 Solution

For one spin channel, let $X_s=a_s^{†}{b}_s$ and $Y_s=X_s^{†}$. The number-operator commutators giveMoving a boson from $B$ to $A$ raises the number difference by two. On a number-difference eigenstate $|d⟩$, $X_s\mathcal{E}(\Delta )|d⟩=\mathcal{E}(d)X_s|d⟩$, while $\mathcal{E}(\Delta -2)X_s|d⟩=\mathcal{E}(d)X_s|d⟩$. ThusSubtracting the reversed products yieldsThis number-difference shift identity fixes the ordering. The placement of the coefficient to the left matters: it acts on the state after the hop. This spectral argument applies to any function defined on the accessible number-difference eigenvalues, without requiring a power-series expansion.