\kappa_2^{N_{\Omega},Q} &
\kappa_2^{N_{\Omega},E} & \kappa_2^{N_{\Omega},p_x} &
\kappa_2^{N_{\Omega},p_y} &  \kappa_2^{N_{\Omega},p_z} \\ 
\kappa_2^{Q,N_{\Omega}} & \kappa_2^{Q,Q} & \kappa_2^{Q,E} & 0 & 0 &  0 \\ 
\kappa_2^{E,N_{\Omega}} & \kappa_2^{E,Q} & \kappa_2^{E,E} & 0 & 0 & 0 \\ 
\kappa_2^{p_x,N_{\Omega}} & 0 & 0 & \kappa_2^{p_x,p_x} & 0 & 0 \\ 
\kappa_2^{p_y,N_{\Omega}} & 0 & 0 & 0 & \kappa_2^{p_y,p_y} & 0 \\ 
\kappa_2^{p_z,N_{\Omega}} & 0 & 0 & 0 & 0 &  \kappa_2^{p_z,p_z} 
\end{pmatrix}~. \label{matrix}\end{aligned}$$ Please note that by construction, Eq.(\[kappa\_n\]), the matrix (\[matrix\]) is symmetric, hence $\kappa_2^{N_{\Omega},Q} =
\kappa_2^{Q,N_{\Omega}}$, etc.

The second matrix $\tilde \kappa_2$, now related to the partition function $\mathcal{Z}^{Q,E,\vec  P}$, is obtained from $\kappa_2$, Eq.(\[matrix\]), by crossing out the first row and first column. In the following we are going to make use of the fact that one can express the determinant of a matrix $A$ by: $$\label{calcdet}
\det |A| ~=~ \sum \limits_{j=1}^n \left( -1\right)^{j+k}
A_{j,k}  M_{j,k} ~,$$ where $A_{j,k}$ is the matrix element $j,k$ of a general non-singular $n\times
n$ matrix $A$, and $ M_{j,k}$ is its complementary minor. A simple consequence of Eq.(\[calcdet\]) is: $$\label{normdet}
\det |\tilde \kappa_2| = \kappa_2^{p_x,p_x} ~\kappa_2^{p_y,p_y} ~\kappa_2^{p_z,p_z}
\left[ \kappa_2^{E,E}~\kappa_2^{Q,Q}- \left(\kappa_2^{E,Q}\right)^2 \right] ~=~
\left(  \kappa_2^{p_x,p_x}  \right)^3 \det |\hat \kappa_2|,$$ where $\kappa_2^{p_x,p_x} =\kappa_2^{p_y,p_y} =\kappa_2^{p_z,p_z} $, 