 Solution

For each opposite-momentum pair use the Bogoliubov transformationEquivalently $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, with $\tanh (2{\theta }_k)=B_k/A_k$. The identity $u_k^2-v_k^2=1$ preserves the canonical commutation relation. Vanishing anomalous terms require $B_k(u_k^2+v_k^2)-2A_k{u}_k{v}_k=0$, and hence the Bogoliubov quasiparticle dispersion isSubstitution and normal ordering giveThis is the requested form after putting each $E_k/2$ inside the oscillator term. By momentum-pair counting in bosonic diagonalization, all sums cover the full nonzero-momentum set; no further factor of two is needed. The bosonic Bogoliubov zero-point shift combines the two vacuum sums before taking a continuum limit.

If $V_k\to V_0>0$ smoothly, the low-energy excitations are gapless phonons:For contact repulsion, the healing length $\xi =\hslash /\sqrt{2mn{V}_0}$ separates this regime from the particle-like regime $E_k\simeq {ϵ}_k+n{V}_0$. Stability requires ${ϵ}_k+2n{V}_k>0$ for every nonzero mode. A general momentum-dependent interaction need not share the contact model's entire curve. An unregularized contact model also makes the vacuum-energy shift ultraviolet divergent; a physical finite-range interaction or scattering-length renormalization is needed for that absolute ground-state energy.