# Resistance in quantum transport

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?

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.