The harmonic oscillator spectrum by ladder operators

Statement

The energy spectrum of the quantum harmonic oscillator is evenly spaced and starts at half a quantum above the classical minimum — obtained entirely from operator algebra, with no differential equation solved.

Why it matters

By D-003, every system near stable equilibrium is a set of oscillators. So this spectrum is the spectrum of almost everything: molecular vibrations, phonons in PU-303, photons in PU-402. The operators â and â^† introduced here are later reinterpreted as annihilating and creating particles, and that reinterpretation is the conceptual core of quantum field theory.

Assumptions

Derivation

Result

Limiting cases

• Large n: spacing ℏω becomes small relative to E_n, and the discrete ladder blurs into the continuous classical energy — the correspondence limit.
• Ground state n = 0: energy is not zero but ½ℏω, the smallest excursion the uncertainty relation permits.
• ℏ → 0: the ½-quantum offset and the level spacing both vanish, recovering the classical oscillator with a continuous energy.

Breaks when

Failure modes

• Treating â and â^† as observables. Neither is Hermitian; neither corresponds to anything measurable. Only N̂ = â^†â does.
• Assuming ⟨x⟩ oscillates in an energy eigenstate. It is zero in every one. Oscillation requires a superposition — which is why coherent states, not eigenstates, are the classical limit.
• Dismissing the zero-point term as a constant offset. It is measurable: the Casimir effect and the persistence of liquid helium at absolute zero both depend on it.

Worked number

The CO molecule vibrates at ω ≈ 4.08 × 10¹⁴ rad s⁻¹. Level spacing ℏω ≈ 4.3 × 10⁻²⁰ J ≈ 0.27 eV, corresponding to infrared radiation near 4.6 μm — which is exactly where CO absorbs, and why it is detectable by infrared spectroscopy. At room temperature k_BT ≈ 0.026 eV, roughly ten times smaller than the spacing, so essentially every molecule sits in n = 0. That single comparison is the whole explanation of why vibrational modes are "frozen out" of the heat capacity — cashed in at D-020.

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Discussion

The whole content of the result E_n = (n + ½)ℏω is packed into two features that look mundane and are not. First, the levels are evenly spaced: the gap between n and n+1 is ℏω for every n. Second, the ladder rests on a pedestal ½ℏω that no choice of coordinate origin can remove. Even spacing is the reason we are allowed to stop thinking about "the n-th excited state" and start thinking about "n identical quanta, each carrying ℏω." Because every quantum costs the same, they are indistinguishable bookkeeping tokens rather than distinct rungs — and that reinterpretation, forced on us by the equal spacing, is precisely what turns one oscillator into a container for identical bosons.

Why is the spectrum forced to have exactly this shape? Trace it back to the single input beyond the Hamiltonian: [x̂, p̂] = iℏ. That one relation makes [â, â^†] = 1, which makes the raising operator a rigid step — [Ĥ, â^†] = +ℏωâ^† says it moves the energy up by exactly ℏω, never more, never less, independent of where you are on the ladder. Uniform spacing is therefore not a coincidence of the parabola; it is the operator statement that the parabola's steepness is the same everywhere. The half-quantum has the same origin: it is the leftover ½ in Ĥ = ℏω(â^†â + ½) that appears only because â and â^† refuse to commute. Set ℏ → 0 and both the spacing and the pedestal vanish together — the classical continuum is recovered, and the discreteness is revealed as a pure interference effect of order ℏ.

The deepest reading is that this derivation is a template, not a special case. Any Hamiltonian you can factor as (something)^†(something) plus a constant, with the two factors obeying a commutator that closes on a number, has an equally spaced spectrum obtained by the identical argument. Angular momentum falls to the same method (Ĵ_± are ladder operators for m); supersymmetric quantum mechanics generalises the factorisation to a whole family of solvable potentials; and the promotion of â, â^† from operators-on-one-particle to operators-that-create-and-destroy-particles is the entire conceptual move of second quantisation. The number operator N̂ = â^†â, whose eigenvalues we proved must be non-negative integers, becomes the occupation number of a field mode: a photon in PU-402 is one rung of one such ladder. The zero-point ½ℏω, summed over the infinitely many modes of the electromagnetic field, is the vacuum energy whose differences are the measurable Casimir force.

One subtlety worth stating precisely: the eigenstates |n⟩ are stationary, so nothing about them oscillates — ⟨x̂⟩ = 0 in every one of them (see below). The classical, visibly-swinging oscillator is not any eigenstate but the coherent state |α⟩, the eigenstate of the (non-Hermitian) lowering operator â|α⟩ = α|α⟩. It is a Poisson superposition of number states, saturates the uncertainty bound ∆x∆p = ℏ/2 for all time, and its centroid traces the classical ellipse in phase space. So the ladder we built is the quantum skeleton; the classical motion is a particular dressing of it.

Common misconceptions.

• n is a count of quanta, not an amplitude. Doubling n does not double a displacement; it adds one more identical energy packet.
• ⟨x̂⟩ = 0 and ⟨p̂⟩ = 0 in every energy eigenstate, because x̂ and p̂ are linear in â and â^†, which change n by one and so have zero diagonal matrix element. Nothing wiggles in a stationary state.
• The half-quantum is not a removable constant. Its mode-summed value shifts with boundary conditions, and that shift is the Casimir effect; it is also why liquid helium does not freeze under its own vapour pressure at T = 0.
• Because x̂ is linear in the ladder operators, it connects only neighbouring rungs: ⟨m|x̂|n⟩ ≠ 0 only if m = n ± 1. This Δn = ±1 selection rule is why a harmonic mode absorbs and emits at the single frequency ω and nowhere else.

Worked examples

Example 1 — Thermal population of the CO stretch: why it is "frozen out." Carbon monoxide vibrates at ω = 4.08 × 10¹⁴ rad s⁻¹. Treating the stretch as a single quantum oscillator in equilibrium at temperature T, find the mean number of quanta n̄, the fraction of molecules in the ground state, and the mean vibrational energy, at T = 300 K and at T = 1500 K. The population of rung n follows the Boltzmann weight P_n ∝ e^−E_n/k_BT.

Example 2 — Zero-point spread of the CO bond, and the virial split, from the ladder operators. For the same molecule (reduced mass μ = m_Cm_O/(m_C+m_O) = 6.86 u = 1.139 × 10⁻²⁶ kg), find the root-mean-square stretch of the bond in the ground state, express it as a fraction of the 113 pm equilibrium bond length, and show that the ground-state energy splits equally between kinetic and potential parts. Invert step 1 to write position in terms of the ladder operators.

Problems

1. A vibrational mode has level spacing ℏω = 0.50 eV. (a) What is its zero-point energy? (b) What is the energy, and vacuum wavelength, of the photon emitted in the transition n = 3 → 2?
   Solution

   (a) E₀ = ½ℏω = 0.25 eV. (b) Because the spectrum is evenly spaced, every single-step transition releases exactly ℏω = 0.50 eV, regardless of which rungs it joins (this is why the mode has one absorption/emission frequency). Wavelength λ = hc/E = (1240 eV·nm)/(0.50 eV) = 2480 nm — mid-infrared.
2. A nanomechanical resonator oscillates at f = 1.0 GHz and is cooled to T = 10 mK. (a) Find ℏω and the zero-point energy. (b) Find the mean phonon number n̄ and comment on whether the resonator is in its quantum ground state.
   Solution

   (a) ω = 2πf = 6.28 × 10⁹ rad s⁻¹; ℏω = hf = (6.626 × 10⁻³⁴)(10⁹) = 6.63 × 10⁻²⁵ J. Zero-point energy ½ℏω = 3.31 × 10⁻²⁵ J (≈ 2.1 μeV). (b) x = ℏω/k_BT = 6.63 × 10⁻²⁵/(1.381 × 10⁻²³ × 0.010) = 4.80; n̄ = 1/(e^4.80 − 1) = 1/120 = 0.0083. With fewer than one phonon on average per hundred cycles, the resonator spends ≈ 99.2% of its time in n = 0 — close to, but not perfectly in, the quantum ground state, which is exactly the regime achieved in cavity-optomechanics ground-state-cooling experiments.
3. Using only â|n⟩ = √n |n−1⟩ and â^†|n⟩ = √(n+1) |n+1⟩, evaluate (a) â^†â|3⟩, (b) ââ^†|3⟩, (c) ⟨2|â^†2|0⟩. Then use (a) and (b) to read off [â, â^†] acting on |3⟩.
   Solution

   (a) â|3⟩ = √3 |2⟩, then â^†√3|2⟩ = √3√3 |3⟩ = 3|3⟩. So â^†â|3⟩ = 3|3⟩ = N̂|3⟩ ✓. (b) â^†|3⟩ = √4 |4⟩, then â√4|4⟩ = √4√4 |3⟩ = 4|3⟩. (c) â^†|0⟩ = |1⟩; â^†|1⟩ = √2 |2⟩; so â^†2|0⟩ = √2 |2⟩ and ⟨2|â^†2|0⟩ = √2. Finally (ââ^† − â^†â)|3⟩ = (4 − 3)|3⟩ = |3⟩, confirming [â, â^†] = 1.
4. A molecular oscillator has ℏω = 0.10 eV and is held at T = 300 K (k_BT = 0.0259 eV). Find the mean occupation n̄, the mean energy ⟨E⟩, and the fraction of molecules in an excited state. Compare qualitatively with the CO result of Example 1.
   Solution

   x = ℏω/k_BT = 0.100/0.0259 = 3.87. n̄ = 1/(e^3.87 − 1) = 1/(47.9 − 1) = 0.0213. ⟨E⟩ = ℏω(½ + n̄) = 0.100 × 0.521 = 0.0521 eV. Fraction excited = 1 − P₀ = e^−x = e^−3.87 = 0.021, i.e. about 2%. Because this mode is softer than CO (0.10 eV vs 0.27 eV), x is smaller and the mode is less completely frozen — a few percent of molecules are already vibrationally excited at room temperature, so this mode begins to contribute to the heat capacity where the CO stretch does not.
5. Using x̂ = √(ℏ/2mω)(â + â^†) and p̂ = i√(mℏω/2)(â^† − â), prove that in the eigenstate |n⟩ the energy splits as ⟨T̂⟩ = ⟨V̂⟩ = ½E_n (the virial theorem), and evaluate the uncertainty product ∆x ∆p. Which state saturates the Heisenberg bound?
   Solution

   By the same argument as Example 2, the off-diagonal â², â^†2 terms vanish in a number state, leaving ⟨x̂²⟩ = (ℏ/2mω)(2n+1) and ⟨p̂²⟩ = (mℏω/2)(2n+1). Then ⟨V̂⟩ = ½mω²⟨x̂²⟩ = ¼ℏω(2n+1) = ½(n+½)ℏω = ½E_n, and ⟨T̂⟩ = ⟨p̂²⟩/2m = ¼ℏω(2n+1) = ½E_n as well — equal split, as the classical virial theorem for a quadratic potential also predicts. Since ⟨x̂⟩ = ⟨p̂⟩ = 0, the variances equal the mean squares, so ∆x ∆p = √(⟨x̂²⟩⟨p̂²⟩) = √[(ℏ/2mω)(mℏω/2)] (2n+1) = (n + ½)ℏ. The ground state n = 0 gives ∆x ∆p = ℏ/2 — it saturates the Heisenberg bound and is a minimum-uncertainty state; every excited state has a strictly larger product.