What is the maximum conductance of a single quantum channel?

The surprising theoretical results of the previous section lead one to question the conductance limit $2e^2/h$ for a single quantum channel of the general 1D-Fermi-gas type described in Section 7.2. For this gedanken-experiment we will consider conventional electron waveguides which are single-mode and long enough that evanescent waves are negligible, but $\ll l_\phi$. We try to encourage more current to pass down a single-mode channel (E) by connecting it to a reservoir via multiple routes (A,B,C,D), as shown in Fig 7.4a, where two routes are used on each side. It is possible to match the junctions so that a wave entering down A,B,C, or D has no reflection back along the same lead. In this case we might guess that the hypothetical left-side observer (from the previous section) would see the single-mode entrances to guides A and B as two `black dots', giving twice the effective absorption cross section, and therefore infer a conductance of twice $2e^2/h$. We might also justify this by saying that waves travelling down A and B will meet and continue down E, and since they have no particular phase relation, their currents will add to give a doubled current through E, as would be necessary.

However there is a fundamental flaw in the above reasoning. The ABE junction can be designed so that if waves come down A and B in phase, they will be adiabatically transformed into the lowest transverse mode of E, so will propagate through to the right side without reflection, carrying a current of twice that of a usual single-mode guide. However, if A and B are $\pi$ out of phase, the same adiabatic transformation must result in the second transverse mode, which is evanescent. So this latter wave will reflect perfectly back out of the left side, and carry no current. Plane waves are impinging from the left reservoir uniformly over all angles, and because of the $> \lambda$ separation of the entrances, an average over angles gives an average over relative phase in A and B. Thus we are left with no increase above the single channel conductance. This property of the ABE junction is not merely practical; rather, it is easy to show that its $3\times3$ $S$-matrix cannot be unitary if a junction is to couple both A$\rightarrow$E and B$\rightarrow$E with unity transmissions. Such an appealing junction is therefore ruled out on the grounds of flux conservation. A consequence is that the entrances to A and B can at most appear `half black' to the observer, due to waves which enter A then exit B and vice versa.

This suggests another way to try and defeat the conductance limit: direct the incoming plane waves in a narrow enough angular distribution so that waves always come down A and B in phase, and this will double the conductance. (This is similar to experiments [198] where the series resistance of two QPCs was found to be less than the sum of the individual QPC resistances, because collimation at the exit of the first QPC illuminated the second with a narrow beam, increasing its conductance). However, this beam is no longer a thermal occupation of incoming states. This illustrates the inextricable link between thermal Fermi occupation of reservoir states and the universal quantum of conductance. At $T{=}0$, thermal occupation at a given chemical potential difference implies that all quantum states lying in the appropriate energy range are filled in the left reservoir and empty in the right. Semiclassically, this corresponds to a uniform distribution in phase space, or when projected into momentum states, uniform in angle, as exemplified by Eq.(7.1). The semiclassical viewpoint allows one to see that since transformations in phase space cannot change the phase space density (Liouville's theorem), neither can the universal conductance per quantum channel be changed. This reminds us that unitarity in quantum mechanics is analogous to Liouville's theorem in classical mechanics.