Symmetry for the exchange of particles

Let's say we have two particles like in this picture - Here it is argued that in 3D after taking the left particle around trajectory 1 or 2 the system should be in the same state since trajectory 1 can be deformed continuously to create the other, which from the same reason is like leaving the particle in place (trajectory $0$), giving -

$$\left|\psi \left( 0 \right) \right>=\left| \psi \left( 1 \right) \right>=\left| \left( 2 \right)\right>$$ So by repeating this twice we get a phase who's square is 1.

But in 2D the different tracks cannot be continuously deformed to create each other so it is possible (but not necessary?) -
$$\left|\psi \left( 0 \right) \right>=\left| \psi \left( 1 \right) \right>\neq\left| \left( 2 \right)\right>$$

My questions are:

1. If only the final state determines the change of the system, why do we care if the paths can or cannot be continuously deformed to create each other? And if we care about the whole path, then how come necessarily in 3D $\left| \psi \left( 1 \right) \right>=\left| \left( 2 \right)\right>$?
2. Why is it said that path 2 is like encircling the other particle twice? (p.4 under eq. 1)

I'm going to answer your second question first because it is the easier of the two. When looking at the picture and looking at path $\lambda_2$, all you see is that particle $1$ moves around particle $2$. In this frame, $2$ is stationary. However, as the figure states, you should switch to the center of mass frame. In this frame, half way around path $\lambda_2$ $1$ is at the position of $2$ and $2$ is at the position of 1. This is to say that the two particles have been exchanged as would be defined in statistical mechanics. Finishing the path exchanges them again. Here, encircling is not a synonym for exchanging. The wording in the article is specific for that point. $1$ encircling $2$ corresponds to a pair of exchanges.

Now, to get a hold on thinking about the first question is to start with conservation principles. As physicists, we like conservation. Define your system properly and count everything involved carefully and the total energy/momentum/charge of the system doesn't change. It's the law for good reason.

If we've just got two (isolated) particles then we've got conservation of momentum by definition. To say that energy is conserved is just to say that we're going to count everything, but let's go further and assume that we have a slightly stricter condition: conservation of mechanical energy, which is to say that no energy is lost to spontaneous radiation or thermal interactions with the background, or anything like that. Let's make things simple: two particles.

Now, conservation of mechanical energy is the statement that all forces acting on the system are conservative. Specifically, remember that a force is conservative if and only if the work done is path independent. If we're talking about moving a particle around obviously we're talking about the existence of a force and since it must be conservative, only the initial and final states matter.

Now, let's switch to thinking about things in the phase space of the system. Here, we can dispense with forces, but we still need a way to keep the idea of path independence since it was central to our assumption of conservation of mechanical energy.

Now, if I have your picture above, every point around $\lambda_1$ represents the location of particle $1$ at some time in the evolution of the system. If I change it just a little bit, make it a little longer or shorter around the end then it doesn't really change what I'm doing with my particle there, right? If you think it does, then imagine smaller changes until you convince yourself that there is some tiny little change that, if you were to look at the system before and after, you wouldn't be able to tell whether or not it had happened. In this case, what you've done is convinced yourself that these two paths are equivalent and you can get the second one by a tiny nudge from the first. The appropriate way to say this is that the second is an infinitesimal deformation of the first. This leads easily to the following definition: two paths are equivalent if and only if one can be obtained from the other by a continuous deformation.

We can now answer the first part of your first question: That the final state determines the change in the system is a statement that the mechanical energy of the system is conserved, which is true if and only if the work done is independent of the path which is true if and only if all paths through the phase space can be continuously connected. It also answers the second part of your first question: In 3-D every path must be equivalent to doing nothing because I can always push them to slightly different and smaller paths until eventually I get to where nothing has gone anywhere.

If we stop here for a moment we can see, qualitatively, why there are only two possible kinds of exchange statistics in 3-D. Since all we've done is evolve the system around a closed loop to get exactly back where we started we must have that the unitary operator corresponding to this operation is the identity. Going back above, we saw that this corresponds to a pair of exchange operations, and so we must have then that $P_{12}^2 = 1$. If we assume (as we should be) that the two particles involved are identical, then $P_{12}$ corresponds to multiplication of $\vert \psi \rangle$ by a scalar, and so we need a number $P$ such that $P^2 = 1$. This of course gives us that the state vector must be either symmetric or anti-symmetric under the exchange of particles.

Thinking about things this way also illustrates why two dimensions are so weird. In two dimensions deforming $\lambda_2$ so that it does not encircle $2$ requires $\lambda_2$ be equivalent to a path going straight through $2$, which is to say that for a moment $1$ literally occupies the same physical space as $2$, which is totally verboten! Thus, $\lambda_1$ cannot be equivalent to $\lambda_2$!

An easy (and fun!) way to see this for yourself at home is to grab two pennies and a piece of string. The string represents the path $\lambda$ and path deformations are handled by holding down the ends and then pushing the string around. It is trivial to reproduce the paths $\lambda_1$ and $\lambda_2$. In particular, on a flat surface, take the string and place both ends underneath one penny, forming a loop, and place the second penny inside the loop (so we have a path equivalent to $\lambda_2$. In this way we can literally see that in 3-D every path is trivial - just pull the loop out of the plane and shrink it.

We can also see the oddity of two dimensions: if everything stays in the plane then the string can't come around the penny - $\lambda_2$ is not equivalent to $\lambda_1$. If I repeat $\lambda_2$ I now have two loops of string around the second penny. Finally, if I take an end of the string and trace it backwards along itself until the two ends are side-by-side again (at the first penny) I have a loop, but now that loop does not encircle the second penny and must be equivalent to the trivial path (and $\lambda_1$.

We can now finish the exercise. $\lambda_2$ and $\lambda_1$ are not equivalent, but $\lambda_2$ still corresponds to a pair of exchanges of identical (quasi-)particles. To see what kind of thing $\lambda_2$ does to the state vector we need to look at the group of rotations in 2-D, which is $SO(2)$. $SO(2)$ is commutative and the group of rotations in 3-D, $SO(3)$, is not. From this we obtain that the operator corresponding to $\lambda_2$ can, in general, be any phase $e^{\imath \phi}$. This makes apparent that if I trace $\lambda_2$ twice I'm going to get $e^{2 \imath \phi}$ but if I go the other way around and take $\lambda_2$ backwards, now I have a path in which $1$ does not encircle $2$ and this must be the identity. Everything else stays the same except that $\lambda_2^2$ is not the identity. This immediately gives us that for a system of $n$ particles the particle exchange operators must yield a representation of the braid group $B_n$ rather than the symmetric group $S_n$.

• Thank! I have two questions: 1. Is it possible to implement the exchange in the lab using only conservative forces? 2. Regarding the last paragraph, why is going backward on path $\lambda_2$ not like going forwards? – proton Dec 17 '17 at 8:10
• 1. Yes, though in this case the appropriate question is to ask "can the exchange be implemented adiabatically?" Quantum mechanics is built off of the Hamiltonian and Lagrangian formalisms and so really we're interested in the energetics. Bringing up conservative forces was to provide a foot hold on the explanation for path independence. 2. I've provided an explanation which addresses this in the new 3rd- and 2nd to last paragraphs. – Matthew Titsworth Dec 18 '17 at 6:22
• In the last paragraph, don't you imply that $\lambda_2^{-1}=1$? – proton Dec 18 '17 at 20:51
• No. The implication is that if I take $\lambda_2$ and then $\lambda_2^{-1}$ I get the identity. – Matthew Titsworth Dec 19 '17 at 3:07