For example, both the $\nu=1/3$ Laughlin state and the Moore-Read state has a simple interpretation in terms of composite fermions, which are bound states of an electron and two fluxes.

Both the Laughlin states and the Moore-Read state also have anyons, since they are both topologically ordered. Laughlin states have Abelian $ne/m$ anyons, with $m=1/\nu$ and $n<m$, and Moore-Read state hosts non-Abelian anyons $\sigma$ with charge $e/4$ and a neutral fermion $\chi$.

However, composite fermions themselves do not appear in the anyon contents of either state, despite being such an important step in describing these states. My question is why.


Anyons are physical excitations in FQH states (by definition, something that can be trapped locally by a potential well), while composite fermions are just symbols in a theory of FQH states. Such symbols usually do not correspond to physical excitations. So they are not included as the physical excitations (the anyons) since they are not.

  • $\begingroup$ Well, this answer is basically “they are not included because they are not”. My question is why the composite fermion does not correspond to physical excitations. In IQHE physical electrons form Landau levels and in FQHE composite fermions form Landau levels. Yet the former are in the spectrum of IQHE while the latter are not in the spectrum of FQHE. Are composite fermions confined, infinitely gapped, or something like that? $\endgroup$ – pathintegral Mar 17 '19 at 0:24
  • $\begingroup$ Composite fermions are not physical excitations, since composite fermions are always fermions while physical excitations are usually anyons. $\endgroup$ – Xiao-Gang Wen Mar 18 '19 at 1:37
  • $\begingroup$ But in general anyons can be fermions. Fermions do appear in the anyon content in many topological orders, such as Laughlin state, toric code, and Pfaffian state. $\endgroup$ – pathintegral Mar 18 '19 at 3:07
  • $\begingroup$ Fermion has Fermi statistics and anyon has fractional statistics. Fermion =/= anyon in general. In general anyons cannot be fermions. $\endgroup$ – Xiao-Gang Wen Mar 18 '19 at 13:44
  • 2
    $\begingroup$ No, this state is not in the physical Hilbert space, since it is not gauge invariant. If you apply composite fermion operator after gauge fixing, then the resulting state is some complicated superposition of the Anyons. Composite fermion picture can be misleading and fail to capture the essence of topological order. $\endgroup$ – Xiao-Gang Wen May 20 '19 at 13:08

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