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For a bosonic mode satisfying the canonical commutation relation $[a,a^{†}]=1$, the number operator $n=a^{†}a$ obeys $n^2=a^{†}a{a}^{†}a=(a^{†}{)}^2{a}^2+n$. Thus a two-body on-site interaction is proportional to $n(n-1)$, with no self-interaction. In a fixed-number two-mode condensate with probability $q$ of occupying this mode, $⟨n(n-1)⟩=N(N-1)q^2$ by the binomial distribution.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

Put $C=\cos (\chi /2)$ and $S=\sin (\chi /2)$. Commutativity of the two creation operators and the binomial theorem giveThusThe state is a fixed-number two-mode condensate with binomial mode occupations; it is not a product of two independent Poisson-distributed Glauber coherent states.

Two localized orbitals coupled by quantum tunnelling and on-site interactions give the smallest Bose-Hubbard model. A symmetric double-well potential with lowest one-particle energies $E_0,E_1$ has $t=(E_1-E_0)/2$ and common diagonal energy $(E_0+E_1)/2$. At fixed total particle number, that common diagonal term is constant. A fixed-number two-mode condensate gives a simple variational method for this model, while stronger interactions can require occupation-correlated trial quantum states.

Two independent boson species occupy uniform single-particle modes on $L$ sites, with fixed occupation numbers $n_A,n_B$. This differs from a fixed-number two-mode condensate whose particles all occupy one superposition of the two modes. Here $⟨a_i^{†}{a}_j⟩=n_A/L$, $⟨b_i^{†}{b}_j⟩=n_B/L$, and the local total number operator has variance $(n_A+n_B)L^{-1}(1-L^{-1})$. The latter follows from two independent binomial distributions with success probability $1/L$. This state satisfies a prescribed Schwinger boson constraint only in the mean unless the local variance vanishes.