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Composite Fermion Approach to Quantum Hall

Physics Asked by yankyl on May 7, 2021

I am following David Tong’s notes on the Quantum Hall Effect (https://arxiv.org/abs/1606.06687). One of the approaches he takes to the FQHE is the composite fermion approach (Section 3.3.2). There are two things I am struggling with.

First of all he says that a vortex is something around which a wavefunction picks up a phase of 2$pi$. He then says that a single electron in the a laughlin state with angular momentum $= m$ can be seen as an electron with $(m-1)$ vortices attached to it. This interpretation is based on the fact that in the Laughlin wavefunction the terms are of the form $(z_i-z_j)^m$ for which he says that the first $(z_i-z_j)$ is needed for that fermi statistics, whereas the remaining m-1 terms are just vortices.

Why this distinction? Something of the form $(z_i-z_j)^m$ should have a wavefunction pick up a phase of 2$pi$m regardless of what you call it. i.e. why isnt what he is calling an electron a vortex as well?

And this matters, in equation 3.34 when he is computing the berry phase he has the Aharonov-Bohm term and then an additional 2$pi$(m-1) for each each electron as though only the $m-1$ vortices contributed a phase and not the "electron".

The other thing I am struggling with is the $nu$ = 1/2 Landau level. He derives (Eqn. 3.35) that the effective magnetic field is $B* = B – (m-1)nPhi_0$ where n is the density. The density is given by $n = nu B/Phi_0$. Therefore B* = B(1-$nu$(m-1)). A Laughlin state with angular momentum = m has filling fraction $nu$ = 1/m so we get that B* = B(1-(m-1)/m) = B/m. How then does the $nu$ = 1/2 filled landau level give you B* = 0 in equation 3.40?

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