Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Anyons only in 2+1 spacetime dimensions - better explanation

+ 4 like - 0 dislike
1067 views

Regrading why anyons exist only in 2+1 spacetime dimensions (which have an arbitrary phase on exchange), I read the reason that the paths for exchange in 3D are deformable into each other while in 2D, they may not be deformed into each other. So what is next? How to prove from this, that there can be arbitrary phase generated on exchange? I understand that in 2D, we have braids, but can I get a proof for an arbitrary phase on exchange which specifically takes into account the topological equivalence of paths in 3D and not in 2D.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user cleanplay
asked Apr 29, 2014 in Theoretical Physics by cleanplay (80 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

Please, let me first refer you to the original paper by Leinaas and Myrheim where the existence of anyonic statistics was first predicted before its actual discovery. All the ingredients of the understanding of the special properties of the two dimensional case already exist in this old paper, however, I'll try to cast it in a more modern terminology:

By passing to a polar center of mass coordinates, the configuration space of two scalar identical particles in $\mathbb{R}^n$ can be represented as:

$$\frac{\mathbb{R}^n \times \mathbb{R}^n}{\sim } = \mathbb{R}^n \times \mathbb{R}^+\times \frac{S^{n-1}}{\mathbb{Z}_2}$$

Where: $\sim$ is the identical particles equivalence relation, the $\mathbb{R}^n$ on the right hand side is the center of mass coordinates, $\mathbb{R}^+$ is the radial coordinate, and $ \frac{S^n}{\mathbb{Z}_2}$ are the angular coordinates. The equivalence relation corresponding to the exchange in the angular coordinates is simply the identification of the antipodal points on the sphere's surface, which accounts for the factor ${\mathbb{Z}_2}$ in the denominator. The central of mass and the radial coordinates are transparent to the exchange, thus we may concentrate on the angular coordinates, which actually consist of the real projective spaces:

$$ RP^n = \frac{S^n}{\mathbb{Z}_2}$$

These spaces are not simply connected, their fundamental groups can be easily deduced from the contractibility of the closed loops on the sphere with the identification of the antipodal points as indicated in the question:

$$\pi_1(RP^n ) = \mathbb{Z}_2, n>1$$

$$\pi_1(RP^1 ) = \mathbb{Z}$$

The usual way to quantize on a non-simply connected manifold is to construct wave functions on the universal covering with appropriate transformation properties upon the exchange:

$$ \psi(x^a) = e^{i\phi} \psi(x) $$

Where $x^a$ is the antipodal point to $x$. Of course, the transformation can be only a phase multiplication, because the choice of the point or its antipode should not change the expectation values of the observables since both points correspond to the same physical point on the configuration space.

Moreover, since $S^n$ is simply connected then, the map $S^n \rightarrow RP^n$ is a covering map, therefore, $ RP^n = \frac{S^n}{\pi_1(RP^n ) }$. The phase transformation must be a representation of $\pi_1(RP^n )$. In our case, when $n>2$, $\mathbb{Z}_2$ has only two phase representation, the trivial representation and the alternating representation.

However, in the two dimensional case, we may choose a representation $\gamma$ in which the generator $\Gamma$ of $ \mathbb{Z}$ is represented by an arbitrary constant phase $e^{i\phi}$, then the representation of an arbitrary element in $ \mathbb{Z}$ will be:

$$\gamma(\Gamma^n) = e^{in\phi}$$

Now, please remember that in the geometric picture of the fundamental group, the generators are represented by closed loops, thus the representation $\gamma$ assigns a phase to every closed loop (such that the composition law of the loops is reflected in the multiplication of the phases). This assignment can be described as the existence inequivalent quantizations corresponding to the set of maps:

$$ \mathrm{Hom}(\pi_1(RP^n ), U(1))$$

When $n>2$, this set contains only two classes, (Bosons and fermions), while when $n=2$, this set is infinite.

In summary, there will be a set of possible quantizations, in each set the angular wave equation should be solved with a different condition upon the exchange of the antipodal points on the sphere. The solution will correspond to a different particle type.

Leinaas and Myrheim, solved the problem of the identical two dimensional isotropic harmonic oscillator with wave functions transforming with an arbitrary phase upon the antipodal identification and found that the spectrum depends on the transformation phase factor.

Now, according to the classification theorem of flat connections, every (projective) representation of the fundamental group can be associated to a flat connection $A$ such that:

$$\gamma(\Gamma) = e^{\int_{\Gamma} A_{\gamma}}$$

This connection is essential for the construction of the quantum operators corresponding to the classical functions on the phase space according the Koopman - Van Hove representation:

$$ \hat{O}_{\gamma} = O - i \hbar X_{O} -A_{\gamma} (X_{O} )$$

Where $O$ is a function on the phase space, $ \hat{O}_{\gamma}$ is the corresponding quantum operator, and $X_O$ is the corresponding Hamiltonian vector field.

By the way, when the configuration space is $S^1$, this flat connection is the famous Aharonov-Bohm connection.

For further reading on the classification of inequivalent quantizations, please see the following two articles by N.P. Landsman and by Doebner Šťovíček, and Tolar . Actually, the first author (Landsman) has reservations (footnote 13 in the article) over the customary explanation using parallel transport and prefers the induced representation reasoning that I tried to follow.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user David Bar Moshe
answered Apr 29, 2014 by David Bar Moshe (4,355 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...