 Solution

Use a real, even interaction $V_{-\mathbf{k}}=V_{\mathbf{k}}$. Let $\mathcal{V}=L^3$ and $N_{\mathrm{ex}}={\sum }_{\mathbf{k}\ne 0}{b}_{\mathbf{k}}^{†}{b}_{\mathbf{k}}$. In the Bogoliubov theory expansion, keep terms with zero or two noncondensed operators. The zero-momentum term gives $V_0{N}_0(N_0-1)/(2\mathcal{V})$. The terms with two noncondensed operators giveHere $V_0$ is the direct condensate interaction and $V_{\mathbf{k}}$ is its exchange contribution. At fixed total particle number, $N_0=N-N_{\mathrm{ex}}$, soThe second term cancels the direct normal term. Adding the kinetic energy yields the number-conserving Bogoliubov quadratic HamiltonianThe constant is the condensate mean-field theory interaction energy; the normal term describes moving a particle out of the condensate, while the anomalous term creates or annihilates an opposite-momentum pair by exchanging two particles with the condensate. Total particle number remains fixed in the underlying theory; the condensate amplitudes in this notation are treated as classical amplitudes.

For condensate fluctuation power counting, put $\delta =N_{\mathrm{ex}}/N\ll 1$. Formal interaction-amplitude counting gives $H_2-E_{\mathrm{mf}}\sim E_{\mathrm{mf}}\delta$, cubic terms $\sim E_{\mathrm{mf}}{\delta }^{3/2}$ and quartic terms $\sim E_{\mathrm{mf}}{\delta }^2$; these are not uniform bounds on individual modes. Thus the latter are down by $\sqrt{\delta }$ and $\delta$ relative to the quadratic interaction terms. The $N_0-1$ correction is relative $1/N$. Homogeneity eliminates terms linear in nonzero-momentum operators. Small depletion and weak interactions, not low temperature alone, justify the truncation. ^[ref]