= Solution
For one <spin> channel, let $X_s=a_s^\dagger b_s$ and $Y_s=X_s^\dagger$. The number-operator <commutators> give
$$
[\Delta,X_s]=2X_s,\qquad[\Delta,Y_s]=-2Y_s.
$$
Moving a <boson> from $B$ to $A$ raises the number difference by two. On a number-difference <eigenstate> $|d\rangle$, $X_s\mathcal E(\Delta)|d\rangle=\mathcal E(d)X_s|d\rangle$, while $\mathcal E(\Delta-2)X_s|d\rangle=\mathcal E(d)X_s|d\rangle$. Thus
$$
X_s\mathcal E(\Delta)=\mathcal E(\Delta-2)X_s,\qquad
Y_s\mathcal E(\Delta)=\mathcal E(\Delta+2)Y_s.
$$
Subtracting the reversed products yields
$$
\boxed{[X_s,\mathcal E(\Delta)]
=[\mathcal E(\Delta-2)-\mathcal E(\Delta)]X_s,}
$$
$$
\boxed{[Y_s,\mathcal E(\Delta)]
=[\mathcal E(\Delta+2)-\mathcal E(\Delta)]Y_s.}
$$
This <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.
Back to article page
