No, Quanta is fine -- you're making a subtle error, which is in fact common even in graduate textbooks.

There are two distinct things going on here: formally "swapping" the arguments in the wavefunction, and physically moving the particles around. A priori, there is no reason the phase shifts from these two actions should be related.

You're focusing on the swap, but it's not interesting: your argument that the swap phase squares to 1 applies to both 2D and 3D equally well. When people talk about anyons, they're talking about the phase from moving the particles. That has more options in 2D because the paths one can take when moving particles around can have a greater range of topologies (https://en.wikipedia.org/wiki/Braid_group).

The reason that textbooks mix these things up is because in relativistic quantum field theories these two phases are always the same, by virtue of the spin-statistics theorem. But the anyons of condensed matter are nonrelativistic.

I maintain that Quanta is definitely wrong. Quoting them again:

> To understand the quantum kingdoms, think of a drawing of loops. Imagine two indistinguishable particles, like electrons. Take one, then loop it around the other so that it ends up back where it started. Nothing seems to have changed. And indeed, in the mathematical language of quantum mechanics, the two wave functions describing the initial and final states must be either equal or off by a factor of −1. ... If the wave functions are identical, your quantum particles are bosons. If they’re off by a factor of −1, you have fermions.

This is nonsense. In 3D, you can't even define what it means to loop one particle around another, and there is nothing at all that says that fermions behave in a particular way when you loop them around some other particle.

Okay, now I see what you mean. They were going in the right direction, but they did mess up. Everywhere they talk about moving a particle in a loop around the other, they really should have talked about moving two particles to exchange them. An exchange is of course topologically distinct from doing nothing. A loop is two exchanges in a row -- which is, as you say, topologically trivial in 3D (so the loop phase must be 1, and the exchange phase is +1 or -1, allowing for bosons and fermions).

I'm a little bit curious here. My understanding is that the wave function has complex values and the "state" is described by the wave function times its complex conjugate. Couldn't they have meant the factor of -1 to be on the imaginary portion and then things still work out to have the same probability distribution?

Suppose you have a very simple system with two indistinguishable particles in one dimension and no spin or any other complications. The wavefunction is just a complex valued function of two variables:

f(x0, x1)

where x0 and x1 are coordinates of the two particles.

The fact that the particles are indistinguishable means that no possible experiment can tell them apart. After a bit of math, this means that f(x0, x1) = k * f(x1, x0) where k is a constant. All wavefunctions are normalized, so |k|^2=1, which constrains k to be on the unit circle.

But you can go farther: it's possible, at least in principle, to come up with an experiment that measures k. So you start with the particles in their initial state, do your experiment, and learn k. (Maybe you need to repeat it a bunch of times -- no big deal.) Now you repeat the same experiment but start with the two particles having their roles swapped. Since the particles are indistinguishable, you had better measure the same value for k. This means:

f(x0, x1) = k * f(x1, x0)

and also

f(x1, x0) = k * f(x0, x1) = k^2 * f(x1, x0)

So k^2 = 1, and that rules out everything except k=1 or k=-1.

Why does the factor of -1 just have to be on the imaginary proportion? seems like this would be true if you multiplied the whole thing by -1.