the average temperature, so linear-response [@Mahan-Sofo1996] works well, even though $(T_L-T_R)/T_L$ is of order one.

In contrast, for quantum systems ($L \ll l_{\rm rel}$), the whole temperature drop occurs on the scale of a few nanometres or less, and so linear-response theory is inapplicable whenever $(T_L-T_R)/T_L$ is not small.

Carnot efficiency {#Sect:Carnot}
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A system must be reversible (create no entropy) to have Carnot efficiency; proposals exist to achieve this in bulk [@Mahan-Sofo1996] or quantum [@Humphrey-Linke2005; @Kim-Datta-Lundstrom2009; @Jordan-Sothmann-Sanchez-Buttiker2013] thermoelectric. It requires that electrons only pass between reservoirs L and R at the energy where the occupation probabilities are identical in the two reservoirs [@Humphrey-Linke2005]. Thus, a thermoelectric requires two things to be reversible. Firstly, it must have a $\delta$-function-like transmission [@Mahan-Sofo1996; @Humphrey-Linke2005; @Kim-Datta-Lundstrom2009; @Jordan-Sothmann-Sanchez-Buttiker2013; @Sothmann-Sanchez-Jordan-Buttiker2013], which only lets electrons through at energy ${\epsilon}_0$. Secondly,[@Humphrey-Linke2005] the load’s resistance must be such that ${ e^{\operatorname{-}} }V = {\epsilon}_0 (1-T_R/T_L)$, so the reservoirs’ occupations are equal at ${\epsilon}_0$, see Fig. \[Fig:Fermi\].

By definition this means the current vanishes, and thus so does the power output, $P_{\rm gen}$. However, one can see how $P_{\rm gen}$ vanishes by considering a quantum system which lets electrons through in a tiny energy window $\Delta$ from ${\epsilon}_0$ to 