Now we treat specifically the double quantum dot system in which one part of the system plays the role of “detector” whereas the other dot is the so-called “device”. By doing this we model the particular \(P(E)\) function \cite{Aguado_2000}. Let us assume the following Hamiltonian for the leads \[H=\sum_{k\alpha\in \{1,2\}\sigma} \epsilon_{\alpha k\sigma} c^\dagger_{\alpha k\sigma} c_{\alpha k\sigma}\] for the central interacting site \[H=\sum_{\sigma} \epsilon_{d\sigma} n_{d\sigma} + E_C (n_{d\sigma}-n)^2\] where \(n_d\) is the dot charge and \(n\) is the polarization charge. For the coupling between the leads and the dot \[H=\sum_{\alpha k\in \{1,2\} \sigma} t_{\alpha k\sigma} e^{\varphi(t)} c^\dagger_{\alpha k\sigma} d_{\sigma} + h.c)\] where \(e^{\varphi(t)}\) contains the operator \(\hat\varphi\) that shifts a well defined charge state by an amount of one, i.e., \[e^{i\hat\varphi} |Q-e\rangle = |Q\rangle\] \(\hat\varphi\) operator is conjugate variable of the charge operator \(Q\), i.e., \([Q,\varphi]=ie\). Besides, \(\varphi=\varphi(0)-e/ \hbar V t\), where \(V\) is the voltage source. Note that in circuits containing capacitors or inductors phases differences are added up as \(eVt/\hbar\). Indeed \(\varphi=\int^t_\infty U(t´)dt´=Q/C\). Therefore when we treat a transport regime described by Markovian processes where tunneling events are modeled by transition rates within a Master equation framework, the fact of having such tunneling terms with the operator \(\varphi\) in it, it is translated into a Fermi-Golden rule for the rates as \[\Gamma(V)=\frac{1}{e^2R_K}\int_{\infty}^\infty dE dE' f(E)[1-f(E')] \int_{\infty}^\infty \frac{dt}{2\pi\hbar} e^{\frac{i}{\hbar}(E-E'+eV)t}\langle e^{i\phi(t)e^{-i\phi(0)}}\rangle\] Note that the product \(f\times (1-f)\) is proportional to the Bose distribution function and denote the distribution of electron-hole excitations. Defining \(J(t)\) as the Fourier transform of the phase correlator that appears in \(\Gamma(V)\) we have that \[\Gamma(V)=\frac{1}{e^2R_K}\int_{\infty}^\infty dE dE' f(E)[1-f(E'+eV)] P(E-E')\] with \[P(E)=\int_{\infty}^\infty dE e^{[J(t)+i/\hbar E t]}\] and \[J(t)=\frac{2\pi}{\hbar R_K}\int_{\infty}^\infty \frac{|Z(\omega)|^2}{\omega^2} S_{I}(\omega)(e^{i\omega t}-1)\] Note that, voltage fluctuations are related to current noise throught the transimpedance \(Z(\omega)\). Here \(R_K=h/2e^2\)

Therfore for our particular case of a double dot, all that we need is to compute the \(P(E)\) properly considering that the current noise produced in the “drive” system yields the \(P(E)\) function that define the new transition rates in the detector or dragged system. Now the question is the following. By implementing this scheme we can follow two different strategies.

• Consider that the drive system is weakly uncoupled from the dragged dot and just replace the noise produced in a single dot in the \(P(E)\) expression.

• Consider the whole system and make the calculation in a kind of “self-consistent” manner