Gaps in FQHE bulk spectrum

Question

In David Tong's note on Quantum Hall Effect, Tong states on page 85 that the incompressible Laughlin wavefunction is "responsible for the gap in the bulk spectrum" of FQHE (Fractional quantum Hall effect). It is not very clear to me how this statement is being made.

In particular, my questions are:

1. What is the relationship between the toy Hamiltonian in 3.17 and the actual FQHE Hamiltonian? Why do we need the toy Hamiltonian in the first place?

2. The toy Hamiltonian will have Laughlin wavefunction in the bottom as ground state, with a finite energy gap to the first excited state, due to its incompressibility. But how does this relate to the spectrum gap in the actual Hamiltonian?

Answer 1

1. The answer to this question involves Haldane's pseudopotentials.

The "actual" Hamiltonian is the projection of the Coulomb interaction to the lowest Landau level (LLL) plus some confining term. For the case of two particles and an isotropic potential, you can write the interaction in terms of Haldane pseudopotentials $v_{m^\prime}$ as $$ H = \sum_{m^{\prime}} v_{m^\prime} P_{m^\prime}, $$ where $P_{m^\prime}$ is the projector onto states with relative angular momentum $m^\prime$. You can also generalize this to more than two particles.

In particular, for the Coulomb potential you find (see e.g. Jain's book on Composite Fermions, Chap. 3.11.2) $$ v_{m^\prime} = v_0 \frac{\Gamma(m^\prime+\frac{1}{2})}{\Gamma(m^\prime+1)}. $$ Note, that this is a monotonically decreasing function of $m^\prime$.

Adding the confining term (and using that high angular momentum states are radially more extended) you find the Hamiltonian in Tong's notes.

Now, for spin-polarized electrons as in the FQHE you can only have odd $m^\prime$ because of the anti-symmetry of the wavefunction. Therefore, it is sufficient, to evaluate the pseudopotentials for these odd $m^\prime$s.

2. Trial wave functions from the toy Hamiltonian

You might now want to try and find a simple, exact ground state of a truncated toy Hamiltonian involving only some of the pseudopotentials ($m^\prime \leq m$) and hope that they already capture the essential physics, while giving you analytical trial wave functions.

For sure, these wave functions won't be exact eigenstates for the true interaction, but in the case of Laughlin's wave functions already give you an excellent approximation of the real physics (as can be confirmed numerically).

In particular, one might hope for the spectra of the two Hamiltonians to share the essential features as for example the number of degenerate ground states, gapped excited states, as well as even physical properties of the excited states.