orthogonality of angular modes makes the one-body term

E a_{1}^{†} a_{1}

, where

E = \frac{ℏ ^{2}}{2 m R ^{2}} .

The contact-interaction integral supplies

(U /4 π) δ_{l_{1} + l_{2}, l_{3} + l_{4}}

, expressing angular momentum conservation. Within modes zero and one, the four mixed permutations each reduce to

a_{1}^{†} a_{0}^{†} a_{0} a_{1}

. The two remaining terms have all four indices equal. Thus

H = E N_{1} + \frac{U}{4 π} [N_{0} (N_{0} - 1) + N_{1} (N_{1} - 1) + 4 N_{0} N_{1}] .

This is the projected two-mode Quantum Hamiltonian; the full interaction could scatter into excluded modes, so the truncation is an assumed approximation rather than an exact invariant subspace of the complete ring problem.

b

Solution

The occupation state is an eigenstate of the projected Quantum Hamiltonian. Since

a_{j}^{†} a_{j}^{†} a_{j} a_{j} = N_{j} (N_{j} - 1)

E_{N_{0}, N_{1}} = E N_{1} + \frac{U}{4 π} [N_{0} (N_{0} - 1) + N_{1} (N_{1} - 1) + 4 N_{0} N_{1}] .

Equivalently, the bracket is

N (N - 1) + 2 N_{0} N_{1}

, where

N = N_{0} + N_{1}

; there is no self-interaction for a single particle.

Put

C = cos (χ /2)

and

S = sin (χ /2)

∣ χ ⟩ = \sum_{j = 0}^{N} (j N) C^{N - j} S^{j} ∣ N - j, j ⟩ .

Thus

P (N_{1} = j) = (j N) (1 - p)^{N - j} p^{j}, p = S^{2}, \overline{N}_{1} = Np, \overline{N}_{0} = N (1 - p) .

The 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.