## Appendix D Effects of Exchange Symmetry on the Total Current Many Body Operator  
In the text, we have assumed that the many-particle wave function in Equation (1) has no exchange symmetry. Here, we discuss the physical soundness of such approximation. Let us evaluate the effects of exchange symmetry on the squared total current of Equation (2) for two particle case, which can be rewritten as:  
\[\langle I_{T}^{2}\rangle=\frac{1}{4}\sum_{i=1}^{2}\langle l_{i}I_{i}\rangle+ \frac{1}{4}\sum_{i=1,j\
eq i}^{2}\langle l_{i}I_{j}\rangle, \tag{11}\]  
where indexes \(i,j\) refers to the particle with coordinate \(x^{i}\) in the phase space. We assume the system wave function described by the two-fermion state with exchange symmetry  
\[\Psi(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi^{1}(x^{1})\psi^{2}(x^{2})-\psi^{2}(x ^{1})\psi^{1}(x^{2})), \tag{12}\]  
where the superindex \(k\) of the state \(\psi^{k}\) refers to the injection time (e.g., \(\psi^{1}\) is injected at \(t_{in}\), while \(\psi^{2}\) is injected at \(t_{in}+\tau_{in}\)). Within this definition of the state, the diagonal average values of the first value in Equation (11) read:  
\[\langle I^{1}I^{1}\rangle = \frac{1}{4}(\langle\psi^{1}|(I^{1})^{2}|\psi^{1}\rangle\langle \psi^{2}|\psi^{2}\rangle+\langle\psi^{2}|\psi^{2}\rangle\langle\psi^{1}|(I^{1 })^{2}|\psi^{1}\rangle \tag{13}\] \[- \langle\psi^{1}|(I^{1})^{2}|\psi^{2}\rangle\langle\psi^{2}|\psi^ {1}\rangle-\langle\psi^{2}|(I^{1})^{2}|\psi^{1}\rangle\langle\psi^{1}|\psi^{2 }\rangle),\]  
for \(i=1,2\). We reasonably assume an initial negligible overlap of the wave packets,  
\[\langle\psi^{\pm}|\psi^{\pm}\rangle\approx\delta_{\alpha,\beta}, \tag{14}\]  
which is conserved during the evolution of the two electrons, carrying out the same procedure for the second diagonal element we get:  
\[\langle I^{1}I^{1}\rangle+\langle I^{2}I^{2}\rangle=\frac{1}{2}(\langle\psi^{ 1}|(I^{1})^{2}|\psi^{1}\rangle+\langle\psi^{2}|(I^{2})^{2}|\psi^{1}\rangle), \tag{15}\]  
as for distinguishable particles. Similarly, the first non-diagonal term of Equation (11) is computed:  
\[\langle I^{1}I^{2}\rangle = \frac{1}{4}(\langle\psi^{1}|I^{1}|\psi^{1}\rangle\langle\psi^{2 }|I^{2}|\psi^{2}\rangle+\langle\psi^{2}|I^{1}|\psi^{2}\rangle\langle\psi^{1}|I ^{2}|\psi^{1}\rangle \tag{16}\] \[- \langle\psi^{1}|I^{1}|\psi^{2}\rangle\langle\psi^{2}|I^{2}|\psi^ {1}\rangle-\langle\psi^{2}|I^{1}|\psi^{1}\rangle\langle\psi^{1}|I^{2}|\psi^{2 }\rangle),\]  
Computing the same for the second non-diagonal element we get,  
\[\langle I^{1}I^{2}\rangle+\langle I^{2}I^{1}\rangle=\frac{1}{2}\langle\psi^{ 1}|I^{1}|\psi^{1}\rangle\langle\psi^{2}|I^{2}|\psi^{2}\rangle. \tag{17}\]  
So using Equations (15) and (17) in Equation (11) we get,  
\[\langle I_{T}^{2}\rangle = \frac{1}{4}\sum_{i=1}^{2}\langle l_{i}I_{i}\rangle+\frac{1}{4} \sum_{i=1,j\
eq i}^{2}\langle l_{i}I_{j}\rangle \tag{18}\] \[\Psi(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi^{1}(x^{1})\psi^{2}(x^{2})-\psi^{2}(x ^{1})\psi^{1}(x^{2})), \tag{12}\]0The mean value of the squared of the total current for distinguishable particles is therefore recovered by assuming \(\langle\psi^{i}|I|\psi^{j}\rangle\approx\delta_{i,j}\). This constraint is fulfilled by Equation (144), together with the definition of the current operator, which is related to momentum (proportional to a spatial derivative in the real space) and position operators. Indeed, we expect that for a weak measurement of the current discussed in the text, the state \(|\psi^{j}\rangle^{\prime}\) produced by its application to a wave packet \((|\psi^{j}\rangle^{\prime}=I|\psi^{j}\rangle)\) is characterized by a spatial localization that does not significantly differ from the unperturbed state \(|\psi^{j}\rangle\), so that \(\langle\psi^{i}|I|\psi^{j}\rangle\approx 0\) is valid. Appendix E The Ontological Meaning of the Total Measured Current \(\hat{I}_{T}\) and the Classical Central Limit Theorem  
We discuss here a simple interpretation of the main result given in Section 2.2.