation from the ground level: D₀ = Dₑ − ½ħω. For H₂, 4.747 − 0.273 = 4.474 eV; for D₂, 4.747 − 0.193 = 4.554 eV.
- Difference: D₀(D₂) − D₀(H₂) = 0.2728 − 0.1929 = 0.0799 eV, about 0.080 eV or 7.7 kJ mol⁻¹. Deuterium is harder to pull apart only because it sits lower in a well of the same depth.
- Comparison: the observed difference is 4.556 − 4.478 = 0.078 eV, so the harmonic prediction is high by about 2 meV, and each absolute D₀ is low by a few meV.
- The residual is anharmonicity. The true zero-point energy is ½ωₑ − ¼ωₑ xₑ, a little below the harmonic value and by different amounts for the two isotopologues, since ωₑ xₑ also scales with μ. The half-quantum accounts for the isotope effect to within about 2%.
Answerωₑ(D₂) = 3112 cm⁻¹; zero-point energies 0.273 eV and 0.193 eV; D₀ = 4.474 eV and 4.554 eV, a difference of 0.080 eV against 0.078 eV observed, the 2 meV residual being anharmonicity.