The surprising theoretical results of the previous section lead one to
question the conductance limit 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
.
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
.
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 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
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
-matrix
cannot be unitary
if a junction is to couple both A
E and B
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 , 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.