What exactly does an exchange of particle labels for identical particle wave functions mean *physically*?

Question

We know that the wavefunction of identical particles behaves as follows: $$\Psi(1,2)=\begin{cases}-\Psi(2,1) & \text{for fermions} \\ +\Psi(2,1) & \text{for bosons} \end{cases}$$

Now, what exactly does exchanging particle labels mean physically? The above relation, as far as I know comes due to the observables remaining the same for two identical particle system under the exchange of particle labels and thus- $$\Psi(1,2)=e^{i\delta}\Psi(2,1)$$ or $$\Psi(1,2)=e^{2i\delta}\Psi(1,2)$$giving us $\exp(i\delta)=\pm1$ which respectively accounts for bosonic and fermionic wavefunctions.

Answer 1

Yes the phase argument is a bit too simple.

For starters, it would not explain anyons n 2D. Secondly, why should $\Psi(1,2)$ be exactly equal to $e^{2i\delta}\Psi(1,2)$ and not just proportional to it? Quantum mechanics is defined over a projective Hilbert space, so any phase factor would not affect the physical observable anyway.

The first thing you need to realise is that the permutation of the particles' labels is a continuous operation over a parameter space, which cannot reliably be captured by an instantaneous operator that just slaps on a phase factor:

If the particles return to the same original position, they trace out a closed loop $\gamma$ (about a point $a$) in this topological space $X$. The description of loops requires basic topology -- in particular, we can use the fundamental homotopy group $\pi_1$ which comprises all topologically distinct loops, that is the different ways one can permute the particles labels.

The smooth and continuous label exchange operation being a proper rotation in $d$ dimensions, it is represented by $SO(d)$.

So you just need to know the fundamental homotopy group of $SO(d)$:

• $\pi_1$ of $SO(2) = \mathbb{Z}$,
• $\pi_1$ of $SO(3) = \mathbb{Z}_2.$

So, exchanging the particles means tracing out a trajectory in parameter space. A phase is picked up: $$ \Psi(1,2) \rightarrow \Psi(1,2) = e^{i\theta} \Psi(1,2),$$ which can be acquired in $\pi_1$ different ways:

• $2\pi / \mathbb{Z}_2$ in $d = 3$ (and $d>3$ for that matter), which means that the phase can be $2\pi / {1,2}$ that is $2\pi$ or $\pi$. Bosons and fermions;
• $2\pi / \mathbb{Z}$ in $d = 2$, hence anyons.

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Addition after comment.

2D

An easy way to visualise why, in 2D, you have a countably infinite way of permuting the particles. Say you want to go around $a$ (e.g. one particles is at $a$, and the other one is being rotated). You are in 2D so confined to a plane. You cannot avoid $a$. So the only way to go "round" $a$ is to cross it. The number of times you cross $a$, the winding number, is what distinguishes that path:

3D

For 3D it's similar albeit slightly more complicated.

Let's say that a rotation of $\theta$ around $\mathbf{n}$ is identical to one of $\pi-\theta$ about $-\mathbf{n}$ so we can consider a hemisphere (with opposite ends identified). What distinguishes the path now is whether or not they can be shrunk down to a point:

Answer 2

The exchange is an artifact of the formalism. When we write it down, we tend to label the particles as 1 and 2 (or something else), but physically there is no difference between the particles. They are not "labeled" in any way in the physical world. Therefore, we need rules for our formalism when we exchange the particles with the different labels. Note, the signs do not come from a phase. It is produced by the statistics, i.e., commutation or anti-commutation relations.