I know that this question has been submitted several times (especially see How are anyons possible?), even as a byproduct of other questions, since I did not find any completely satisfactory answers, here I submit another version of the question, stated into a very precise form using only *very elementary general assumptions of quantum physics*. In particular I will not use any operator (indicated by $P$ in other versions) representing the swap of particles.

Assume to deal with a system of a couple of identical particles, each moving in $R^2$. Neglecting for the moment the fact that the particles are indistinguishable, we start form the Hilbert space $L^2(R^2)\otimes L^2(R^2)$, that is isomorphic to $L^2(R^2\times R^2)$. Now I divide the rest of my issue into several elementary steps.

**(1)** Every element $\psi \in L^2(R^2\times R^2)$ with $||\psi||=1$ defines a state of the system, where $|| \cdot||$ is the $L^2$ norm.

**(2)** Each element of the class $\{e^{i\alpha}\psi\:|\; \psi\}$ for $\psi \in L^2(R^2\times R^2)$ with $||\psi||=1$ defines the same state, and a state *is* such a set of vectors.

**(3)** Each $\psi$ as above can be seen as a complex valued function defined, up to zero (Lebesgue) measure sets, on $R^2\times R^2$.

**(4)** Now consider the "swapped state" defined (due to (1)) by $\psi' \in L^2(R^2\times R^2)$ by the function (up to a zero measure set):

$$\psi'(x,y) := \psi(y,x)\:,\quad (x,y) \in R^2\times R^2$$

**(5)** The physical meaning of the state represented by $\psi'$ is that of a state obtained form $\psi$ with the role of the two particles interchanged.

**(6)** As the particles are identical, the state represented by $\psi'$ must be the same as that represented by $\psi$.

**(7)** In view of (1) and (2) it must be:
$$\psi' = e^{i a} \psi\quad \mbox{for some constant $a\in R$.}$$

**Here physics stops.** I will use only mathematics henceforth.

**(8)** In view of (3) one can equivalently re-write the identity above as

$$\psi(y,x) = e^{ia}\psi(x,y) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [1]\:.$$

**(9)** Since $(x,y)$ in [1] is *every* pair of points up to a zero-measure set, I am allowed to change their names obtaining

$$\psi(x,y) = e^{ia}\psi(y,x) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [2]$$

(Notice the zero measure set where the identity fails remains a zero measure set under the reflexion
$(x,y) \mapsto (y,x)$, since it is an isometry of $R^4$ and Lebesgues' measure is invariant under isometries.)

**(10)** Since, again, [2] holds almost everywhere for every pair $(x,y)$, I am allowed to use again [1] in the right-hand side of [2] obtaining:

$$\psi(x,y) = e^{ia}e^{ia}\psi(x,y) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\:.$$

(This certainly holds true outside the union of the zero measure set $A$ where [1] fails and that obtained by reflexion $(x,y) \mapsto (y,x)$ of $A$ itself.)

**(11)** Conclusion:

$$[e^{2ia} -1] \psi(x,y)=0 \qquad\mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [3]$$

Since $||\psi|| \neq 0$, $\psi$ cannot vanish everywhere on $R^2\times R^2$.
If $\psi(x_0,y_0) \neq 0$, $[e^{2ia} -1] \psi(x_0,y_0)=0$ implies $e^{2ia} =1 $ and so:

$$e^{ia} = \pm 1\:.$$

And thus, apparently, anyons are not permitted.

Where is the mistake?

**ADDED REMARK.** (10) is a completely mathematical result. Here is another way to obtain it. (8) can be written down as $\psi(a,b) = e^{ic} \psi(b,a)$ for some **fixed** $c \in R$ and **all** $(a,b) \in R^2 \times R^2$ (I disregard the issue of negligible sets). Choosing first $(a,b)=(x,y)$ and then $(a,b)=(y,x)$ we obtain resp. $\psi(x,y) = e^{ic} \psi(y,x)$ and $\psi(y,x) = e^{ic} \psi(x,y)$. They immediately produce [3] $\psi(x,y) = e^{i2c} \psi(x,y)$.

So the *physical* argument (4)-(7) that we have permuted again the particles and thus a further new phase may appear does not apply here.

**2nd ADDED REMARK.** It is clear that as soon as one is allowed to write

$\psi(x,y) = \lambda \psi(y,x)$ for a **constant** $\lambda\in U(1)$ and **all** $(x,y) \in R^2\times R^2$

the game is over: **$\lambda$ turns out to be $\pm 1$ and anyons are forbidden.**
This is just mathematics however. My guess for a way out is that the true configuration space is not $R^2\times R^2$ but some other space whose $R^2 \times R^2$ is the universal covering.

An idea (quite rough) could be the following. One should assume that particles are indistinguishable from scratch already defining the configuration space, that is something like $Q := R^2\times R^2/\sim$ where $(x',y')\sim (x,y)$ iff $x'=y$ and $y'=x$. Or perhaps subtracting the set $\{(z,z)\:|\: z \in R^2\}$ to $R^2\times R^2$ before taking the quotient to say that particles cannot stay at the same place. Assume the former case for the sake of simplicity. There is a (double?) covering map $\pi : R^2 \times R^2 \to Q$. My guess is the following. If one defines wavefunctions $\Psi$ on $R^2 \times R^2$, he automatically defines many-valued wavefunctions on $Q$. I mean $\psi:= \Psi \circ \pi^{-1}$. The problem of many values physically does not matter if the difference of the two values (assuming the covering is a double one) is just a phase and this could be written, in view of the identification $\sim$ used to construct $Q$ out of $R^2 \times R^2$: $$\psi(x,y)= e^{ia}\psi(y,x)\:.$$
Notice that the identity cannot be interpreted literally because $(x,y)$ and $(y,x)$ are the same point in $Q$, so my trick for proving $e^{ia}=\pm 1$ cannot be implemented. The situation is similar to that of $QM$ on $S^1$ inducing many-valued wavefunctions form its universal covering $R$. In that case one writes $\psi(\theta)= e^{ia}\psi(\theta + 2\pi)$.

**3rd ADDED REMARK** I think I solved the problem I posted focusing on the model of a couple of anyons discussed on p.225 of this paper matwbn.icm.edu.pl/ksiazki/bcp/bcp42/bcp42116.pdf suggested by Trimok. The model is simply this one:
$$\psi(x,y):= e^{i\alpha \theta(x,y)} \varphi(x,y)$$

where $\alpha \in R$ is a constant, $\varphi(x,y)= \varphi(y,x)$, $(x,y) \in R^2 \times R^2$ and $\theta(x,y)$ is the angle with respect to some fixed axis of the segment $xy$. One can pass to coordinates $(X,r)$, where $X$ describes the center of mass and $r:= y-x$. Swapping the particles means $r\to -r$. Without paying attention to mathematical details, one sees that, in fact:
$$\psi(X,-r)= e^{i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{i \alpha \pi} \psi(y,x)\quad (A)$$
for an anti clock wise rotation. (For clock wise rotations a sign $-$ appears in the phase, describing the other element of the braid group $Z_2$. Also notice that, for $\alpha \pi \neq 0, 2\pi$ the function vanishes for $r=0$, namely $x=y$, and this corresponds to the fact that we removed the set $C$ of coincidence points $x=y$ from the space of configurations.)

However a closer scrutiny shows that the situation is more complicated:
The angle $\theta(r)$ is not well defined without fixing a reference axis where $\theta =0$. Afterwards one may assume, for instance, $\theta \in (0,2\pi)$, **otherwise $\psi$ must be considered multi-valued**. With the choice $\theta(r) \in (0,2\pi)$, (A) does not hold everywhere. Consider an anti clockwise rotation of $r$. If $\theta(r) \in (0,\pi)$ then (A) holds in the form
$$\psi(X,-r)= e^{+ i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{+ i \alpha \pi} \psi(y,x)\quad (A1)$$
but for $\theta(r) \in (\pi, 2\pi)$, and **always for a anti clockwise rotation** one finds
$$\psi(X,-r)= e^{-i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{- i \alpha \pi} \psi(y,x)\quad (A2)\:.$$
Different results arise with different conventions. In any cases it is evident that the phase due to the swap process is **a function of $(x,y)$** (even if locally constant) and not a constant. This invalidate my "no-go proof", but also proves that the notion of anyon statistics is deeply different from the standard one based on the groups of permutations, where the phases due to the swap of particles is **constant** in $(x,y)$. As a consequence the **swapped state is different from the initial one**, differently form what happens for bosons or fermions and **against the idea that anyons are indistinguishable particles.** [Notice also that, in the considered model, swapping the *initial pair* of bosons means $\varphi(x,y) \to \varphi(y,x)= \varphi(x,y)$ that is $\psi(x,y)\to \psi(x,y)$. That is, **swapping anyons does not mean swapping the associated bosons**, and it is correct, as it is another physical operation on different physical subjects.]

Alternatively one may think of the anyon wavefunction $\psi(x,y)$ as a **multi-valued** one, again differently from what I assumed in my "no-go proof" and differently from the standard assumptions in QM. This produces a truly constant phase in (A). However, it is not clear to me if, with this interpretation the swapped state of anyons is the same as the initial one, since I never seriously considered things like (if any) Hilbert spaces of multi-valued functions and I do not understand what happens to the ray-representation of states.
**This picture is physically convenient, however, since it leads to a tenable interpretation of (A) and the action of the braid group turns out to be explicit and natural.**

Actually a last possibility appears. One could deal with (standard complex valued) wavefunctions defined on $(R^2 \times R^2 - C)/\sim$ as we know (see above, $C$ is the set of pairs $(x,y)$ with $x=y$) and we define the swap operation in terms of phases only (so that my "no-go proof" cannot be applied and the transformations do not change the states):

$$\psi([(x,y)]) \to e^{g i\alpha \pi}\psi([(x,y)])$$

where $g \in Z_2$. This can be extended to many particles passing to the braid group of many particles. Maybe it is convenient mathematically but is not very physically expressive.

In the model discussed in the paper I mentioned, it is however evident that, up to an unitary transformation, the Hilbert space of the theory is nothing but a standard bosonic Hilbert space, since the considered wavefunctions are obtained from those of that space by means of a unitary map associated with a singular gauge transformation,
and just that singularity gives rise to all the interesting structure!
However, in the initial bosonic system the singularity was pre-existent: the magnetic field was a sum of Dirac's delta.
I do not know if it makes sense to think of anyons independently from their dynamics.
And I do not know if this result is general. I guess that moving the singularity form the statistics to the interaction and *vice versa* is just what happens in path integral formulation when moving the external phase to the internal action, see Tengen's answer.

This post imported from StackExchange Physics at 2014-04-11 15:20 (UCT), posted by SE-user V. Moretti