I've been wondering about the standard argument that the only possible identical particles in three dimensions are bosons or fermions. The argument goes like this:
Consider exchanging the positions of two identical particles in 3D. Because the particles are identical, $|\Psi|^2$ remains the same after having performed the exchange, so we must have $$\Psi \to \Psi e^{i\theta}.$$ If we perform the exchange twice, the path described is homotopic to the trivial path. This is easiest to see by realizing that the double exchange of particles A and B is the same as having particle A encircle particle B completely, and this path can just be lifted into the third dimension and shrunk down to a point. Alternatively, it's because $\pi_1(S^2/\mathbb{Z}_2) = \mathbb{Z_2}$.
Because the phase picked up along a path is a homotopy invariant, and the path of a double exchange is homotopic to the trivial path, $$e^{2 i \theta} = 1$$ which implies that under a single exchange, $\Psi\to\pm\Psi$, corresponding to bosons and fermions.
This argument sounds good, but it sneaks in the crucial physical input without justification: why should the phase be a topological invariant? Why can't it change under a deformation of the path? Not only do I not know how to prove this, it doesn't even appear to be true; for example, if there were a magnetic field, changing the path would change the magnetic flux through it, and hence the phase.
No comments:
Post a Comment