# Fermions, Bosons and Anyons.

• I

## Main Question or Discussion Point

A straightforward argument for showing that indistinguishable particles in 3D can either be bosons or fermions goes as follows.

Consider the wavefunction of identical particles 1 and 2 at positions $\psi (\vec{r}_1, \vec{r}_2)$. If we swap these particles around then this just becomes $\psi (\vec{r}_2, \vec{r}_1)$. Since the particles are indistinguishable we expect our observations not to change, so:
$$\left|\psi (\vec{r}_1, \vec{r}_2)\right|^2 = \left|\psi (\vec{r}_2, \vec{r}_1)\right|^2$$
Implying
$$\psi (\vec{r}_1, \vec{r}_2) = e^{i \phi}\psi (\vec{r}_2, \vec{r}_1)$$
If we swap the particles twice, we expect to get back to the original state, so we get:
$$\psi (\vec{r}_1, \vec{r}_2) = e^{2i \phi}\psi (\vec{r}_1, \vec{r}_2)$$
Which means $e^{i\phi} = \pm 1$. Fermions then correspond to the - sign, and Bosons to the + sign.

I'm aware this only applies in 3D, and not, for example in 2D. A knockdown of what I've read around is that swapping the particles involves moving them such that they never occupy the same position. Equivalently you move from a relative position vector $\vec{r} = \vec{r}_2 - \vec{r}_1$ to $-\vec{r}$ along some path in $\vec{r}$ space that doesn't pass through the origin.

The explanation is then that while all paths in 3D (and higher) space between two points are topologically equivalent, a path in 2D that wraps around the origin cannot be smoothly varied into a path that does not, since we have specified that a path cannot pass through the origin.

What I'm confused by is why does it matter whether the paths are topologically equivalent, if the paths are between the same two states? All I used to argue about the existence of fermions and bosons are the states. Since the wavefunction represents all we can know about a system, why does it matter fundamentally about how we got there? I think essentially I'm asking why topology is the fundamental consideration here, rather than state.

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DrDu
Your argument also ignores the possibility to get a more complex behaviour when more than 2 particles are involved.

Demystifier
As pointed out in
https://www.physicsforums.com/insights/anyon-demystified/
it is really about finding an interaction which makes wave functions obey a strange statistics. It turns out that such interaction is possible only in 2 dimensions. So fundamentally it's not about paths, it's about interactions. But interaction turns out to be a kind of "gauge" interaction, described by a two-point vector potential ${\bf A}({\bf x}_1, {\bf x}_2)$. As in ordinary gauge interaction, the corresponding phase factor of the wave function is something like
$$e^{i\int d{\bf x}\cdot {\bf A}}$$
and this is where the path comes from. Essentially, the path appears in a way similar to that in the Aharonov-Bohm effect.

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• Jezza
DrDu
Thank you for all the responses! I think I'm beginning to see what's going on. The first place I read about it considers paths where $|\vec{r}|$ is kept constant, so that paths lie on a sphere (or circle in 2D) (A similar explanation is here - pg6 onwards: https://arxiv.org/abs/hep-th/9209066). I think I'm mistakenly identifying the phase picked up by the wavefunction with the angular distance between points on the sphere (e.g. in 3D a $\pi$ phase shift is picked when moving between diametrically opposite points on the sphere (one exchange), which are obviously spatially separated by an angle $\pi$).
I think I'm now right in saying that the two quantities bear no relation, and that the global phase picked up depends on the path taken. Since we haven't got just 2 categories of closed path in 2D, we can't place any constraints on the phase picked up on a path and so we can go no further than specifying some arbitrary phase $e^{i\phi}$.