Eₙ = (n + ½)ℏω and zero-point energy

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 — norm positivity 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.

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: in this model CO absorbs at 2170 cm⁻¹ from n = 0 and at exactly the same frequency from n = 1, so hot bands lie on top of the fundamental and only anharmonicity separates them. Real CO absorbs at 2143 cm⁻¹ from n = 0 and about 27 cm⁻¹ lower from n = 1, and that splitting, 2ω̃ₑx̃ₑ, is a direct read of the anharmonic terms. Every superposition is periodic, since |ψ(t)⟩ = e^(−iωt/2) Σ cₙ e^(−inωt)|n⟩ returns to the same physical state after 2π/ω — the ket itself comes back as −|ψ(0)⟩, the sign being the ½ — 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.

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.

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.

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 rarely 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.

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 for perfect mirrors; real metals give less at so small a gap, and the force, once their finite conductivity is included, agrees with measurement. 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.