# Composite Fermion Approach to Quantum Hall

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 $$B* = B - (m-1)n\Phi_0$$ 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?

• I have no idea what the answer is, but I hope this question gets a good answer. I found those notes exceptionally clear until they got to composite fermions, and then absolutely nothing made any sense. Commented May 3, 2019 at 10:52
• @knzhou well, at least I'm not alone here Commented May 3, 2019 at 10:54
• If you don't get answers here, you could also try emailing Tong. He's very concerned about the clarity of the notes and has helped me in the past. Commented May 3, 2019 at 10:55

1. In the CF's view, the Laughlin wave function is made up of two parts. First, it is just the $$\prod_{i which represents the $$\nu=1$$ IQH state for noninteracting electrons. Second, it is the $$\prod_{i which representing attaching $$m-1$$ vortices to each electron. The second term describe the strong correlation of FQH to the first order approximation. To answer your question more specific, think of $$\nu=1$$ IQH state as a special Laughlin state with zero vortex attachment, thus zero correlation between electrons. Here $$m=1$$; and you don't have any attached vortex.(The IQH are formed by noninteracting electrons.)
2. As I already mentioned above, the general idea of CF is mapping strongly-correlated electrons to noninteracting electrons to the first order approximation. The non-interacting electrons only contribute AB phase, which is the first term (try to check it with $$\nu=1$$ again). The second term comes from the attached vortices of the moving CF. The detailed derivation is given on p 143 of Jain's book (Cambridge 2007).
3. The 1/2 state is a CF Fermi-sea state, which is essentially different from Laughlin (CFFS is gapless but Laughlin is gapped.) Here the number of attached vortices for CFFS is 2, not 1! So $$B^*=B-mn\Phi_0$$ instead. The CFFS wave function is a product of slater determinant of plane waves times $$\prod_{i. The SD part describe noninteracting electrons in zero field, and the second part describes the attached vortices. Here maybe you can see the answer to your first question more clear. The noninteracting electron part is not always $$\prod_{i! It is only so when the mapped noninteracting electrons fill the lowest effective LL, which is the case for Laughlin but not for other states. You can have a Laughlin state at 1/2, which is simply $$\prod_{i. Here one of $$(z_i-z_j)$$ comes from noninteracting electrons and the other comes from attached vortice. But that is a bosonic state, not fermionic.

According to the original HLR paper https://journals.aps.org/prb/pdf/10.1103/PhysRevB.47.7312

it is incomplete in Jain's approach to start with Laughlin states: they correspond to $$\nu^*=1$$ IQH states of composite fermions, and states higher in the hierarchy correspond to higher filling IQH states. But what about zero magnetic field composite fermion fermi liquid?

HLR says we should start with $$2k$$ flux quanta attached to a fermion (the number of flux quanta must be an even integer to ensure the composite fermion satisfies fermion statistics), find the magnetic field $$B$$ such that the effective magnetic field for composite fermion $$B^*=B-2k\Phi_0 n_e$$ vanishes $$B^*=0$$ to get the composite fermion Fermi liquid. Then $$v^*=p$$ IQH states of CF has $$B^*=n_e\Phi_0/p$$. Transforming to electrons, we have a magnetic field $$B=\frac{2pk+1}{p}n_e\Phi_0$$, or $$\nu=\frac{p}{2pk+1}$$. For $$k=1$$ this reproduces the Jain sequence, where we have $$2$$ flux quanta attached to each electron in $$\nu=1/3$$ FQHE.