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

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.