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Let $V=L^3$ be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:In the Fourier series for the contact interaction, the two creation operators supply phase $e^{-i({\mathbf{k}}_1+{\mathbf{k}}_2)\cdot \mathbf{r}}$ and the two annihilation operators supply $e^{i({\mathbf{k}}_3+{\mathbf{k}}_4)\cdot \mathbf{r}}$. Fourier orthogonality with periodic boundary conditions givesThe four normalization factors together contribute $V^{-2}$, leaving the momentum-space contact-interaction HamiltonianThe Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer isBoth are already in normal ordering; no extra commutator constant is introduced.

Normalize the specified state by $⟨0|0⟩=1$ and write $⟨A⟩=⟨0|A|0⟩$. Its zero mode is a coherent state, so any normally ordered product of zero-mode creation operators and annihilation operators can be evaluated using ${\psi }_0\mapsto \sqrt{N_0}$ and ${\psi }_0^{†}\mapsto \sqrt{N_0}$. In particular $⟨({\psi }_0^{†}{)}^2{\psi }_0^2⟩=N_0^2$, rather than $N_0(N_0-1)$, which would apply to a zero-mode number state.

Sort the momentum-space contact-interaction Hamiltonian by its number of zero-mode factors. Four such factors give $u{N}_0^2/(2V)$. Three zero-mode factors and one nonzero momentum violate momentum conservation, so no such term survives. With two zero-mode factors there are three possibilities:

The sums over $\mathbf{k}\ne 0$ include both members of each opposite-momentum pair, consistently with the overall factor $1/2$. Terms with one zero-mode factor contain three nonzero-mode factors; their expected values vanish in the centered multimode squeezed vacuum. Indeed each $\psi$ is linear in $\phi ,{\phi }^{†}$, and an odd product has zero vacuum expectation value. Terms with no zero-mode factors are independent of $N_0$ at fixed ${\theta }_k$. ThereforeTo evaluate these vacuum expectation values, use the inverse Bogoliubov transformation and its Hermitian adjoint:Since ${\phi }_{\mathbf{k}}|0⟩=0$, the only nonzero two-factor vacuum expectation value is $⟨{\phi }_{\mathbf{k}}{\phi }_{\mathbf{q}}^{†}⟩={\delta }_{\mathbf{k},\mathbf{q}}$. The bosonic squeezed-vacuum contractions are consequentlyIn particular the three requested quantities are $-\sinh {\theta }_k\cosh {\theta }_k$, $-\sinh {\theta }_k\cosh {\theta }_k$ and ${\sinh }^2{\theta }_k$, respectively. The minus sign in the pair vacuum expectation values comes from the inverse Bogoliubov transformation and is essential to lowering the interaction energy. Substitution yields