If we are working with nontrivial topological insulator with broken time reversal symmetry then we can expect that we have some chiral edge states. Chiral states have the property that the current can go in only one direction, so the current cannot scatter and because of that the current should be dissipationless. However, I want to know if there is really no heat loss in such systems?

For example consider the four-terminal Hall system on the picture (red lines represent chiral edge states):

Topological insulator with chiral edge states

We can examine this system using Landauer-Buetikker formula: $$ \tag{1} I_i = \frac{e^2}{h} \sum_{j=1}^{N} \left( T_{ji} V_i - T_{ij} V_j \right) $$ where

  1. $N$ is the number of terminals. In our case $N=4$.
  2. $T_{ij}$ is transmission probability from contact i to contact j. In our particular example we have that $T_{i+1,i} = 1$ (with the convention that $T_{N+1,N} \equiv T_{1,N}$), whereas all the other $T_{ij}$ are zero since the conducting states are chiral.
  3. $V_i$ is the voltage on terminal $i$.
  4. $I_i$ is the difference between the current coming out the terminal $i$ and the current coming in the terminal $i$.

Now, let's create a voltage drop between terminals 1 and 4 (using a battery for example). Since the terminal 2 is not attached to battery, all the current that comes in has to go out so we have that $I_2 = 0$ (have in mind the definition of $I_i$). Using this equation and (1) we can easily obtain that $V_1 = V_2$. Similarly, using $I_3 = 0$ we obtain $V_3 = V_2$. Finally, using equation (1) on terminal 1, we can obtain: $$ \tag{2} I_1 = \frac{e^2}{h} (V_1 - V_4) $$ From this we can calculate two-terminal resistance: $$ R_{14,14} \equiv \frac{V_1-V_4}{I_1} = \frac{h}{e^2} $$ Does this nonzero resistance imply that we have some kind of heat dissipation in Hall systems like this one? If so, then why do we often call edge states dissipationless?

I emphasize that I'm quite confident that presented example has no mistakes because it can be found in Tkachov as well as in this paper.


1 Answer 1


Resistance does not mean that there is heat dissipated. The resistivity tensor $\rho$ is a matrix that relates the electric field $\mathbf E$ to the current density $\mathbf j$, as $$\mathbf E = \rho \mathbf j.$$ In an ideal quantum Hall system with filling factor $\nu=1$, the resistivity is $$\rho = \frac{h}{e^2}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$

Applying an electric field (voltage) in the vertical direction will induce a current in the horizontal direction and vice-versa. The power dissipated at each point along the edge is given by $$p = \mathbf E \cdot \mathbf j = (\rho\mathbf j) \cdot \mathbf j = 0$$

i.e. There is no dissipation as the voltage across terminals induces a current perpendicular to it.

  • $\begingroup$ I am not so sure that this is actually true. Of course, everything you have written works perfectly fine in the bulk and I agree with it. It is the standard argument that we have no dissipation in the bulk for the case of antisymmetric resistivity tensor .However, that does not mean that it is also true for the edge states. Roth (paper I referenced in my question) explicitly states that Ohm's law does not hold for the edge states because they are not localized! That's why we have to use Landauer-Buetikker formula. This means that your argument fails in the case of edge states. $\endgroup$
    – RedGiant
    Apr 27, 2020 at 12:44
  • $\begingroup$ Ohm's law is not valid in the sense that $\mathbf E \propto \mathbf j$, however, the fact that they are related by a matrix $\rho$ is what it means to work in the realm of linear response theory. This is still true for the edges. $\endgroup$
    – user111476
    May 5, 2020 at 19:40
  • $\begingroup$ I think that n the Landauer-Buttiker formula (2) the energy is dissipated in the leads not in the edges. $\endgroup$
    – mike stone
    Aug 24, 2021 at 22:24

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