## Appendix A Numerical Method SMPIMC
### Smpinc Algorithm for MDFs  
Calculation of the single-particle MDF is based on the sine transform (14) of the \(\xi\)-distribution, which is obtained from the path integral representation (26) via the Monte-Carlo procedure similar to the one for PCFs. The difference is that the first electron is represented as the open poly-line with the vertices \(\mathbf{q}_{1}^{k}=\mathbf{z}_{1}^{k}+(1-k/M)(\mathbf{q}_{1}+\xi/2)+(k/M)( \mathbf{q}_{1}-\xi/2)\). Only the differential coordinate \(\xi\) participates in the calculation of the single-particle MDF, while the other coordinates define only the weight and probability density. The detailed SMPIMC algorithm for the calculation of the single-particle MDF is presented below. 1. Set the run number \(l=0\) and the initial state \(\mathbf{x}_{0}\): the coordinates \(\mathbf{q}_{a}\) are uniformly distributed in the simulation box, while the relative coordinates \(\mathbf{z}_{a}^{k}\) and the differential coordinate \(\xi\) are equal to zero. 2. Set the step number \(i=1\) and the first state \(\mathbf{x}_{1}=\mathbf{x}_{0}\). 3. Select the type of step: \(\delta\xi\)-step with the probability \(P_{\xi}\), \(\delta q\)-step with the probability \(P_{q}\) or \(\delta z\)-step with the probability \(P_{z}=1-P_{\xi}-P_{q}\). If \(\delta\xi\)-step has been chosen, modify \(\xi\rightarrow\xi+\Delta\xi\) with \(\Delta\xi\) uniformly distributed in the volume \(\sim\lambda^{3}\). Otherwise, select the particle number \(a=1,\ldots,N\) randomly; if \(\delta q\)-step has been chosen, modify \(\mathbf{q}_{a}\rightarrow\mathbf{q}_{a}+\Delta\mathbf{q}\) with \(\Delta\mathbf{q}\) uniformly distributed in the volume \(\sim L^{3}/N\); if \(\delta z\)-step has been chosen, select the \"bead\" number \(k=1,\ldots,M-1\) randomly and modify \(\mathbf{z}_{a}^{k}\rightarrow\mathbf{z}_{a}^{k}+\Delta\mathbf{z}\) with \(\Delta\mathbf{z}\) uniformly distributed in the volume \(\sim\lambda^{3}/(4\pi K)^{3/2}\). Take into account the PBC. The resulting state has to be set as the proposed state \(\mathbf{x}_{i}^{\prime}\). 4. Accept the proposed state \(\mathbf{x}_{i}^{\prime}\) with the probability \(A(\mathbf{x}_{i}\rightarrow\mathbf{x}_{i}^{\prime})\) (A4) or reject it. In the case of acceptance, set \(\mathbf{x}_{i+1}=\mathbf{x}_{i}^{\prime}\), in the case of rejection, set \(\mathbf{x}_{i+1}=\mathbf{x}_{i}\). 5. Calculate the absolute value \(\xi=|\xi|\) and build the related histogram \(F_{\Omega}^{n}(\mathbf{x}_{i})\), where \(n=1,\ldots,N_{\xi}\) is the number of the cell with the length \(\Delta_{\xi}\), so \(n=\left[\xi/\Delta_{\xi}\right]+1\). 6. Repeat steps (3)-(5) for \(i=1,\ldots,N_{steps}\). 7. Calculate the average histogram for the obtained sample of \(N_{steps}\) states via the averaging of \(f_{\Omega}^{n}\) with \(g(\mathbf{x})\) as the weight function: \[\langle f_{\Omega}^{n}\rangle_{l}=\frac{\sum_{i=1}^{N_{steps}}f_{\Omega}^{n}( \mathbf{x}_{i})\,g(\mathbf{x}_{i})}{\sum_{i=1}^{N_{steps}}g(\mathbf{x}_{i})}.\]
8. To obtain the histogram of the \(\xi\)-distribution on the \(l\)-th run take into account the angular distribution:\[\langle f^{n}\rangle_{l}=\frac{\langle f_{\Omega}^{n}\rangle_{l}}{\Delta_{n}}, \quad\Delta_{n}=4\pi\Delta_{\xi}^{3}\Big{(}n^{2}-n+1/3\Big{)}.\]
9. Repeat steps (2)-(8) for \(l=1,2,\ldots,N_{runs}\), but instead of the initialization \(\mathbf{x}_{1}=\mathbf{x}_{0}\) use the last state from the previous run: \(\left.\mathbf{x}_{1}\right|_{l}=\left(\mathbf{x}_{N_{steps}}\right)\right|_{l -1}\). 10. As a result, the sample of average histograms of \(\xi\)-distribution \(\langle f^{n}\rangle_{l}\), \(l=0,1,\ldots,N_{runs}\) is obtained. Considering the 0-th run as idle and omitting it to eliminate the influence of the initial state, calculate the resulting average histogram over the sample with the statistical error as follows: \[\langle f^{n}\rangle=\frac{\sum_{l=1}^{N_{runs}}\langle f^{n}\rangle_{l}}{N_ {runs}},\quad\sigma(f^{n})=\sqrt{\frac{\sum_{l=1}^{N_{runs}}\left(\langle f^{ n}\rangle_{l}-\langle f^{n}\rangle\right)}{N_{runs}}}.\]
11.