Why does topology matter in determining fermions and bosons?

In summary, 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,
  • #1
Jezza
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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 [itex]\psi (\vec{r}_1, \vec{r}_2)[/itex]. If we swap these particles around then this just becomes [itex]\psi (\vec{r}_2, \vec{r}_1)[/itex]. Since the particles are indistinguishable we expect our observations not to change, so:
[tex]
\left|\psi (\vec{r}_1, \vec{r}_2)\right|^2 = \left|\psi (\vec{r}_2, \vec{r}_1)\right|^2
[/tex]
Implying
[tex]
\psi (\vec{r}_1, \vec{r}_2) = e^{i \phi}\psi (\vec{r}_2, \vec{r}_1)
[/tex]
If we swap the particles twice, we expect to get back to the original state, so we get:
[tex]
\psi (\vec{r}_1, \vec{r}_2) = e^{2i \phi}\psi (\vec{r}_1, \vec{r}_2)
[/tex]
Which means [itex]e^{i\phi} = \pm 1[/itex]. 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 [itex]\vec{r} = \vec{r}_2 - \vec{r}_1[/itex] to [itex]-\vec{r}[/itex] along some path in [itex]\vec{r}[/itex] 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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  • #2
Your argument also ignores the possibility to get a more complex behaviour when more than 2 particles are involved.
 
  • #3
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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  • #4
You can view it this way: Your wavefunction depends on r1 and r2 or equivalently (r1+r2)/2, the center of mass, and r=r2-r1, the relative position.
Clearly, R is uninteresting and we can ignore it in the following, so we take psi to depend on the vector r. Now in 2d, as particles are indistinguishable, we can restrict the polar angle phi to the range 0 to pi. We now have to specify the boundary conditions on psi when we go back from phi=pi to 0. In 2d, psi can get multiplied by any phase factor. However, in 3d we have to consider both phi and theta.
 
  • #5
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 [itex]|\vec{r}|[/itex] 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 [itex]\pi[/itex] phase shift is picked when moving between diametrically opposite points on the sphere (one exchange), which are obviously spatially separated by an angle [itex]\pi[/itex]).

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 [itex]e^{i\phi}[/itex].

Does this sound like I've got the right idea?
 

1. What are fermions and what are their distinguishing characteristics?

Fermions are a type of subatomic particle that follow the Fermi-Dirac statistics and have half-integer spin. They are the building blocks of matter and are characterized by their unique property of obeying the Pauli exclusion principle, which states that no two fermions can occupy the same quantum state simultaneously.

2. What are bosons and how do they differ from fermions?

Bosons are another type of subatomic particle that follow the Bose-Einstein statistics and have integer spin. They are characterized by their ability to occupy the same quantum state simultaneously, unlike fermions. They are responsible for mediating fundamental forces, such as the photon for electromagnetism and the gluon for the strong nuclear force.

3. What are anyons and how are they different from fermions and bosons?

Anyons are a theoretical type of subatomic particle that have fractional spin. They are neither fermions nor bosons, but exhibit properties of both. They are predicted to exist in two-dimensional systems and are relevant in the study of topological states of matter.

4. How are fermions, bosons, and anyons important in the field of quantum mechanics?

Fermions, bosons, and anyons are important in quantum mechanics because they follow different statistical distributions and have different properties, which play a crucial role in understanding and predicting the behavior of particles at the subatomic level. Their study is essential in developing theories and models for the behavior of matter and energy.

5. What are some real-life applications of fermions, bosons, and anyons?

Fermions, bosons, and anyons have a wide range of applications in different fields. For example, fermions are crucial in the understanding of materials and their properties, while bosons are important in the field of quantum computing. Anyons are being studied for their potential use in building quantum computers and for their role in topological quantum computing.

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