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The rate at which particles cross an oriented surface, measured as particle count per unit time. If a local continuity equation has number density $n$ and current density ${\mathbf{j}}_N$, then ${\partial }_{t}n+\nabla \cdot {\mathbf{j}}_N=0$ and $J_N={\int }_S{\mathbf{j}}_N\cdot d\mathbf{S}$. In a closed two-reservoir system, a left-to-right particle current is $-d{N}_L/dt=d{N}_R/dt$. Neutral particles carry this current without an associated electric current.

The printed identity has the wrong sign for the specified relative phase. Differentiating the energy found above and evaluating the operator commutator giveThus the phase derivative of a two-mode condensate energy isA concrete counterexample to the printed sign is $N=100$, $\theta =\pi /4$, $\chi =\pi /2$ and $t>0$: its left side is $100t$, whereas its printed right side is $-100t$, independently of $w$ and $\mu$.

There is also a direct derivation without inserting the Hamiltonian operator. Since $|\theta ,\chi ;N⟩=e^{i\chi {\hat{n}}_R}|\theta ,0;N⟩$, differentiating this unitary operator gives ${\partial }_{\chi }E=i⟨[\hat{H},{\hat{n}}_R]⟩$. Conservation of the total particle number, $[\hat{H},{\hat{n}}_L+{\hat{n}}_R]=0$, then supplies the minus sign for the left well.

By the Heisenberg equation of motion, $\hslash d⟨{\hat{n}}_L⟩/dt=i⟨[\hat{H},{\hat{n}}_L]⟩$. Define the bosonic Josephson junction particle current from left to right as $J_{L\to R}=-d⟨{\hat{n}}_L⟩/dt$. The corrected identity meansThe energy slope with respect to the relative phase drives particle transfer; an extremum with respect to that phase has zero instantaneous particle current. The printed sign would work with $e^{-i\chi }$ in the right-well orbital, or with $[{\hat{n}}_L,\hat{H}]$ in place of $[\hat{H},{\hat{n}}_L]$.

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.