The Bogoliubov transformation, phonons and depletion

Units are SI throughout; the gas is non-relativistic, so no metric signature enters. Topic 15.07 left the dilute gas, after a₀ → √N₀, as the Hamiltonian quoted here, H = gN²/(2V) + Σ′[(eₖ + gn)a†ₖaₖ + (gn/2)(a†ₖa†₋ₖ + a₋ₖaₖ)], with eₖ = ħ²k²/(2m), g = 4πħ²aₛ/m and the prime excluding k = 0. Every term conserves momentum, so the modes fall into independent pairs (k, −k), but the gn terms move atoms out of the condensate, or back, two at a time, and no state of definite aₖ occupation is an eigenstate. The cure is to mix adding an atom at k with removing one at −k: aₖ = uₖbₖ − vₖb†₋ₖ, with uₖ and vₖ real and even in k. Then [aₖ, a†ₖ] = uₖ²[bₖ, b†ₖ] + vₖ²[b†₋ₖ, b₋ₖ] = uₖ² − vₖ², while [aₖ, a₋ₖ] = −uₖvₖ + uₖvₖ = 0, so the b's are bosons exactly when uₖ² − vₖ² = 1. Writing uₖ = cosh θₖ and vₖ = sinh θₖ builds that in. The b's do not conserve atom number, and need not: the replacement a₀ → √N₀ already made the condensate a reservoir.

Substitute into one pair's Hamiltonian, (eₖ + gn)(a†ₖaₖ + a†₋ₖa₋ₖ) + gn(a†ₖa†₋ₖ + a₋ₖaₖ), and normal-order the b's. The terms in b₋ₖbₖ and its conjugate carry the coefficient gn cosh 2θₖ − (eₖ + gn) sinh 2θₖ, so they vanish when tanh 2θₖ = gn/(eₖ + gn). Then cosh 2θₖ = (eₖ + gn)/Eₖ and sinh 2θₖ = gn/Eₖ, with Eₖ = √((eₖ + gn)² − (gn)²) = √(eₖ(eₖ + 2gn)), and the number terms collect (eₖ + gn) cosh 2θₖ − gn sinh 2θₖ = Eₖ. The result is H = E₀′ + Σ′Eₖb†ₖbₖ, with E₀′ = gN²/(2V) + ½Σ′(Eₖ − eₖ − gn) and vₖ² = sinh²θₖ = ½[(eₖ + gn)/Eₖ − 1]. Dimensions: eₖ, gn and Eₖ are all energies, so θₖ is a pure number, as an angle must be. A real angle needs |tanh 2θₖ| below 1, that is eₖ(eₖ + 2gn) above zero, at every k. That holds for g above zero; an attractive gas has Eₖ² below zero wherever eₖ is less than 2|g|n, so its uniform condensate is unstable.

Expand Eₖ both ways. For eₖ ≪ gn, Eₖ ≈ √(2gn eₖ) = ħck with c = √(gn/m): the lowest quanta are sound waves, in a gas whose atoms would be free without g. For eₖ ≫ gn, Eₖ ≈ eₖ + gn − (gn)²/(2eₖ): an atom added at high k feels 2gn from direct and exchange scattering off the condensate, less the chemical potential μ = gn it is measured from. The crossover eₖ = gn falls at k = √(2mgn)/ħ = 1/ξ, where Eₖ = √3 gn. For sodium with n = 1.0 × 10²⁰ m⁻³ and aₛ = 2.75 nm, gn/h = 1.52 kHz (73 nK), c = 5.1 mm s⁻¹ and ξ = 0.38 μm. Checks: √(J kg⁻¹) is m s⁻¹, and g → 0 sends c → 0 and Eₖ → eₖ, the free-field limit. Measured: Andrews and co-workers (MIT, 1997) timed sound pulses in sodium condensates at millimetres per second, rising as √n, and Bragg spectroscopy of rubidium-87 (Steinhauer and co-workers, 2002) traced Eₖ from its linear to its quadratic regime.

The ground state |0⟩ is defined by bₖ|0⟩ = 0, but it is not empty of atoms outside k = 0: ⟨0|a†ₖaₖ|0⟩ = vₖ²⟨0|b₋ₖb†₋ₖ|0⟩ = vₖ². Written in atoms, it is the condensate dressed with pairs at (k, −k), the amplitude for j pairs going as (−vₖ/uₖ)^j. Vogels and co-workers (MIT, 2002) saw this mixing directly: quasiparticles imprinted at +k in sodium released atoms at −k as well as at +k. Two limits: vₖ² ≈ mc/(2ħk) at small k, large in each mode but harmless once d³k supplies k², and vₖ² ≈ (gn/(2eₖ))² ∝ k⁻⁴ at large k. So n − n₀ = ∫d³k/(2π)³ vₖ² converges at both ends. Put k = x/ξ: n − n₀ = (1/(2π²ξ³))∫x²v²(x) dx, and with w = x² + 2 the integral is elementary and equals √2/6. Since 1/ξ² = 8πnaₛ, (N − N₀)/N = (8/(3√π))√(naₛ³), about one atom per 84 healing volumes. Checks: naₛ³ is dimensionless, and the depletion vanishes as aₛ → 0, the ideal gas. For sodium naₛ³ = 2.08 × 10⁻⁶ and 0.217 per cent of the atoms are out of k = 0 at zero temperature. Lopes and co-workers (2017) confirmed the law in a potassium-39 box, tuning aₛ with a Feshbach resonance.

E₀′ diverges: the summand Eₖ − eₖ − gn tends to −(gn)²/(2eₖ) ∝ k⁻², and ∫k² dk k⁻² grows with the cutoff. The fault lies in g, not in the transformation. Identifying the coupling with 4πħ²aₛ/m, as topic 15.07 did, is the first Born approximation; the second Born term, named here and not derived, makes the bare coupling g + (g²/V)Σ 1/(2eₖ), 2eₖ being the kinetic energy of a pair at ±k. Put into gN²/(2V) it adds back ½Σ′(gn)²/(2eₖ) and cancels the divergence term by term. The finite remainder is quoted: Lee, Huang and Yang's E₀/N = (gn/2)[1 + (128/(15√π))√(naₛ³)], a 0.69 per cent correction for the sodium example, seen in the equation of state of lithium-7 gases and stabilising the quantum droplets of potassium mixtures. Every result here is only the leading term of an expansion in √(naₛ³). Liquid helium-4, its atoms 0.36 nm apart, about 1.4 hard-core diameters, has no small parameter at all; neutron scattering puts about 7 per cent of its atoms in the condensate, far outside any depletion of order √(naₛ³) ≪ 1.