= Solution
In the <Bogoliubov quasiparticle> vacuum, $\alpha_{\mathbf k}|0_\alpha\rangle=0$, but the inverse <Bogoliubov transformation> gives $\langle b^\dagger_{\mathbf k}b_{\mathbf k}\rangle=v_k^2$. Consequently the <quantum depletion of a Bose condensate> is
$$
\boxed{n-n_0=\frac1{\mathcal V}\sum_{\mathbf k\ne0}v_k^2=\frac1{2\mathcal V}\sum_{\mathbf k\ne0}\left(\frac{\epsilon_k+nV_k}{\sqrt{\epsilon_k(\epsilon_k+2nV_k)}}-1\right).}
$$
\b[This expression assumes zero temperature.] At nonzero <temperature>, thermal <Bogoliubov quasiparticles> add to the depletion:
$$
n-n_0=\frac1{\mathcal V}\sum_{\mathbf k\ne0}\left[v_k^2+\frac{A_k}{E_k}\frac1{e^{\beta E_k}-1}\right].
$$
Thus the printed formula is the quantum contribution, or a low-temperature approximation when the thermal contribution is negligible. It is nonnegative for a stable quadratic <Hamiltonian> and vanishes in the noninteracting zero-temperature limit.
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