= Solution
Put $A_k=\epsilon_k+nV_k$ and $B_k=nV_k$, and assume the homogeneous condensate is stable, so $A_k>|B_k|$ for nonzero modes. A real <Bogoliubov transformation> is
$$
b_{\mathbf k}=u_k\alpha_{\mathbf k}-v_k\alpha_{-\mathbf k}^\dagger,\qquad u_k^2-v_k^2=1.
$$
The canonical condition preserves the bosonic <canonical commutation relations>. Cancellation of pair terms requires $2A_ku_kv_k=B_k(u_k^2+v_k^2)$. Therefore
$$
\boxed{E_k=\sqrt{A_k^2-B_k^2}=\sqrt{\epsilon_k(\epsilon_k+2nV_k)},}
$$
$$
u_k^2=\frac12\left(\frac{A_k}{E_k}+1\right),\qquad
v_k^2=\frac12\left(\frac{A_k}{E_k}-1\right),\qquad u_kv_k=\frac{B_k}{2E_k}.
$$
The sign of $v_k$ follows that of $B_k$. <Normal ordering> the transformed operators gives the <bosonic Bogoliubov zero-point shift>:
$$
\boxed{H=\frac{\mathcal V}{2}V_0n^2+\frac12\sum_{\mathbf k\ne0}(E_k-A_k)+\sum_{\mathbf k\ne0}E_k\alpha_{\mathbf k}^\dagger\alpha_{\mathbf k}.}
$$
Rearranging the last two terms is exactly the requested $-\frac12\sum A_k+\sum E_k(\alpha_k^\dagger\alpha_k+1/2)$ 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 $V_k\to V_0>0$:
$$
\boxed{E_k\simeq\hbar c_s|k|,\qquad c_s=\sqrt{nV_0/m}.}
$$
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> $\xi=\hbar/\sqrt{2mng}$ controls crossover to the particle regime $E_k\simeq\epsilon_k+ng$. 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-81-bogoliubov-dispersion.png]
{title=Gapless phonon-to-particle crossover of the Bogoliubov dispersion for a repulsive contact interaction}
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