University Physics V · The Quantum Harmonic Oscillator · 7.4

Eₙ = (n + ½)ℏω & Zero-Point Energy

Two claims share one formula, and they need separating. The gap ℏω is what a spectrometer reads; the ½ is what it never reads. Derive the floor from the commutator instead of asserting it, then meet the experiments — isotope shifts, Casimir plates — where the half-quantum finally moves a dial.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Eₙ = (n + ½)ℏω welds two claims together, and they carry different evidential weight. The first is the spacing: because Ĥ = ℏω(â†â + ½) and â†â has the integer spectrum the ladder argument forced, every rung sits exactly ℏω above the last — no n², no −1/n², no crowding at a limit — which is why a harmonic bond absorbs in one band rather than a series, and why the ½ never appears in a transition frequency. The second is the floor.

Setting â|0⟩ = 0 gives ⟨0|Ĥ|0⟩ = ℏω/2 in one line, but that is no artefact of how the factorisation was ordered: minimise ⟨Ĥ⟩ = σₚ²/2m + ½mω²σₓ² over every normalised state subject only to σₓ σₚ ≥ ℏ/2 and ℏω/2 comes back, attained at σₓ = √(ℏ/2mω) and σₚ = √(mℏω/2) with the kinetic and potential shares equal. Non-commutation, not the ladder, is what forbids sitting still at the bottom of the well. The cost is sharp.

For one oscillator at fixed ω the ½ is a c-number you may legally subtract, invisible in every commutator and every spectrum. What you may not do is call it a constant, because ½ℏ√(k/μ) moves whenever the force constant, the reduced mass or a boundary condition moves — which is why deuterium shifts a dissociation energy by tens of meV, and why the same ½ summed over the field modes between two mirrors is a pressure you can measure.

Simple definition: The oscillator's energy eigenvalues are Eₙ = (n + ½)ℏω: an infinite ladder of equally spaced rungs whose lowest, the zero-point energy ℏω/2, lies above the bottom of the well and cannot be lowered at fixed ω.
Example: CO's stretch has ω̃ = 2170 cm⁻¹, so ℏω = 269 meV and E₀ = 134.5 meV. Since ℏω/k_B = 3.12×10³ K, at 300 K only 3 molecules in 10⁵ sit above the floor — and every one of the rest still carries that 134.5 meV.

The eigenvalue, read off the ladderĤ = ℏω(â†â + ½) ⇒ Eₙ = (n + ½)ℏω, n = 0, 1, 2, …

One line fixes the whole spectrum: a ladder with no top, and a floor that is not the bottom of the well.

ω = √(k/m) in rad s⁻¹, ℏ = 1.055×10⁻³⁴ J s; every level non-degenerate in one dimension

Uniform spacingEₙ₊₁ − Eₙ = ℏω, for every n

A harmonic bond shows one fundamental band, not a series: CO absorbs at 2170 cm⁻¹ from every rung alike.

Contrast the infinite well, Eₙ ∝ n², and the Coulomb well, Eₙ ∝ −1/n²

Zero-point energyE₀ = ⟨0|ℏω(â†â + ½)|0⟩ = ℏω/2, since â|0⟩ = 0

It comes from the ordering inside [â, â†] = 1, the same commutator that built the ladder.

Equivalently ½ℏ√(k/μ) — a function of force constant and reduced mass, not a constant

The same floor from uncertainty alone⟨Ĥ⟩ = σₚ²/2m + ½mω²σₓ² ≥ ω σₓ σₚ ≥ ℏω/2

No ladder assumed: non-commutation by itself forbids rest at the well bottom, and ⟨T̂⟩ = ⟨V̂⟩ = ℏω/4.

AM–GM, then Robertson; equality at σₓ = √(ℏ/2mω), σₚ = √(mℏω/2)

Isotope shift of a bond energyD₀ = Dₑ − ½ℏω, ω = √(k/μ) ⇒ D₀(D₂) − D₀(H₂) = ½ℏ(ω_H − ω_D)

The whole 0.078 eV by which D₂ outbinds H₂ is half-quantum — the ½ weighed, not inferred.

Same electrons, so k and Dₑ are unchanged; ω̃ falls 4401 → 3116 cm⁻¹ from H₂ to D₂

The same ½ summed over field modesP = π²ℏc / (240 d⁴)

Σ ½ℏωₖ diverges, but its dependence on a boundary condition is finite and measured.

Ideal mirrors at T = 0, gap d in metres; 13 Pa at d = 100 nm, 1.3×10⁵ Pa at 10 nm

Read the spectrum straight off â†â

The whole result is one substitution. The factorisation gave Ĥ = ℏω(â†â + ½), and the ladder argument gave the number operator â†â the spectrum n = 0, 1, 2, … . Since Ĥ is ℏω times that operator plus a c-number, the two share eigenvectors, and acting on |n⟩ returns Ĥ|n⟩ = ℏω(n + ½)|n⟩. So Eₙ = (n + ½)ℏω, in the same basis (|n⟩), with no differential equation anywhere and no boundary condition imposed by hand — normalisability did that work when the ladder was forced to terminate. Two things the algebra does not supply are worth naming now. It does not prove that each rung is one-dimensional: non-degeneracy follows from a separate theorem about normalisable solutions of a second-order ODE in one dimension, and it fails outright in three, where the isotropic oscillator's level n carries (n+1)(n+2)/2 states. And it says nothing about whether this quadratic Ĥ is the right Hamiltonian — that question belongs to the cubic term you dropped at the minimum.

Even rungs: what a constant gap buys

Eₙ₊₁ − Eₙ = ℏω for every n, and that uniformity is unusual. An electron in a 0.30 nm infinite well has E₁ = 4.18 eV and gaps of 3E₁ = 12.5 eV, then 5E₁ = 20.9 eV, widening without limit; hydrogen's gaps instead shrink like 1/n² − 1/(n+1)² and pile up at the ionisation limit. Only the oscillator repeats itself. Three consequences follow. A harmonic bond has a single fundamental band: CO absorbs at 2170 cm⁻¹ from n = 0 and, in this model, at exactly the same frequency from n = 1, so hot bands lie on top of the fundamental and only anharmonicity separates them — the few cm⁻¹ of splitting you actually measure is a direct read of the cubic term. Every superposition is strictly periodic, since |ψ(t)⟩ = e^(−iωt/2) Σ cₙ e^(−inωt)|n⟩ returns to itself after 2π/ω, which is what later lets a coherent state follow a classical orbit without spreading. And in that expression the ½ appears only as a global phase, so no probability can see it.

The floor, and exactly how arbitrary it is

Put the ground state into the Hamiltonian: ⟨0|Ĥ|0⟩ = ℏω⟨0|â†â|0⟩ + ½ℏω, and â|0⟩ = 0 kills the first term, leaving E₀ = ℏω/2. Now ask what that number is worth. Add a constant C to any Hamiltonian and no eigenvector, no commutator, no transition frequency and no probability changes; a global phase e^(−iCt/ℏ) is the whole effect. So for a single oscillator at fixed ω you may write Ĥ′ = ℏω â†â and lose nothing, and in that strict sense E₀ = ℏω/2 is a choice of zero. What makes it physics is that it is not a constant. Written out it is ½ℏ√(k/μ), a function of the force constant and the reduced mass, so anything that changes either changes the floor by a definite, calculable amount — and differences of floors are precisely what an experiment can reach. Keep the two statements apart: an absolute zero-point energy is a convention, a difference of zero-point energies is data.

Uncertainty returns the same ℏω/2, with no ladder

Suppose you had never met â. Take any normalised state and write ⟨Ĥ⟩ = ⟨p̂²⟩/2m + ½mω²⟨x̂²⟩, then split each second moment into a spread and a mean: ⟨Ĥ⟩ = σₚ²/2m + ⟨p̂⟩²/2m + ½mω²σₓ² + ½mω²⟨x̂⟩². The two mean terms are non-negative, so drop them — displacing the state or setting it moving only costs energy. For what remains, A + B ≥ 2√(AB) gives σₚ²/2m + ½mω²σₓ² ≥ ω σₓ σₚ, and Robertson gives σₓ σₚ ≥ ℏ/2, so ⟨Ĥ⟩ ≥ ℏω/2 for every vector in the space. Both inequalities tighten at once when σₚ = mωσₓ and σₓ σₚ = ℏ/2, which forces σₓ = √(ℏ/2mω) and σₚ = √(mℏω/2), with kinetic and potential each holding ℏω/4. The ground state is therefore not a state that happens to have this energy; it is the minimiser. Squeeze σₓ and σₚ must grow, and the potential energy saved is more than repaid in kinetic energy.

How big is ℏω/2, and where it stops mattering

Put numbers on both ends of the scale. CO's stretch at ω̃ = 2170 cm⁻¹ gives ℏω = 269 meV and E₀ = 134.5 meV. In temperature units that is ℏω/k_B = 3.12×10³ K, so at 300 K the mean occupancy is 1/(e^10.41 − 1) = 3.0×10⁻⁵: essentially every molecule in the room sits on the ground rung, and every one of them carries 134.5 meV that no cooling can remove. The spread that energy corresponds to is σₓ = √(ℏ/2μω) = 3.37 pm for μ = 6.856 u — about 3% of the 112.8 pm bond, which is why zero-point motion matters for hydrogen-bond geometry and tunnelling rates but never for the shape of a molecule. At the other end, a 2 s pendulum has ω = π rad s⁻¹, so its rungs are ℏω = 3.3×10⁻³⁴ J apart and its floor is 1.0×10⁻¹⁵ eV, while a 1 mJ swing sits at n = 3×10³⁰. The ladder is still there; nothing in a laboratory resolves a step that fine.

Weighing the half-quantum

Three places where the ½ stops being invisible. Isotopes: H₂ and D₂ share one Born–Oppenheimer curve, so they share k and the well depth Dₑ, yet their measured bond energies differ by 0.078 eV — which is, to a quarter of a percent, the difference of their zero-point energies. That is the ½ on a balance. Condensed matter: helium's van der Waals well is so shallow that ℏω/2 exceeds it, and ⁴He will not solidify at any temperature below about 25 atmospheres. Fields: every mode of the electromagnetic field is one of these oscillators, so the vacuum between two mirrors carries Σ ½ℏωₖ. That sum diverges, but its dependence on the gap does not, and P = π²ℏc/240d⁴ predicts 13 Pa at 100 nm, which is measured. Where the accounting still fails is the absolute total: cut the same sum off at the Planck scale and it exceeds the observed vacuum energy density by some 120 orders of magnitude. Differences are trustworthy; the absolute is an open problem.

Change one variable at a time

Make the relationship visible.

Interactive model

Reduced mass μ0.93 u

Lower rung n of the marked gap0

Slide μ from 0.93 u (a C–H bond) to 1.72 u (the same bond deuterated) and watch the parabola sit still while every rung drops: ω̃ falls from about 3020 to 2220 cm⁻¹ and the floor from 187 to 138 meV. Then move n and check that the gap arrow never changes length.

LEVEL Eₙ187 meV

GAP Eₙ₊₁ − Eₙ375 meV

ZERO-POINT E₀187 meV

WAVENUMBER ω̃3021 cm⁻¹

Live interpretationLEVEL Eₙ: 187 meV. GAP Eₙ₊₁ − Eₙ: 375 meV. ZERO-POINT E₀: 187 meV. WAVENUMBER ω̃: 3021 cm⁻¹

Catch the common trap

Explain before calculating.

Misconception check

The half-quantum is neither an additive constant you may quietly set to zero nor stored energy you may extract, and both errors come from reading ½ℏω as a number instead of as ½ℏ√(k/μ). For one oscillator at fixed ω it really is a c-number: it cancels from every transition frequency, commutes with everything, and enters e^{(−iĤt/ℏ)} only as a global phase, so subtracting it changes no prediction. Change the parameters and it moves — deuterating a C–H bond drops the floor by about 50 meV, and moving two mirrors changes the mode sum between them. Nor is it fuel: ℏω/2 is the minimum of ⟨Ĥ⟩ over every normalised state, so no process at fixed ω takes a system below it. And it is not thermal jitter. At T = 0 the oscillator is in the pure state |0⟩, and ⟨x̂²⟩ ≠ 0 there because [x̂, p̂] = iℏ, not because anything is being kicked.

A C–H stretch absorbs at 3000 cm⁻¹. Deuteration leaves the electronic curve, and so k and Dₑ, unchanged. What happens to the band and to D₀?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyCarbon monoxide's stretching mode has ω̃ = 2170 cm⁻¹. Find the level spacing ℏω and the zero-point energy E₀ in meV and in joules, find the temperature at which k_BT equals ℏω, and say what fraction of the CO molecules in a 300 K room sit above the ground rung.

Convert once: 1 cm⁻¹ = 0.12398 meV, so ℏω = 2170 × 0.12398 = 269.0 meV = 4.311×10⁻²⁰ J. This is the gap between every neighbouring pair of rungs, not just the first.
E₀ = ½ℏω = 134.5 meV = 2.155×10⁻²⁰ J. The absorption sits at ℏω, i.e. λ = 1/2170 cm = 4.61 μm in the infrared, and that line carries no information about the ½.
ℏω/k_B = 4.311×10⁻²⁰ / 1.381×10⁻²³ = 3.12×10³ K. Room temperature is a fortieth of that, so this mode is deep in the frozen regime.
Boltzmann on the ladder: n̄ = 1/(e^(ℏω/k_BT) − 1) with ℏω/k_BT = 3122/300 = 10.41, so e^10.41 = 3.3×10⁴ and n̄ = 3.0×10⁻⁵. Three molecules in 10⁵ are excited; the other 99.997% sit on the ground rung — each still carrying 134.5 meV.

Answerℏω = 269 meV = 4.31×10⁻²⁰ J; E₀ = 134.5 meV = 2.16×10⁻²⁰ J; ℏω/k_B = 3.12×10³ K; n̄ = 3.0×10⁻⁵ at 300 K. Cooling to absolute zero would remove the 0.003% of excitation and none of the 134.5 meV.

MediumWithout using the ladder at all, show that ⟨Ĥ⟩ ≥ ℏω/2 for every normalised state of the oscillator, find the widths that attain the bound and the kinetic and potential shares there, then evaluate the ground-state position spread for CO (μ = 6.856 u, ω = 4.088×10¹⁴ rad s⁻¹, bond length 112.8 pm).

Split each second moment into a spread and a mean: ⟨Ĥ⟩ = (σₚ² + ⟨p̂⟩²)/2m + ½mω²(σₓ² + ⟨x̂⟩²) ≥ σₚ²/2m + ½mω²σₓ², with equality only when ⟨x̂⟩ = ⟨p̂⟩ = 0. Displacing the state or setting it moving can only cost more.
Apply A + B ≥ 2√(AB) to the two surviving terms: σₚ²/2m + ½mω²σₓ² ≥ 2√(ω²σₓ²σₚ²/4) = ω σₓ σₚ.
Now apply Robertson, σₓ σₚ ≥ ℏ/2, giving ⟨Ĥ⟩ ≥ ℏω/2. No differential equation and no ladder entered — only [x̂, p̂] = iℏ.
Equality needs both steps tight: σₚ = mωσₓ from AM–GM and σₓ σₚ = ℏ/2 from Robertson. Together mωσₓ² = ℏ/2, so σₓ = √(ℏ/2mω) and σₚ = √(mℏω/2). Then ⟨T̂⟩ = σₚ²/2m = ℏω/4 and ⟨V̂⟩ = ½mω²σₓ² = ℏω/4 — an equal split.
CO: μ = 6.856 × 1.6605×10⁻²⁷ = 1.138×10⁻²⁶ kg, so 2μω = 9.307×10⁻¹² and σₓ = √(1.0546×10⁻³⁴ / 9.307×10⁻¹²) = √(1.133×10⁻²³) = 3.37×10⁻¹² m.

Answer⟨Ĥ⟩ ≥ ℏω/2, saturated by the Gaussian with σₓ = √(ℏ/2mω), σₚ = √(mℏω/2) and ⟨T̂⟩ = ⟨V̂⟩ = ℏω/4. For CO, σₓ = 3.37 pm — 3.0% of the 112.8 pm bond, and it does not shrink however far the gas is cooled.

HardH₂ has ω̃ₑ = 4401 cm⁻¹. Deuterium substitution leaves the Born–Oppenheimer curve, and hence k and the well depth Dₑ, untouched. Predict ω̃ₑ(D₂), then predict the difference in dissociation energies D₀ and compare with the measured D₀(H₂) = 4.478 eV and D₀(D₂) = 4.556 eV. Take μ = 0.5039 u for H₂ and 1.0071 u for D₂, 1 cm⁻¹ = 1.2398×10⁻⁴ eV, and anharmonicity constants ω̃ₑx̃ₑ = 121.3 cm⁻¹ (H₂) and 61.8 cm⁻¹ (D₂).

Same k, different μ, so ω ∝ 1/√μ. The mass ratio is 1.0071/0.5039 = 1.9985, hence ω̃ₑ(D₂) = 4401/√1.9985 = 4401/1.4137 = 3113 cm⁻¹, against a measured 3116 cm⁻¹ — the 0.1% gap is the breakdown of the clamped-nuclei approximation, not of the scaling.
Harmonic floors, using the measured wavenumbers: ½ω̃ₑ = 2201 cm⁻¹ for H₂ and 1558 cm⁻¹ for D₂. Since D₀ = Dₑ − E₀ with a common Dₑ, D₀(D₂) − D₀(H₂) = 2201 − 1558 = 643 cm⁻¹ = 0.0797 eV.
Measured: 4.556 − 4.478 = 0.078 eV, i.e. 629 cm⁻¹. The harmonic prediction is about 2% high, and the reason is that the true floor is G(0) = ½ω̃ₑ − ¼ω̃ₑx̃ₑ + …, with a larger correction for the lighter, more anharmonic H₂.
Correct both floors: H₂ gives 2201 − 30.3 = 2170 cm⁻¹ and D₂ gives 1558 − 15.5 = 1542 cm⁻¹, a difference of 628 cm⁻¹ = 0.0779 eV against the measured 0.078 eV — agreement to a quarter of a percent.
Read the result: the two molecules have identical electronic binding, so the entire 0.078 eV by which D₂ is the stronger bond is a difference of half-quanta. The ½ that cancels out of every spectrum has just been weighed.

Answerω̃ₑ(D₂) ≈ 3113 cm⁻¹ (measured 3116). The harmonic zero-point difference is 643 cm⁻¹ = 0.080 eV; with the first anharmonic term it is 628 cm⁻¹ = 0.0779 eV, against a measured D₀ split of 0.078 eV.