We start by applying the P(E)-Theory to the case of a quantum dot which is unequally coupled to a resistive envinroment. For such calculation we utilize the following expression \[\Gamma_{i\rightarrow j}^{l}= 2\pi \int |t_{l}|^2 \rho(\epsilon_i-\mu_i) \bar\rho(\epsilon_j-\mu_j)P_{l}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j\,\] Let us consider a processes in which we fill up a quantum dot with one electron. In this case we need to compute \(\Gamma_{0\rightarrow 1}^{1}\). We denote the leads with labels \(l=\{1,2\}\) for left and right contacts. The dot occupation in this case changes from \(n=0\) to \(n=1\). Therefore \(\rho(\epsilon_i-\mu_i)=f(\epsilon_i-\mu_1)\) and \(\bar\rho(\epsilon_j-\mu_j)=\delta(\epsilon_j-\mu_d)\) \[\Gamma_{0\rightarrow 1}^{1}= 2\pi \int |t_{1}|^2 f(\epsilon_i-\mu_1) \delta(\epsilon_j-\mu_d)P_{1}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j\] On the other hand when the dot becomes empty, an electron in the dot leaves it through barrier \(2\), the transition rate is \[\Gamma_{1\rightarrow 0}^{2}= 2\pi \int |t_{2}|^2 \rho(\epsilon_i-\mu_i) \bar\rho(\epsilon_j-\mu_j) P_{2}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j\] where now \(\rho(\epsilon_i-\mu_i)=\delta(\epsilon_i-\mu_d)\), and \(\bar\rho(\epsilon_j-\mu_j)=[1-f(\epsilon_j-\mu_2)]\). The transition rate read \[\Gamma_{1\rightarrow 0}^{2}= 2\pi \int |t_{2}|^2 [1-f(\epsilon_j-\mu_2)] \delta(\epsilon_i-\mu_d) P_{2}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j\] We now perform the product of these two rates, i.e., \(\Gamma_{0\rightarrow 1}^1\Gamma_{1\rightarrow 0}^2\) and the integration using the delta functions, then we get \[\begin{gathered} \Gamma_{0\rightarrow 1}^1\Gamma_{1\rightarrow 0}^2= (2\pi)^2 \int \int |t_{1}|^2 |t_{2}|^2 [1-f(\epsilon_j-\mu_2)] \\P_{2}(\mu_d-\epsilon_j) f(\epsilon_i-\mu_1) P_{1}(\epsilon_i-\mu_d) d\epsilon_id\epsilon_j\end{gathered}\] However, the current is proportional to the balance between the two processes \(\Gamma_{0\rightarrow 1}^1\Gamma_{1\rightarrow 0}^2 \) (which is already computed) and its reversal sequence, from right to left \(\Gamma_{0\rightarrow 1}^2\Gamma_{1\rightarrow 0}^1\). Using previous result we have \[\begin{gathered} \Gamma_{0\rightarrow 1}^2\Gamma_{1\rightarrow 0}^1= (2\pi)^2 \int \int |t_{1}|^2 |t_{2}|^2 [1-f(\epsilon_j-\mu_1)] \\ P_{1}(\mu_d-\epsilon_j)f(\epsilon_i-\mu_2) P_{2}(\epsilon_i-\mu_d) d\epsilon_id\epsilon_j\end{gathered}\] As mentioned \(I\propto \Gamma_{0\rightarrow 1}^1\Gamma_{1\rightarrow 0}^2- \Gamma_{0\rightarrow 1}^2\Gamma_{1\rightarrow 0}^1\) Then we have \[\begin{gathered} I\propto (2\pi)^2 \int \int |t_{1}|^2 |t_{2}|^2 \Biggr\{ [1-f(\epsilon_j-\mu_1)] P_{1}(\mu_d-\epsilon_j)f(\epsilon_i-\mu_2) P_{2}(\epsilon_i-\mu_d)\\ -[1-f(\epsilon_j-\mu_2)] P_{2}(\mu_d-\epsilon_j)f(\epsilon_i-\mu_1) P_{1}(\epsilon_i-\mu_d) \Biggr\} d\epsilon_id\epsilon_j\end{gathered}\] Let us made a quick check when there is no environment, then \(P_{1,2}(\epsilon_{i,j}-\mu_d)=\delta(\epsilon_{i,j}-\mu_d)\), that we can use for integration in \(\epsilon_{i,j}\), then \[\begin{gathered} I\propto (2\pi)^2 |t_{1}|^2 |t_{2}|^2 \left\{ [1-f(\mu_d-\mu_1)] f(\mu_d-\mu_2)-[1-f(\mu_d-\mu_2)]f(\mu_d-\mu_1) \right\}\end{gathered}\] (Note that we have assumed that \(t_1\), and \(t_2\) are energy-independent). Then, \[\begin{gathered} I\propto (2\pi)^2 |t_{1}|^2 |t_{2}|^2 \left\{f(\mu_d-\mu_1)] -f(\mu_d-\mu_2) \right\}\end{gathered}\] Note that at, under equilibrium conditions, \(\mu_1=\mu_2\) the current vanishes identically as expected.

We are back to the full calculation. In order to made progress analytically we impose equilibrium condition where \(\mu_1=\mu_2=\mu\), then \[\begin{gathered} I\propto (2\pi)^2 \int \int |t_{1}|^2 |t_{2}|^2 \Biggr\{ [1-f(\epsilon_j-\mu)] P_{1}(\mu_d-\epsilon_j)f(\epsilon_i-\mu) P_{2}(\epsilon_i-\mu_d)\\ -[1-f(\epsilon_j-\mu)] P_{2}(\mu_d-\epsilon_j)f(\epsilon_i-\mu) P_{1}(\epsilon_i-\mu_d) \Biggr\} d\epsilon_i d\epsilon_j\end{gathered}\] We factorize the Fermi functions inside the integrals \[\begin{gathered} I\propto (2\pi)^2 \int \int |t_{1}|^2 |t_{2}|^2 [1-f(\epsilon_j-\mu)] f(\epsilon_i-\mu)\\ \Biggr\{ P_{1}(\mu_d-\epsilon_j) P_{2}(\epsilon_i-\mu_d)- P_{2}(\mu_d-\epsilon_j) P_{1}(\epsilon_i-\mu_d) \Biggr\} d\epsilon_i d\epsilon_j\end{gathered}\] Considering the limit in which only one of the contacts is coupled to the environment, let us say lead 1, then we have \(P_2=\delta(\epsilon_i-\mu_d)\) \[\begin{gathered} I\propto (2\pi)^2 \int |t_{1}|^2 |t_{2}|^2 \Biggr\{ [1-f(\epsilon-\mu)] f(\mu_d-\mu) P_{1}(\mu_d-\epsilon)\\ -[1-f(\mu_d-\mu)] f(\epsilon-\mu) P_{1}(\epsilon-\mu_d)\Biggr\} d\epsilon \\ \end{gathered}\] Simplifying \[\begin{gathered} I\propto (2\pi)^2 |t_{1}|^2 |t_{2}|^2 \Biggr\{ f(\mu_d-\mu)\int [1-f(\epsilon-\mu)] P_{1}(\mu_d-\epsilon)d\epsilon \\ -[1-f(\mu_d-\mu)]\int f(\epsilon-\mu) P_{1}(\epsilon-\mu_d) d\epsilon \Biggr\} \end{gathered}\] Remarkably, this integral, and therefore the current \(I\) is in general nonzero with an asymmetric environment.