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## 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 [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.

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.