Rosa edited Now_we_treat_specifically_the__.tex  almost 9 years ago

Commit id: 5475053cbfc7ccc9c74ba3b3670b674308301a27

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\begin{equation}  H=\sum_{\sigma} \epsilon_{d\sigma} n_{d\sigma} + E_C (n_{d\sigma}-n)^2  \end{equation}  where $n_d$ is the dot charge  and for $n$ is the polarization charge. For  the coupling between the leads and the dot \begin{equation}  H=\sum_{\alpha k\in \{1,2\} \sigma} t_{\alpha k\sigma} e^{\varphi(t)} c^\dagger_{\alpha k\sigma} d_{\sigma} + h.c)  \end{equation} 

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  \begin{equation}  \Gamma(V)=\frac{1}{e^2R_T}\int_{\infty}^\infty \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)}} e^{i\phi(t)e^{-i\phi(0)}}\rangle  \end{equation}  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  \begin{equation}  \Gamma(V)=\frac{1}{e^2R_T}\int_{\infty}^\infty \Gamma(V)=\frac{1}{e^2R_K}\int_{\infty}^\infty  dE dE' f(E)[1-f(E'+eV)] P(E-E') \end{equation}  with   \begin{equation}  P(E)=\int_{\infty}^\infty dE e^{J(t)+i/\hbar e^{[J(t)+i/\hbar  E t} t]}  \end{equation}  and   \begin{equation}  J(t)=\frac{2\pi}{\hbarR_K}\int_{\infty}^\infty 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) \end{equation}  Note that, voltage fluctuations are related to current noise throught the transimpedance $Z(\omega)$. Here $R_K=h/2e^2$