Question

Find the equation of right circular cylinder having the line

−2

2

=

−1

1

=



3

and passes through the point (0, 0, -1).

Answer 1

You have just seen that Laughlin's argument explains the quantization of the Hall conductance in terms of a pump which moves electrons through the bulk of a Hall cylinder, from one edge to the other of the cylinder.

Compare the situation with the simple electron pump which you studied earlier in the lecture. There, the pump moved electrons from one metallic lead to the other. Clearly the pump worked thanks to the availability of electronic states at the Fermi level in the two metallic leads. Otherwise, it would have no electrons to take and no place to drop them. Without the metallic leads, the pump would be like an empty carousel.

When applied to the Hall cylinder, this simple reasoning shows that Laughlin's argument necessarily implies the presence of electronic states localized at the edges of the sample.

It is in fact very easy to convince ourselves that such states must exist. We just need to think again about the classical trajectory of an electron with velocity vv moving in a perpendicular magnetic field BB. This trajectory is a circular orbit with radius given by the cyclotron radius.

What happens to the classical trajectory of an electron when the center of the orbit is too close to the edge of the cylinder, say closer than a cyclotron radius? It is easier drawn than said: