The question is, in short, where in the composite fermion argument are electron-electron interactions used?

I know that interactions, namely Coulomb repulsion, between electrons are crucial in explaining the FQHE. Without them we simply fill the known single particle eigenstates and we are certain that at non integer filling fraction the system is not incompressible. In other words it is not gapped. Such a gap, seen at integer filling, along with disorder which pins the Fermi energy within this gap, is the reason for the IQHE. With this pinning we can guarantee that extended edge states are at the Fermi energy.

However, where in the composite fermion theory is the fact we have electron-electron interactions used? I know that adding flux quanta is analagous to adding higher order zeros in the wavefunction, which makes sense as this can be understood as electrons not wanting to be near each other. But one can still seemingly go through the composite fermion argument without expressing the desire to drive electrons further apart. I'll now go through the composite fermion argument, I'd appreciate comments on where electron interactions are needed for the argument to work.

  • Firstly we establish that adding an even number of flux quanta, antiparallel to the applied field, to each electron does not change the physics of our system. One does this by checking Aharonov–Bohm phases. The composite particles are still fermions and the spectra of eigenstates and eigenenenergies are unchanged. (I'm a little sketchy on this. I don't like the fact that even for particles on a ring, adding a flux quanta through the ring does keep the same eigenstates and eigenvalues, but eigenstates shift energies. Meaning the physics has changed. But this isn't too related to my main question. Comments would be appreciated however.)

  • One then takes a mean field approach. We add the new flux to the magnetic field to get an effective magnetic field. Typically this is less than the original magnetic field. Famously it can be made zero for filling fraction $1/2$. We then use this magnetic field to produce an effective filling fraction. For example, if our actual filling fraction is $v=1/3$ then, after adding two flux quanta to each electron, our effective magnetic field is $B_{eff}=B-2\phi_0 n$. Here $B$ is the original field, $\phi_0$ is the flux quanta and n is the 2D electron density. As filling factor is given by $v=n\phi_0/B$ the effective filling factor of our composite fermions is $v_{eff}=\frac{n\phi_0}{B-2\phi_0n}=\frac{1}{\frac{B}{n \phi_0}-2}=\frac{1}{3-2}=1$. Hence we have a filled Landau level of composite fermions. The system is gapped and we expect behaviour analogous to the IQHE.

  • Finally, we use Laughlin's corbino ring to deduce the Hall resistance. We adiabatically turn on a flux through the ring. Keeping with the filling fraction $1/3$ example, we must turn on three flux quanta to net transfer one electron to the outer edge of the ring. For IQHE with filling fraction one we would need just one flux quanta, but here we need another two as we are shifting two flux quanta as well as one electron. So to keep the magnetic field at the correct value, we add another two flux quanta to the centre of the ring. Let it take time $T$ and let it change at constant rate. The current is $e/T$. The voltage induced around the ring, due to changing flux, is $\frac{3h}{eT}$. Thus the Hall resistance is $\frac{3h}{e^2}=\frac{h}{\frac{1}{3}e^2}=\frac{h}{v_{eff}e^2}$.

So, what have I missed. Where are the electron interactions needed? On one hand I know that they are needed else we would just fill our single particle eigenstates and there is no gap at fractional filling fraction. On the other hand, I've just ran through an argument for FQHE without mentioning them. I must be missing something!

  • $\begingroup$ How do you expect vortices to attach to the electrons without interactions? I mean sure, without interactions you can write down a state of electrons with vortices attached, but it's the interactions that make this state energetically favorable. $\endgroup$
    – d_b
    Commented Oct 11, 2020 at 21:49
  • $\begingroup$ I was under the impression that these vortices are unobservable, the physics of the system remains the same with them on or off. Meaning we can just replace the electrons with the composite fermions and it is the same problem. $\endgroup$ Commented Oct 11, 2020 at 21:55
  • $\begingroup$ Obviously I'm misunderstanding something here $\endgroup$ Commented Oct 11, 2020 at 22:08
  • $\begingroup$ You mention choosing states but when we choose to add flux/vortices we are changing the problem. And we do so in such a way so as not to actually change it. Then we solve and there are a a spectrum of eigenstates. $\endgroup$ Commented Oct 11, 2020 at 22:22

1 Answer 1


The interaction is needed for the validity of the mean field approximation, which assumes that the gap remains intact. It is clear that without interaction the argument fails for reasons you have mentioned (no gap). It is unfortunately not possible to derive analytically for which interaction the gap will survive, so the argument only gives a possible way to understand the fractional quantum Hall effect, which must be confirmed by detailed calculation for a given interaction. The key prediction is that the spectrum of interacting electrons at B has the same structure as that of non-interacting electrons at B*, which has been confirmed by detailed calculations for electrons in the lowest Landau level interacting with the Coulomb interaction for a range of filling factors. The correspondence, and thus the mean field theory, breaks down not only for non-interacting electrons, but also for interacting electrons in, say, the 10th Landau level.


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