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The Bogoliubov vacuum has original-boson occupation $v_k^2=({\alpha }_k/{ϵ}_k-1)/2$, reducing the ordered spin by its average over momenta. For the zero-field planar ferromagnetic chain, $v_k^2\sim 1/(2\sqrt{2}|k|)$. Even after excluding the exact uniform zero mode, the sum grows as $(2\pi \sqrt{2}{)}^{-1}\log L$. It invalidates a thermodynamic fixed-orientation expansion with finite dilute depletion; it does not represent a negative physical magnetization.