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I know that due to quantisation condition of conductance: $$G = \frac{2e^{2}}{h}$$ The minimum resistance possible in nano scale channels is around 13k$\Omega$. This conductance assumes ballistic transport and hence electron does not undergo any collision, then why does such a huge resistance shows up? is it because the number of electrons passing through the channel is very small or there is something more fundamental to this high value of resistance?

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In the context of transport through point contacts/narrow leads, the actual formula for ballistic conductance is \begin{equation*} G=\frac{2e^2}{h}N_c \end{equation*}

where $N_c$ is the 'number of conduction channels'. What this means is electrons passing through a narrow channel is effectively in an infinite square well in the lateral direction and is therefore constrained to have momentum in the lateral direction as $\hbar n\pi/W$, where $n$ is some integer and $W$ is the width of the channel. In other words, not all electrons approaching a narrow constriction can pass through it due to this condition.

Given that the transverse momentum can atmost be $\hbar k_F$, where $k_F$ is the Fermi wavenumber, $N_c \pi/W \lesssim k_F \implies N_c \sim k_FW/\pi$, for $k_FW\gg 1$.

Note that in the classical limit $k_F W\rightarrow \infty$, the ballistic conductance is infinite, so the resistance is 0.

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