Comments on: 195. Integer quantum Hall effect – plateaus in Hall resistivity http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/#utm_source=feed&utm_medium=feed&utm_campaign=feed For physicts by physicists Tue, 31 Jan 2012 20:24:03 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Dmitry http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/comment-page-1/#comment-5657 Dmitry Sat, 24 Jan 2009 21:13:24 +0000 http://www.nonequilibrium.net/?p=1160#comment-5657 <blockquote>Does that sound correct?</blockquote> Thanks, I think so. It seems that I was a bit overwhelmed by the beauty of my explanation ;-) Cheers, Dmitry.

Does that sound correct?

Thanks, I think so. It seems that I was a bit overwhelmed by the beauty of my explanation ;-)

Cheers,
Dmitry.

]]> By: tg http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/comment-page-1/#comment-5647 tg Sat, 24 Jan 2009 01:07:23 +0000 http://www.nonequilibrium.net/?p=1160#comment-5647 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. 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.

]]> By: 198. Fractional quantum Hall effect - a few words about theory http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/comment-page-1/#comment-5642 198. Fractional quantum Hall effect - a few words about theory Fri, 23 Jan 2009 17:06:34 +0000 http://www.nonequilibrium.net/?p=1160#comment-5642 [...] after going through the integer quantum Hall effect, we are a kind of ready to discuss one of the biggest puzzles in condensed matter theory: the [...] [...] after going through the integer quantum Hall effect, we are a kind of ready to discuss one of the biggest puzzles in condensed matter theory: the [...]

]]> By: Dmitry http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/comment-page-1/#comment-5639 Dmitry Fri, 23 Jan 2009 12:14:32 +0000 http://www.nonequilibrium.net/?p=1160#comment-5639 You are welcome :) Regarding your explanation of vanishing [tex]\rho_{xx}[/tex] - 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 [tex]E_F[/tex] (and recall that all the states below [tex]E_F[/tex] are filled) cannot jump to lower levels, and the current flows unaffected by longitudinal resistivity. What happens if [tex]E_F[/tex] approaches [tex]E_n[/tex]? The states in the closest vicinity of [tex]E_n[/tex] are delocalized, and electrons <em>can</em> jump from one state to another in the narrow region near [tex]E_n[/tex]. 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 [tex]E_n[/tex]. 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). You are welcome :)

Regarding your explanation of vanishing \rho_{xx} – 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 E_F (and recall that all the states below E_F are filled) cannot jump to lower levels, and the current flows unaffected by longitudinal resistivity.

What happens if E_F approaches E_n? The states in the closest vicinity of E_n are delocalized, and electrons can jump from one state to another in the narrow region near E_n. 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 E_n.

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).

]]> By: tg http://www.nonequilibrium.net/195-integer-quantum-hall-effect-diagonal-resistivity/comment-page-1/#comment-5634 tg Thu, 22 Jan 2009 23:28:05 +0000 http://www.nonequilibrium.net/?p=1160#comment-5634 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? 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?

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