the matrix $G$ defined by Eq. (\[equ:DefG\]). $N = 1331$, $\alpha = 0.37$, 100 realizations. Gray bars represent positive entries and white bars negative ones.[]{data-label="fig:GAveraged"}](G1 "fig:"){width=".9\linewidth"} ![First $20 \times 20$ entries of the matrix $G$ defined by Eq. (\[equ:DefG\]). $N = 1331$, $\alpha = 0.37$, 100 realizations. Gray bars represent positive entries and white bars negative ones.[]{data-label="fig:GAveraged"}](G3 "fig:"){width=".9\linewidth"}

To elucidate the the influence of the different spectra in the 1D and 3D case, we simply substitute the eigenvalues of the 1D case by the ones of the 3D case (i.e. use 1D $G$-matrix together with the corresponding 3D eigenvalues): $$\begin{aligned}
   \langle \tilde{Y}_2^{(1D)}\rangle_{p_0, \tau} &\simeq \sum_{ij} \left\langle G^{(1)}_{ij}\right\rangle_\tau \left( e^{-\tilde{\lambda}^{(3)}_i}\right)^t \left( e^{-\tilde{\lambda}^{(3)}_j}\right)^t,
   \label{equ:1Dwith3Deigenvalues}
   \intertext{but still take}
   \langle \tilde{Y}_2^{(3D)}\rangle_{p_0, \tau} &\simeq \sum_{ij} \left\langle G^{(3)}_{ij}\right\rangle_\tau \left( e^{-\tilde{\lambda}^{(3)}_i}\right)^t \left( e^{-\tilde{\lambda}^{(3)}_j}\right)^t.\nonumber\end{aligned}$$ Thus, in this approximation, the only difference between the 1D and the 3D case is the shape of the matrix $G$ defined by Eq. (\[equ:DefG\]). The result of the approximation Eq. (\[equ:1Dwith3Deigenvalues\]) is also presented in Fig. \[fig:Factorized\]. Although the plateau of the case Eq. (\[equ:1Dwith3Deigenvalues\]) lies lower than the one of Eq. (\[equ:Factorized\]), it is still there and pronounced. Thus, the exact form of the spectrum is only responsible for the height and 