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If $|ψ\rangle$ is the state of a system of two indistinguishable particles, then we have an exchange operator P which switches the states of the two particles. Since the two particles are indistinguishable, the physical state cannot change under the action of the exchange operator, so we must have $P|ψ\rangle=λ|ψ\rangle$ where $|λ|=1$. Obviously switching the states of the two particles, and then switching them back, leaves the particles with their original states, so $P^2|ψ\rangle=(λ^2)|ψ\rangle=|ψ\rangle$, so $λ=±1$, and thus the state of the system must be either symmetric or anti-symmetric with respect to exchange.

Now I've heard that this reasoning does not hold for two dimensions, leading to the possibility of anyons, for which you can have $λ$ be something other than 1 or -1. How in the world is that possible? Where is the flaw or oversight in the above reasoning, that makes it exclude the 2D case? Where in the above proof are we assuming that space is three-dimensional?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: Let me present the proof in a step-by-step fashion, so the error can be more easily identified:

  1. For any states $|ψ_1\rangle$ and $|ψ_2\rangle$, define $P|ψ_1\rangle|ψ_2\rangle$ to be $|ψ_2\rangle|ψ_1\rangle$

  2. For identical particles, $P |\psi \rangle$ and $|\psi \rangle$ correspond to the same physical state (i.e. ray), so we must have $P |\psi \rangle = \lambda |\psi \rangle$ for some complex number $\lambda$.

  3. Applying the definition of $P$ in step 1 twice, we have $P^2|ψ_1\rangle|ψ_2\rangle=PP|ψ_1\rangle|ψ_2\rangle=P|ψ_2\rangle|ψ_1\rangle = |ψ_1\rangle|ψ_2\rangle$, so for any two-particle state $|\psi\rangle$, we have $P^2|\psi \rangle = |\psi \rangle$.

  4. Applying step 2 twice, we have $P^2|ψ\rangle = PP|ψ\rangle = P \lambda |ψ\rangle = \lambda P |ψ\rangle = \lambda^2 |ψ\rangle$

  5. By steps 3 and 4, we have $\lambda^2 = 1$ and thus $\lambda = ±1$

I assume the problem is with step 3 somehow, but I'm not sure what the problem is, since it follows directly from the definition in step 1. Is the problem with the definition in step 1, then? But how can a definition be wrong?

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Your argument is not correct, after double exchange, one need only $P^2|\psi\rangle = e^{i\alpha}|\psi\rangle$. This is because we are interested in physical states (and a phase factor does not change the physical state). The fundamental difference is topological. The fundamental group of $S0(n)$ is $Z_2$, for $n\geq3$, and $Z$ for $n=2$. The structure of this fundamental group is in direct relation with the possible statistics. –  Trimok Nov 11 '13 at 10:46
@Trimok Why would only the physical state need to stay the same? In one dimension, if the potential is symmetric about $x=0$, then the energy eigenstates must be eigenstates of a parity operator, and the square of that parity operator is 1. So what is the justification for that operator having a square of 1, and why doesn't it work in showing that the exchange operator we have here must also have a square of 1? –  Keshav Srinivasan Nov 11 '13 at 14:02
@Trimok: I think the point of Keshav is valid. In fact, your answer is based on what I called the "physical operation" on the wavefunction, that is, on take on particle and you make do a circle around the other. Then you need rotations and thus SO(n), and you can use your argument. –  Adam Nov 11 '13 at 14:33
@Trimok I hope it's now clear from my step-by-step proof that the fact that $P^2 |\psi \rangle = |\psi \rangle$ follows directly from the definition of $P$. –  Keshav Srinivasan Nov 11 '13 at 16:59
@KeshavSrinivasan : With your definition of $P$ in step $1$, it is obvious that the eigenvalues are $\pm 1$. But the physical operation $P_{phys}$ is in fact this : considering one particle fixed, and making a rotation of $\pi$ on the second particle. And the physical operation $P^2_{phys}$ is keeping one particle fixed, and making a rotation of $2 \pi$ on the second particle. And this involves topological properties of $SO(n)$ which may give phase factors which are different for $n=2$ and $n \geq 3$ –  Trimok Nov 12 '13 at 16:41

1 Answer 1

The thing is that the operation "exchange of two particles" has to be defined properly. What is the meaning of $P$ ?

We can imagine the operator $P$ is not physical (in the sense that is does not correspond to a physically possible operation). For instance, $P\psi(x_1,x_2)=\lambda\psi(x_2,x_1)$ in the sense that it only exchange the argument of the mathematical object $\psi(x_1,x_2)$. In that case, dimension of space does not play a role, and thus statistics should only be fermions or bosons.

But is it really reasonable to consider this mathematical operation $P$ that we can not do in the lab (and thus test) ? If not, then we should describe $P$ as a (complicated) operation on the particles, that really takes them and physically exchange them (that is : we apply a force on the particles to move them in space, until we go from $\psi(x_1,x_2)$ to $\psi(x_2,x_1)$). In this case, we directly see that dimensionality is important.

For instance, in dimension one, the particles have to go through each other. This is fine if they don't interact. But if they do interact, then the exchange might be ill-defined (it might completely change the state of the system). This is an heuristic reason why bosons and fermions are much the same in one dimension (which allow bosonization of fermions in 1D).

In 2D and 3D, the exchange of particle is well defined even if they interact, since it is sufficient that they are sufficiently far from each other to neglect the interaction. But then in 2D, there is this non-trivial issue that making a loop around something is not (necessarily) equivalent to doing nothing. You can count the number of loops, and there is no way to smoothly transform the loop into a point (but you can do that in any dimension greater than two).

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Why must an exchange operator correspond to a physically possible operation? Why can't it just be a purely mathematical operation? –  Keshav Srinivasan Nov 11 '13 at 13:54
This is a very good question. And people have thought that the mathematical operation was enough for a long time. I don't have a definite answer (I'd be glad if someone knows one), but maybe we can think of this problem that way : if the two approach (mathematical and physical) give the same answer (as in 3D), pick the one you want. If they give different answer, pick the one that can be realized in the lab. –  Adam Nov 11 '13 at 14:25
Also, it might be related to the fact that most known anyons are "relational anyons", meaning that the two particles that are exchanged are different particles, like electrons and magnetic flux. The only way to do a meaningful operation of exchange is then by making the electron do a loop around the flux "physically". –  Adam Nov 11 '13 at 14:28
@Keshav: I'll turn your question around: why must a purely mathematical exchange operator be well-defined? It certainly could be, and it is in some theories of physics. But must it be? –  Peter Shor Nov 11 '13 at 19:19
What makes you think $P |\psi\rangle = e^{i \theta} |\psi\rangle$ for some definite $\theta$? What does this mean? You're saying that the mathematical operation of swapping two identical particles corresponds to applying some definite phase to the state. But pulling the particles out of the universe, swapping them, and putting them back in doesn't correspond to anything physical, so why does there have to be a specific phase? To correspond to something physical, you need to say how you are going to swap them. –  Peter Shor Nov 11 '13 at 20:00

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