R}
(z,\vec{K})$ given by (\[rdef\]).

A few comments are in order about this expression. For bosons the state $|\vec{k}_1\,\vec{k}_2\rangle $ is the same as $|\vec{k}_2\,\vec{k}_1\rangle$. So the relative momenta $\vec{k}$ and $-\vec{k}$ should be identified with each other. Both $\vec{k}_1$ and $\vec{k}_2$ run through a Brillouin zone appropriate for the structure of the lattice, and this in principle defines the ranges of $\vec{K}$ and $\vec{k}$. But as was mentioned before, we have chosen the Brillouin zone such that there are no points whose values of $\vec{K}$ differ by a reciprocal lattice vector.

From the definition (\[rdef\]) of ${\cal R}(z,\vec{K})$ and the expression (\[states\]) for $E_0(\vec{K},\vec{k})$ one sees that $2\Delta$ occurs as a shift in the energy variable and the $\rho_2$ are functions of $E-2\Delta$. Taking $E-2\Delta$ as an integration variable in (\[b2e\]) one finds $$b_2^{\rm int} =\frac{1}{N} \int dE' \left[ \rho_2(E') -
\rho_2^0(E')\right]
e^{-\beta E'},
\label{b2e2}$$ with $$\rho_2(E')-\rho_2^0(E')= -\frac{1}{\pi}\mathop{\rm Im}\nolimits
\left[ \frac{\partial}{\partial z}
\sum_{\vec{K}} \ln {\cal R}'(z,\vec{K}) \right]_{z=E'+i\epsilon} ,
\label{rhoe2}$$ and ${\cal R}'(z,\vec{K})$ given by $${\cal R}'(z,\vec{K}) = \frac{1}{N} \sum_{\vec{k}} \frac{1}
{z-\varepsilon(
\frac{\vec{K}}{2}+\vec{k}) -\varepsilon(\frac{\vec{K}}{2} -\vec{k})} .
\label{r2}$$ In this formula $\Delta$ is eliminated, as it should be. Using the fact that $\mathop{\rm Im}\nolimits \ln z = \mathop{\rm arg}\nolimits
z$, we 