 Solution

In the Bogoliubov quasiparticle vacuum, ${\alpha }_{\mathbf{k}}|0_{\alpha }⟩=0$, but the inverse Bogoliubov transformation gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Consequently the quantum depletion of a Bose condensate isThis expression assumes zero temperature. At nonzero temperature, thermal Bogoliubov quasiparticles add to the depletion: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.