Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 81 / 2

Write $\mathcal{V}=L^3$ and $N_0=N-N_{\mathrm{ex}}$, where $N_{\mathrm{ex}}={\sum }_{\mathbf{k}\ne 0}{b}_{\mathbf{k}}^{†}{b}_{\mathbf{k}}$. Macroscopic occupation permits $b_0,b_0^{†}\simeq \sqrt{N_0}$, with relative commutator corrections of order $1/N_0$. Keep the fixed-total-number correction before replacing $n_0$ by $n$.

The four-condensate term is $V_0{N}_0(N_0-1)/(2\mathcal{V})$. Its leading expansion gives $V_0{N}^2/(2\mathcal{V})-n{V}_0{N}_{\mathrm{ex}}$. Terms with two condensate operators give normal scattering $n(V_0+V_{\mathbf{k}})b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}$ and conversion of two condensate particles into opposite-momentum excitations, with coefficient $n{V}_{\mathbf{k}}/2$. The $n{V}_0$ normal contribution cancels against the change of condensate energy at fixed $N$. Hence the number-conserving Bogoliubov quadratic Hamiltonian isUse a real even interaction transform, $V_{\mathbf{k}}=V_{-\mathbf{k}}$, as for a real inversion-symmetric pair potential. The constant is the condensate interaction energy; the diagonal part combines particle kinetic energy with condensate scattering; the anomalous pair terms create and annihilate correlated opposite momenta while transferring particles to or from the condensate.

At fixed volume, counting replaced condensate operators gives quadratic vertices proportional to $N_0$, omitted cubic vertices proportional to $\sqrt{N_0}$ and quartic vertices with no condensate factor. There are also the finite-$N_0$ correction in $N_0(N_0-1)$ and corrections from the condensate-number expansion. Their physical control parameter is small depletion, $N_{\mathrm{ex}}/N\ll 1$, together with weak interaction; simply dropping them is not justified by low temperature in a strongly interacting gas. This is the weakly interacting Bose gas approximation.

Put $A_k={ϵ}_k+n{V}_k$ and $B_k=n{V}_k$, and assume the homogeneous condensate is stable, so $A_k>|B_k|$ for nonzero modes. A real Bogoliubov transformation isThe canonical condition preserves the bosonic canonical commutation relations. Cancellation of pair terms requires $2A_k{u}_k{v}_k=B_k(u_k^2+v_k^2)$. ThereforeThe sign of $v_k$ follows that of $B_k$. Normal ordering the transformed operators gives the bosonic Bogoliubov zero-point shift:Rearranging the last two terms is exactly the requested $-\frac{1}{2}\sum A_k+\sum E_k({\alpha }_k^{†}{\alpha }_k+1/2)$ form. Each momentum appears in the sums, so opposite-momentum pairs are not counted again separately.

The Bogoliubov quasiparticle dispersion is linear at low momentum when $V_k\to V_0>0$:It is a gapless phonon of the condensate, unlike the free-particle quadratic spectrum. For a contact repulsion $V_k=g>0$, the healing length $\xi =\hslash /\sqrt{2mng}$ controls crossover to the particle regime $E_k\simeq {ϵ}_k+ng$. A momentum-dependent potential can modify the higher-energy curve; the sketch uses the contact case and does not assume every interaction has that entire shape.

In the ground state of the diagonal Hamiltonian, ${\alpha }_{\mathbf{k}}|0_B⟩=0$. The Bogoliubov transformation therefore gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Summing all nonzero momenta proves the quantum depletion of a Bose condensate:The nonzero depletion arises from correlated pairs in the interacting quasiparticle vacuum, not thermally populated quasiparticles.

The printed expression is the zero-temperature result. At a nonzero temperature the complete leading-order expression instead hasThe additional term is thermal depletion and cannot be discarded merely because the gas is condensed.