= Solution
In the <ground state> of the diagonal Hamiltonian, $\alpha_{\mathbf k}|0_B\rangle=0$. The <Bogoliubov transformation> therefore gives $\langle b_{\mathbf k}^\dagger b_{\mathbf k}\rangle=v_k^2$. Summing all nonzero momenta proves the <quantum depletion of a Bose condensate>:
$$
\boxed{n-n_0=\frac1{2L^3}\sum_{\mathbf k\ne0}\left(\frac{\epsilon_{\mathbf k}+nV_{\mathbf k}}{E_{\mathbf k}}-1\right).}
$$
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 has
$$
\langle b_{\mathbf k}^\dagger b_{\mathbf k}\rangle=v_k^2+(u_k^2+v_k^2)n_B(E_k),\qquad
n_B(E)=\frac1{e^{\beta E}-1}.
$$
The additional term is thermal depletion and cannot be discarded merely because the gas is condensed.
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