The weakly interacting Bose gas and Bogoliubov's replacement

Units are SI, with energies also quoted as temperatures E/k_B; the gas is non-relativistic, so the course's metric signature (+, −, −, −) never enters. Quantum Mechanics II's first Born amplitude, quoted, is f = −(mᵣ/2πħ²)ṽ(q), with ṽ the Fourier transform of the potential and mᵣ the reduced mass. Take v(r) = g δ(r): its transform is the constant g, and two atoms of mass m have mᵣ = m/2, so f = −mg/(4πħ²) at every angle and energy. The scattering length is defined by f → −aₛ at low energy, hence aₛ = mg/(4πħ²) and g = 4πħ²aₛ/m, derived. Dimensions: ħ²aₛ/m carries J² s² m kg⁻¹ = J m³, an energy times a volume, as the coefficient of δ(r) must. For ⁸⁷Rb, m = 1.443 × 10⁻²⁵ kg and the measured aₛ = 5.3 nm, about 100 Bohr radii, give g = 5.13 × 10⁻⁵¹ J m³, repulsive. Match the unsymmetrised amplitude: the bosonic two-body operator generates the exchange amplitude itself, so symmetrising first would count it twice.

The real Rb–Rb potential is anything but weak. Its well binds dozens of molecular vibrational states, and a potential that binds states is never perturbative: its exact aₛ swings through ±∞ each time the well deepens enough to hold one more, which no first-order formula can follow. What rescues the model is a separation of scales. Beyond the potential's range r₀, a few nanometres for alkali atoms, the zero-energy wavefunction is ψ ∝ 1 − aₛ/r whatever the potential (quoted), so when kr₀ ≪ 1 and kaₛ ≪ 1 every observable depends on aₛ alone. g δ(r) is not an approximation to the potential but a different potential with the same aₛ at first order. Pushed to second order it fails, the sum over intermediate states of 1/k² diverging at large k; the next-order Lee-Huang-Yang energy, named here, is finite only because that divergence cancels. In a gas the model also needs n aₛ³ ≪ 1, so the spacing far exceeds aₛ and three atoms rarely meet: ⁸⁷Rb at 1.0 × 10²⁰ m⁻³ gives 1.5 × 10⁻⁵ and a spacing of 215 nm, against 0.36 nm in liquid ⁴He.

Without interactions the ground state at T = 0 is |N, 0, 0, …⟩, every atom in the k = 0 plane wave. A weak repulsion leaves most of them there: N₀ stays close to N, and the excited population N − N₀ = Σ′⟨a†ₖaₖ⟩ is a small fraction that the next topic computes. On the condensate, a₀|N₀⟩ = √N₀|N₀ − 1⟩ and a₀†|N₀⟩ = √(N₀ + 1)|N₀ + 1⟩. The two amplitudes differ by the factor √(1 + 1/N₀), and the commutator [a₀, a₀†] = 1 is smaller than a₀†a₀ = N₀ by 1/N₀, which is 10⁻⁵ for a hundred thousand atoms. Bogoliubov therefore treated a₀ and a₀† as commuting numbers, both equal to √N₀ once the condensate's phase is chosen real. Physically the condensate becomes a reservoir: taking an atom out or putting one in leaves its amplitude unchanged to order 1/N₀. The price shows in the one place it could matter, a₀†a₀†a₀a₀ = N₀(N₀ − 1), which the replacement turns into N₀², a relative error 1/N₀ that is negligible for every gas treated here.

Quote Unit 10's Hamiltonian with v(p) = g: H = Σₖ εₖ a†ₖaₖ + (g/2V) Σ a†ₖ₊ₚ a†ₖ′₋ₚ aₖ′aₖ, with εₖ = ħ²k²/2m. Classify each interaction term by how many of its four operators carry nonzero momentum. None: (g/2V)N₀². One: impossible, since three zero momenta and one nonzero cannot balance. Two: six choices. Both creators give (gN₀/2V)a†ₖa†₋ₖ, both annihilators its adjoint, and the four mixed choices each give (gN₀/2V)a†ₖaₖ, two direct and two exchange, so 2gn₀ in all, with n₀ = N₀/V. Three or four: smaller by further powers of the excited fraction, and dropped. Now eliminate N₀ with N₀ = N − Σ′a†ₖaₖ, so (g/2V)N₀² ≈ gN²/(2V) − gn Σ′a†ₖaₖ, and set N₀ ≈ N in coefficients that already multiply excited operators. The result, derived, is H ≈ gN²/(2V) + Σ′[(εₖ + gn)a†ₖaₖ + ½gn(a†ₖa†₋ₖ + aₖa₋ₖ)]. An excited atom feels 2gn from the condensate and takes gn from the condensate's own energy.

In the bare condensate every excited occupation vanishes, so the leading energy is the constant: E₀ = gN²/(2V) and E₀/N = gn/2, derived. Differentiating gives μ = ∂E₀/∂N = gn, the scale beside εₖ in the quadratic terms, and P = −∂E₀/∂V = gn²/2, stable only for positive aₛ; with negative aₛ the uniform gas collapses, and trapped ⁷Li and ⁸⁵Rb condensates survive only below a critical number. Dimensions: gn is J m³ × m⁻³ = J. Free limit: g → 0 gives the ideal condensate's E₀ = 0. One-particle limit: the exact pair count gives gN(N − 1)/(2V), zero for one atom. For ⁸⁷Rb at 1.0 × 10²⁰ m⁻³, gn/2 = 2.57 × 10⁻³¹ J = k_B × 18.6 nK and μ/h = 775 Hz. The sharpest measured check is the counting itself. Two atoms in one mode cost g/V, in different modes 2g/V, so a thermal gas has twice the condensate's interaction energy per atom; for three atoms the factor is 3! = 6, and JILA found three-body loss in ⁸⁷Rb condensates 7.4 ± 2.0 times slower than in thermal clouds of equal density.

With a₀ a number, H no longer commutes with N̂: the pair terms change the excited population by two with no condensate atoms to pay. The natural object is then K = H − μN̂ with μ = gn, and the atom number is right only on average. Number-conserving formulations restore exact conservation, from Girardeau and Arnowitt in 1959 to Gardiner and to Castin and Dum in the late 1990s, using operators such as a₀†aₖ/√N̂₀ that move an atom between mode k and the condensate at fixed N; they reproduce the same quadratic Hamiltonian at this order and are named, not derived. The approximation is a chain of smallness: N₀ ≫ 1 for the replacement, n aₛ³ ≪ 1 for the contact interaction and the dropped cubic and quartic terms, and T well below the condensation temperature. What it delivers is a quadratic form, not a spectrum. The pair coupling is nearly half the diagonal coefficient wherever εₖ ≪ gn, that is well below k_c = √(8πnaₛ), 3.65 μm⁻¹ for ⁸⁷Rb at 10²⁰ m⁻³, and only the next topic's transformation can diagonalise it there.