traces $\tr\widehat{\mathcal U}_m$ and $\tr\widehat{\mathcal V}_{m}$ (see below). In the momentum representation, $\hat{\varepsilon}_{0}$ is a diagonal matrix of the eigen-energies of the $XX$ model at $h=0$ [@21]. Formally expanding ${\widetilde F}_{m+1;1}({\lambda})$ in the powers of $\widehat{\mathcal U}_m$ we shall obtain the answer in two lowest orders: $$\begin{aligned}
&{\widetilde F}_{m+1;1}({\lambda})\approx
F_{m+1;1}({\lambda})+F_{m+1;1}({\lambda})
\tr\widehat{\mathcal U}_m-\tr\widehat{\mathcal V}_{m},
\nonumber
\\
&\tr\widehat{\mathcal U}_m=(M-2m)F_{1;1}({\lambda}),
\label{45}
\\
&\tr\widehat{\mathcal V}_{m}=F_{m+1;1}(2{\lambda})-
2\sum_{l=1}^{m}F_{m+1;l}({\lambda})F_{l;1}({\lambda}),
\nonumber\end{aligned}$$ where the notation $F_{j;l}({\lambda})$ implies the relations . Although $M$ and $m$ are chosen to be large enough, the ratio $m/M$ is assumed to be finite. Equation  is fulfilled in each order separately by the terms presented in ${\widetilde F}_{m+1;1}$ .

By an analogy with the ferromagnetic case, let us act on ${\widetilde F}_{m+1,1}({\lambda})$  by the operator $\mathcal D^K_{\lambda/2}$. Then, in the first order we shall obtain the relation . In the second order, the answer is of the following form: $$MC_K^LC_K^L-C_K^L\sum_{l=0}^KC_K^l+2\sum_{l=1}^m
\begin{vmatrix}
C_K^{L+l-1}&C_K^L
\\
C_K^L&C_K^L
\end{vmatrix}.
\label{46}$$ By virtue of , the result of the application of $\mathcal D^K_{\lambda/2}$ to the second order function $F_{j_1,j_2;l_1,l_2}(\lambda)$ is connected, as a particular case of , with the number of the two-dimensional paths $\mathcal T_K$, and is expressed through the corresponding determinant. It means that it will be appropriate 