3
$\begingroup$

In three spatial dimensions there are only two possible statistics: Bose-Einstein and Fermi-Dirac. This is the fact related with the statement that first homotopic group of 3-dimensional configuration space is permutations group. In a case of two spatial dimensions there are many types of statistics, which is related to fact that first homotopic group in two dimensions is braid group.

These are the words from Weinberg QFT Vol. 1. Can someone explain how explicitly appear many types of statistic in a case of two dimensions as the consequence of the statements above?

$\endgroup$
3
$\begingroup$

The standard treatment of identical particles starts with one-particle states and then imposes symmetry conditions. This is kind of backwards. If you look at a formula like $$|x_1,x_2\rangle = \frac{1}{\sqrt 2}(|x_1\rangle | x_2\rangle \pm |x_1\rangle| x_2\rangle)$$ you are saying that the state is a linear combination of states where particle A is at $x_1$ and particle B is at $x_2$ or vice versa. But what you want to implement is that there is no such thing as particle A and particle B, there's just two particles.

The more honest way is like this. Our particles move in $\mathbb R^k$, so we have $(\mathbb R^k)^n$ for $n$ particles. To make them indistinguishable, identify all points that just differ by permuting coordinates. This is taking the quotient with $S_n$, the permutation group for $n$ objects. For this quotient to be nice1, we have to delete points where two or more particles are in the same place. That is, our configuration space is $$M = ((\mathbb R^k)^n - \Delta)/S_n.$$

Now suppose that we have two particles in the state $|\psi(t=0)\rangle = |x_1,x_2\rangle$. Remember, this means that one particle is at $x_1$ and one particle is at $x_2$. The state evolves with the path integral, $$\langle \phi|\psi(t=T)\rangle = \int D\psi\, \exp(-i\textstyle\int_0^T \mathcal L \, dt)$$ where the integral is of course over all paths. The normalization of the path integral is set by the initial condition $$\langle\phi|x_1,x_2\rangle = \delta_{\phi,[x_1,x_2]} \tag{1}.$$ Now it is possible to argue that for short times, contributions from paths homotopic to the classical path $\gamma_c$ (i.e. the solution to the Euler-Lagrange equations) dominate. Therefore, the initial condition (1) only tells you about paths homotopic to the classical trajectory. You can define a different propagator by $$\langle \phi|\psi(t=T) \rangle = \int D\psi\, \sum_\alpha \chi(\alpha) \exp(-i\textstyle\int_0^T \mathcal L \, dt)$$ where $\alpha$ runs over the homotopy classes of paths between the initial and final points and $\chi(\alpha)$ are some complex weights. These paths can be labeled (not uniquely, but the ambiguity is inessential) with the fundamental group of $M$. Because of unitarity and the semigroup property of the propagator, we must have $|\chi(\alpha)| = 1$ and $\chi(\alpha)\chi(\beta) =\chi(\alpha\beta)$ the multiplication on the right being in the fundamental group. This is precisely the statement that the $\chi(\alpha)$ are a unitary representation of the fundamental group. This argument is presented in more detail (mostly it's about how to be careful about associating paths with elements of the fundamental group) in Ref. [1].

Since $\Delta$ consists of points verifying conditions like $x_1 = y_1, \ldots, x_k = y_k$, $\Delta$ is the union of spaces of codimension $k$. To affect the fundamental group, you need codimension $2$, therefore $(\mathbb R^k)^n - \Delta$ is simply connected for $k \ge 3$. (See Ref. [2] for a better argument.) Thus in three or more dimensions, $M$ is the quotient of a simply connected space, meaning that its fundamental group is $S_n$. The only unitary representations of $S_n$ are the familiar bosonic $$\chi(\alpha) = 1$$ and fermionic $$\chi(\alpha) = \begin{cases}+ 1 & \alpha \text{ is odd} \\ -1 & \alpha \text{ is even}\end{cases}$$ representations. For $k = 2$, this argument does not work, and the fundamental group of $M$ is a braid group. Again, see Ref. [2] for a full argument that holds also when the particles move on a general manifold.

To get back to the elementary treatment, in more than two dimensions, $F = (\mathbb R^k)^n - \Delta$ is the universal cover of $M$. Any loop in $M$ lifts to one in $F$, unique up to choice of starting point, with endpoint dependent only on the homotopy class. Thus consider moving our particles around in a homotopically non-trivial way but returning to the original state with one particle at $x_1$ and one at $x_2$. Lift this path so that it starts at $(x_1,x_2)$ (ordered pair!) and ends at $(x_2, x_1)$. Since on the actual configuration space $M$, this path enters the path integral with a $\pm 1$, we can get away with not considering $M$ if we have $|(x_1, x_2)\rangle = \pm |(x_2,x_1)\rangle$.

References

  1. Laidlaw, M. G. G. & DeWitt, C. M. Feynman Functional Integrals for Systems of Indistinguishable Particles. Phys. Rev. D 3, 1375-1378 (1971).
  2. Birman, J. S. On Braid Groups. Communications on Pure and Applied Mathematics 22, 41-72 (1969).


1 Here nice means that $M$ is a manifold. If you don't delete $\Delta$ you get something called an orbifold which is a much less nice class of objects.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.