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The question Statistics of bound states of anyons with order pq, and its answer inspires this question.

Suppose you have an anyonic particle with nonintegral spin s. Presumably, if there's an (unstable) bound state between this anyon and its antianyonic particle (an analog of a positronium), the total angular momentum (plus relative orbital angular momentum) of this bound state will have to be integral. This is because this bound state would presumably decay away entirely into say photons or phonons which are bosonic.

What this means is if you encircle an antianyon around an anyon of the same species counterclockwise by $2\pi$, you pick up no phase factor of the sort $e^{i\theta}$, or rather, $\theta=0$. Otherwise, you pick up a nonintegral contribution to the spin of the bound state coming from the relative orbital angular momentum.

OK, but now consider the following three particle system. In region A, you have an anyon. In region B, you have a localized bound state of an anyon and its antianyon of the same species. The bound state is bosonic. Presumably, that means if you encircle the bound state once counterclockwise around the anyon at A by $2\pi$, you pick up no phase factor? At any rate, once it decays away entirely into photons and phonons, it can't pick up any phase factor, and by continuity over time, this means it can't pick up any phase factor before it decays either. But won't this phase factor be the sum of the phase factors you pick up by encircling an anyon around the anyon at A, and encircling an antianyon around the anyon at A? According to the second paragraph, the latter phase has to be zero? So, the former phase also has to be zero? But when you encircle an anyon about another anyon, you pick up a nontrivial phase factor, right?

What am I missing here? This issue doesn't crop up for fermions because a single encirclement leads to a phase factor of $(-1)^2$, which is just 1.

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2 Answers 2

Let me elaborate on goojob's answer. Of course the spin of the antianyon is s, and not -s. This comes from CPT reversal. C doesn't change the spin at all. T reverses the sign of the spin. P is a reflection about an odd number of spatial dimensions. With three spatial dimensions, or an odd number of them, an inversion which reflects all the spatial dimensions changes parity. Such an inversion doesn't flip the signs of any the spin components. Spin is a 2-vector, or in equivalently in 3D, an axial vector. So, with an odd number of spatial dimensions, the sign of the spin flips under CPT. With two spatial dimensions on the other hand, to reverse parity, we should only reflect one spatial dimension, and that flips the sign of the spin. Two flips, one from T and the other from P, means the spin doesn't change sign.

So, if we have a bound state of p anyons and q antianyons, it will have a total spin of $(p-q)^2 s \mod 1$ . What matters is the total anyonic charge p-q. If we have two composite anyons with anyonic charges m and n respectively, under a full rotation of one about another, it will pick up a phase factor of $e^{4\pi i mns}$.

Suppose we have two anyonic "flavors", with spins $s_1$ and $s_2$ respectively. Then, a bound state with anyonic flavor charges of m and n respectively will have a total spin of $m^2 s_1 + n^2 s_2 + mn r \mod 1$ where r is an additional parameter. If we have composite anyons with charges $(m_a,n_a)$ and $(m_b,n_b)$ respectively, the phase factor will be $\exp\{2\pi i [2m_a m_b s_1 + 2n_a n_b s_2 + (m_an_b + n_a m_b)r] \}$. So the phase factor clearly depends upon more than just the spins.

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It is not true that you pick no phase if you drag an anyon along a closed contour encircling its anti-particle. It's pretty obvious that you pick the opposite phase than if you encircle two anyons around each other. Imagine that the phase is due to a "charge" or "flux" at the origin, much like in the Aharonov-Bohm effect, so if you revert the charge, you revert the phase, too.

So if two anyons change their wave function by a factor of $\exp(i\theta)$ after a 360-degree revolution, an anti-anyon/anyon pair doing the same 360-degree rotation will pick $\exp(-i\theta)$ while anti-anyon/anti-anyon pair will pick the same $\exp(i\theta)$ as the original anyon pair.

This is also consistent with the fact that an anyon-anti-anyon bound state behaves as a boson that may encircle anything without changing the phase factors. The phases from the anyon and the anti-anyon cancel.

I think that you got your incorrect guess that the anyon-anti-anyon mutual revolution picks no phase by imagining that anyon has spin $s$ so the antiparticle has the spin $1-s$ modulo one and you added these two spins, getting zero. But this is a wrong calculation of the monodromy because the relevant phase is derived from $s_1 s_2$, the product of the two spins, modulo 1, not from their sum. Moreover, the question whether the antiparticle has spin $s$ or $-s$ is a bit subtle, too.

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