Rosa edited We_start_by_applying_the__.tex  almost 9 years ago

Commit id: 91e193b33d4521e764300107672e3f586e0fdb76

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\end{multline}  Simplifying  \begin{multline}  I\propto (2\pi)^2f(\mu_d-\mu)  |t_{1}|^2 |t_{2}|^2 \Biggr\{ \int 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{multline}  Remarkably, this integral, and therefore the current $I$ is in general nonzero with an asymmetric environment.