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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

2 Answers 2

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

So, the issue with your argument is that you assume the Hilbert space of two particles is the tensor product of two one-particle Hilbert spaces, and then indistinguishability gives you a quotient of the two-particle Hilbert space. This is not the most correct way of thinking about it.

The correct way, going back to the (beautiful) original paper of Leinaas and Myrrheim, is to quantise the two-particles taking indistingushability into account during the quantisation.

I'll just summarise the idea here without talking about fibre bundles, for details it's worth going through that paper. My elementary knowledge of fibre bundles was enough to get through most of it in an afternoon (the part with spin is somewhat more confusing and also seems slightly fishy in a way I can't pin down).

If we postulate that two particles are identical, our quantisation itself must reflect that; that is, if we view quantisation as a process of assigning a basis element to every point in configuration space $(x_1,x_2)$ must go to the same basis element as $(x_2,x_1)$.

However, that is different from saying that $\psi(x_1,x_2) = \psi(x_2,x_1)$, since there's a lot more structure to the Hilbert space than just `assigning a basis element to every point in configuration space.' In particular, we usually require continuity and differentiability (I'll just call this smoothness, out of ignorance of details) of the wavefunction; there shouldn't be any unnatural jumps if we go from $(x_1,x_2)$ to $(x_2,x_1)$.

You should be wondering, why does this smoothness requirement give extra structure? For that, let's take a digression, based on Dirac's magnetic monopoles paper. Consider one particle in 3-d space. The wavefunction is defined only up to a phase, remember? Well, why then do we insist that the phase actually exists? The only thing that we strictly need is phase differences. So, if we take those words seriously, we are forced to conclude that the only reason it makes sense to speak of the absolute phase of a wavefunction is that some phase function that gives the correct phase differences exists. More precisely, there's a physically meaningful vector field $\kappa$ which encodes phase differences. Heuristically, we want this to be $\nabla \phi$ where $\phi$ is a phase function. But, this is only possible globally if $\nabla \times \kappa = 0$. If $\nabla \times \kappa \ne 0$ (as is the case when there's a magnetic flux; another case where this is possible, though not necessary, wherever the probability of finding the particle is $0$), what we can do is pick patches where $\nabla \times \kappa = 0$ and define a phase function within each patch. The restriction on the patches you can choose is given by the facts that $\oint \kappa \cdot dl \ne 0$ whenever the loop is such that any surface which ends on that loop passes through a point where $\nabla \times \kappa \ne 0$ whereas $\oint \nabla \phi \cdot dl = 0$ necessarily; no single patch can have one of these `non-contractible' loops.* With that restriction taken care of, suppose, for definiteness, you have two patches $A$ and $B$ ($\mathbb{R}^3 - (A \cup B)$ is the set of points where $\nabla \times \kappa \ne 0$); then there are two phase functions $\phi_A$ and $\phi_B$ and the consistency condition is that $\nabla \phi_A |_{A \cap B} = \nabla \phi_B |_{A \cap B} = \kappa |_{A \cap B}$. The takeaway is that there's a sense in which the wavefunction can be multivalued, and that's because we imposed smoothness on phase differences; and we need multivalued-ness if there are non-contractible loops.

Now, before we go back to two particles, suppose your space actually had a hole. Then, there would be no reason for $\oint \kappa \cdot dl$ along a non-contractible loop to be $0$. So, multivalued-ness is a general feature of spaces with non-contractible loops.

Now we come back to two particles. To review, we've assigned a basis element (and therefore a $1-$dimensional Hilbert space to each point in the $6-$d configuration space, with the constraint that we assigned the same basis element to $(x_1,x_2)$ and $(x_2,x_1)$. Another way to think about this is that we fold the configuration space itself, so that now we have some weird-looking space in which $(x_1,x_2)$ and $(x_2,x_1)$ are the same point (if the particles are in 1-d, this folded space look like what we get after folding a square piece of paper along the diagonal). Now, this folded space has a boundary -- the $x_1 = x_2$ bit. The main thing I want to show here is that there is a non-contractible loop here, and these non-contractible loops are qualitatively different in 2 and 3 dimensions.

Consider the wavefunction as particle 1 goes around particle 2, which is fixed at $a$ (this is not physical movement -- we're merely tracking the change in the wavefunction). In 3 dimensions, this loop is contractible; just change the plane of the loop till it doesn't enclose the other particle and then reduce its size. Therefore, in three dimensions the wavefunction is single-valued and your argument works. In 2 dimensions, however, it's not, and therefore the wavefunction may well be multi-valued. The particles for which this multvalued-ness turns up are called anyons.

Note: a lot of people will answer this in terms of adiabatic movement of particles around each other. The reason for this is that adiabatic movement approximates the `tracking the change of the wavefunction' that I described in the last paragraph.

*The word non-contractible seems a bit weird in my exposition. The intuition is this: if you took the regions where $\kappa$ has a curl to be holes in the space, then these loops would be ones which can't smoothly be contracted to $0$.

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