= Solution
{c}
\b[The printed identity has the wrong sign for the specified relative phase.] Differentiating the <energy> found above and evaluating the <operator commutator> give
$$
\frac{\partial E}{\partial\chi}=tN\sin(2\theta)\sin\chi,\qquad
[\hat H,\hat n_L]=t(\hat a_L^\dagger\hat a_R-\hat a_R^\dagger\hat a_L),\qquad
i\langle[\hat H,\hat n_L]\rangle=-tN\sin(2\theta)\sin\chi.
$$
Thus the <phase derivative of a two-mode condensate energy> is
$$
\boxed{\frac{\partial E}{\partial\chi}=-i\langle[\hat H,\hat n_L]\rangle
=i\langle[\hat H,\hat n_R]\rangle.}
$$
A 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 $|\theta,\chi;N\rangle=e^{i\chi\hat n_R}|\theta,0;N\rangle$, differentiating this <unitary operator> gives $\partial_\chi E=i\langle[\hat H,\hat n_R]\rangle$. 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>, $\hbar\,d\langle\hat n_L\rangle/dt=i\langle[\hat H,\hat n_L]\rangle$. Define the <bosonic Josephson junction> <particle current> from left to right as $J_{L\to R}=-d\langle\hat n_L\rangle/dt$. The corrected identity means
$$
\boxed{J_{L\to R}=\frac1\hbar\frac{\partial E}{\partial\chi}
=\frac{tN}{\hbar}\sin(2\theta)\sin\chi.}
$$
The <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 $[\hat H,\hat n_L]$.
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