Fixed-number two-mode condensate ($|\chi ⟩=(C{a}_0^{†}+S{a}_1^{†}{)}^N|0⟩/\sqrt{N!}$)

All $N$ bosons occupy the same normalized one-particle superposition with $|C{|}^2+|S{|}^2=1$. Its occupation distribution in the original two modes is binomial, not two independent Poisson distributions. In particular, $⟨N_1⟩=N|S{|}^2$ and $⟨N_0{N}_1⟩=N(N-1)|C{|}^2|S{|}^2$. These factorial moments give finite-particle corrections to interaction energies. The state is also the two-mode realization of a spin coherent state with spin $N/2$.

For $|\chi ⟩=(c{a}_L^{†}+e^{i\chi }s{a}_R^{†}{)}^N|0⟩/\sqrt{N!}$ with real $c,s$ and $c^2+s^2=1$, one has $|\chi ⟩=e^{i\chi n_R}|0_{\mathrm{phase}}⟩$. Differentiating $E=⟨\chi |H|\chi ⟩$ gives ${\partial }_{\chi }E=i⟨[H,n_R]⟩$. If $[H,n_L+n_R]=0$, this equals $-i⟨[H,n_L]⟩$. The Heisenberg equation of motion identifies the left-to-right particle current as $J={\partial }_{\chi }E/\hslash$. Reversing the definition of the relative phase reverses this sign.