# Do anyons obey the exclusion principle?

In general, if we have two indistinguishable particles in states $$\psi_1$$ and $$\psi_2$$, then starting in the combined state $$|\psi_1\psi_2\rangle$$ and then exchanging them will produce the state $$e^{i\theta}|\psi_2\psi_1\rangle$$, or $$e^{i2\pi s}|\psi_2\psi_1\rangle$$. For bosons, the phase factor $$e^{i\theta} e^{i2\pi s} = e^{i2\pi}$$ comes out to one, so given two particles in the same state we get the tautological statement $$|\psi\psi\rangle = 1|\psi\psi\rangle \rightarrow |\psi\psi\rangle = |\psi\psi\rangle$$. For fermions, however, phase factor comes out to negative one, and we get $$|\psi\psi\rangle = -|\psi\psi\rangle$$, which can only be satisfied if the amplitude is zero—i.e., the state has zero probability. And thus, we get the well-know Pauli Exclusion Principle: no two fermions can occupy the exact same quantum state.

For anyons, the phase factor can be any value on the complex unit circle. Thus, for a pair of particles in identical states, we get the generalized expression $$|\psi\psi\rangle = z|\psi\psi\rangle$$.

If $$z$$ happens to be 1, then we're dealing with bosons, and everything is fine. But what if it is neither 1 nor -1? Is the result that anyons end up obeying an exclusion principle just like fermions, with bosons being the unique special case, or is it possible to have states which remain unchanged under certain phase shifts, which would thus be allowed (resulting in a rather more complicated sort of semi-exclusion)?