195. Integer quantum Hall effect – plateaus in Hall resistivity
Save This Post as PDF This post is my attempt to give an answer to tg’s question: Ok, we a kind of proved that the Hall resistivity is quantized in the integer quantum Hall effect, but how to explain plateaus in the Hall resistivity 
and deep minima in diagonal resistivity 
as shown on the Fig.
We have to introduce quite a non-trivial idea to explain both effects – we will say that electronic bound states exist in the material that do not contribute to the current and conductivity. There are several mechanisms for these bound states to appear – for example, Wigner cristallization or Anderson localization of electrons on the fluctuations of effective potential (generated, for example, by impurities in the material). The latter actually allows to explain almost all experimental data.
How does the Anderson localization work? Imagine that there are impurities built into the material of the 2-dimensional sample. In the presence of the random potential generated by them the energy of every Landau level becomes a function of coordinates 
. For every electron, its wave function is localized near equipotential surface
In places where random potential fluctuates strongly (say, near some point 
– I’ll choose it by hands), equipotential trajectories are closed, and wave function of the corresponding electron is also localized near . As a result, this electron cannot contribute to the overall current (at 
) simply because it cannot pass from one edge of the sample to the other.
Now, let us average over the random potential generated by impurities. What would be the result of such averaging? The Landau levels will acquire some width (related to the mean free path and, therefore, the strength of the random potnetial – and we will consider the case of the weak potential), that is, the density of states 
will be peaked near the Landau levels 
, but there also will be (localized) states with energy slightly larger or smaller than 
.
Let us consider the behaviour of the Hall resistivity changing the Fermi energy 
. First, if is located in between Landau levels 
and 
, the contribution into the overall current is given only by delocalized states with energies 
,
, and is given by the formula (5) that we derived in the previous post.
Let us slowly change the value of the Fermi energy . It is clear that as long as is in the interval of localized states near the Landau level 
, the Hall resistivity remains the same, since localized states do not contribute into the current. The situation finally changes when enters the interval of delocalized states near the Landau level . Concentration of carriers will rapidly change (new delocalized states are getting occupied!), and this will correspond to a sudden jump in the value of . If will grow even further, it will again enter the interval of localized states, near the level this time, and the Hall resistivity will remain the same.
How to explain the behaviour of diagonal resistivity ? Actually, I said enough for you to be able to find the answer yourselves – can you do that? 
Update. My explanation of the fact that vanishes is the following (see also in comments). Non-vanishing resistivity comes from dissipation effects. In order for the energy to be dissipated, the electron has to jump from a level of higher energy to a level of lower energy. Then, the difference can be transferred to the fluctuations of the lattice (to heat, in other words).
Now, if the temperature is not completely zero, there are always some empty states below the energy of a given electron. However, all these states (as well as the current state of electron) are localized – that is, separates in space, and the distance between them is way larger than the mean free path for electrons in the sample. So, basically, electrons near (and recall that all the states below are filled) cannot jump to lower levels, and the current flows unaffected by longitudinal resistivity.
What happens if approaches a Landau level ? The states in the closest vicinity of are delocalized, and electrons can jump from one state to another in the narrow region near . This leads to a slight jump in dissipation and diagonal resistivity. As you can imagine, the jump should be related to the width of the band of extended states near .
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Thanks for the post
Lets see, since the extended states are located only at Landau levels E_n and further if the fermi level doesn’t coincide with any of these levels then the particles won’t have any place to scatter too. Does that sound correct? If yes, then it may explain why \rho_{xx} is zero when E_f lies between Landau levels.
Thinking about the case when E_f coincides with one of the E_n’s I am not quite sure. Perhaps, the scattering rate is still small (compared to the H = 0 case) because of the sharpness of the extended states at E_n?
One related thing which I find a bit confusing is that when diagonal resistivity \rho_{xx} is zero and transverse \rho_{xy} isn’t, then the diagonal *conductivity* \sigma_{xx} is also zero. Maths is obvious but physics isn’t to me. Any simple way to look at it?
You are welcome
Regarding your explanation of vanishing
– not quite, if I did understand you correctly. My picture is the following. Non-vanishing resistivity comes from dissipation effects. In order for the energy to be dissipated, the electron has to jump from a level of higher energy to a level of lower energy. The energy surplus can be transferred to the fluctuations of the lattice etc. (to heat, in other words).
Now, if the temperature is not completely zero, there are some empty states below the energy of a given electron. However, all these states (as well as the current state of electron) are localized – that is, separates in space, and the distance between them is way larger than the mean free path for electrons. So, basically, electrons near
(and recall that all the states below 
are filled) cannot jump to lower levels, and the current flows unaffected by longitudinal resistivity.
What happens if
approaches 
? The states in the closest vicinity of 
are delocalized, and electrons can jump from one state to another in the narrow region near 
. This leads to a slight jump in dissipation and diagonal resistivity. As you can imagine, the jump should be related to the width of the band of extended states near 
.
As for the physics behind the relation between resistivity and conductivity, what would you want – they are both tensors, one is inverse of the other. It’s always a bit non-trivial if you are inverting the matrix
Not sure if there is some physics in this relation deeper than that.
Cheers,
Dmitry.
P.S. Heh, I think, it’s a nice explanation
Probably, will republish it as a separate post (or will make an update to this one).
Ummm…may be I wasn’t very clear but yours explanation seems to be similar to what I tried to write (unless I am lost
). Schematically, Dissipation rate for current ~ (backward) scattering rate ~ matrix element for scattering*phase space volume available –> 0 since near Fermi energy there aren’t extended states to connect to. Does that sound correct? Also, when $E_f$ coincides with some $E_n$ then the no. of gapless/extended states are proportional to bandwidth and thus scattering rate and hence resistivity should scale as bandwidth as you wrote.
Though thinking directly in real space as you explained seems more direct and physical.
Thanks, I think so. It seems that I was a bit overwhelmed by the beauty of my explanation
Cheers,
Dmitry.