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  Braiding statistics of anyons from a Non-Abelian Chern-Simon theory

+ 9 like - 0 dislike
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Given a 2+1D Abelian K matrix Chern-Simon theory (with multiplet of internal gauge field $a_I$) partition function:

$$ Z=\exp[i\int\big( \frac{1}{4\pi} K_{IJ} a_I \wedge d a_J + a \wedge * j(\ell_m)+ a \wedge * j(\ell_n)\big)] $$

with anyons (Wilson lines) of $j(\ell_m)$ and $j(\ell_n)$.

One can integrate out internal gauge field $a$ to get a Hopf term, which we interpret as the braiding statistics angle, i.e. the phase gained of the full wave function of the system when we do the full braiding between two anyons:

$$ \exp[i\theta_{ab}]\equiv\exp[i 2 \pi\ell_{a,I}^{} K^{-1}_{IJ} \ell_{b,J}^{}] $$ see also this paper and this paper.

I would like to know the way(s) to obtain braiding statistics of anyons from a Non-Abelian Chern-Simon theory? (generically, it should be a matrix.) How to obtain this braiding matrix from Non-Abelian Chern-Simon theory?

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user Idear
asked Dec 15, 2013 in Theoretical Physics by wonderich (1,500 points) [ no revision ]

2 Answers

+ 8 like - 0 dislike

How to obtain this braiding matrix from Non-Abelian Chern-Simon theory?

To obtain braiding matrix $U^{ab}$ for particle $a$ and $b$, we first need to know the dimension of the matrix. However, the dimension of the matrix for Non-Abelian Chern-Simon theory is NOT determined by $a$ and $b$ alone. Say if we put four particles $a,b,c,d$ on a sphere, the dimension of the degenerate ground states depend on $a,b,c,d$. So even the dimension of the braiding matrix $U^{ab}$ depends on $c$ and $d$. The "braiding matrix" $U^{ab}$ is mot deterimened by the two particles $a$ and $b$.

Bottom line: physically, the Non-Abelian statistics is not described by the "braiding matrix" of the two particles $a$ and $b$, but by modular tensor category.

This post imported from StackExchange Physics at 2014-04-05 03:33 (UCT), posted by SE-user Xiao-Gang Wen
answered Dec 15, 2013 by Xiao-Gang Wen (3,485 points) [ no revision ]
+ 5 like - 0 dislike

The (unitary) "phase" factor for non-Abelian anyons satisfies the (non-Abelian) Knizhnik-Zamolodchikov equation:

$$\big (\frac{\partial}{\partial z_{\alpha}} + \frac{1}{2\pi k} \sum_{\beta \neq \alpha} \frac{Q^a_{\alpha}Q^a_{\beta}}{z_{\alpha} - z_{\beta}}\big )U(z_1, ....,z_N) = 0 $$

Where $z_{\alpha}$ is the complex plane coordinate of the particle $\alpha$ , and $Q^a_{\alpha}$ is the matrix representative of the $a-$th gauge group generator of the particle $\alpha$ and $k$ is the level .

Please, see the following two articles by Lee and Oh (article-1, article-2).

In the first article they explicitly write the solution in the case of the two-body problem:

$$U(z_1, z_2) = exp( i\frac{Q^a_1Q^a_2}{2\pi k} ln(z_1-z_2))$$

The articles describe the method of solution:

The non-Abelian phase factor can be obtained from a quantum mechanical model of $N$ particles on the plane each belonging possibly to a different representation of the gauge group minimally coupled to a gauge field with a Chern-Simons term in the Lagrangian.

The classical field equations of the gauge potential can be exactly solved and substituted in the Hamiltonian. The reduced Hamiltonian can also be exactly solved. Its solution is given by the action of a unitary phase factor on a symmetric wave function. This factor satisfies the Knizhnik-Zamolodchikov equation.

The unitary phase factor lives in the tensor product Hilbert space of the individual particle representations. The wave function is a vector in this Hilbert space valued holomorphic function depending on the $N$ points in the plane.

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user David Bar Moshe
answered Dec 15, 2013 by David Bar Moshe (4,355 points) [ no revision ]
Thanks David. You offer another way I did not learn before. How to connect this viewpoint to a non-Ab Chern-Simons theory? (the way I knew before was the way in Witten's Jones polynomial paper and Wilson loop approach. Is there a connection to this Wilson loop approach?)

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user Idear
@Idear These two approaches are very similar (but not exactly equivalent). Actually, Witten in his Jones polynomial paper (on page 365) refers to this similarity and asserts that the Wilson loop can be "thought of" as the trajectory of a particle in 2+1 dimensions. Witten refers to a famous paper by Polyakov adopting the strategy that Lee and Oh used later. This is only one of the numerous issues that Witten only talked about in his Jones polynomial paper (even without giving a single formula), which proved to be very fruitful for subsequent research.

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user David Bar Moshe
@Ideat cont. Witten's approach is more "thermodynamical" and he prefers to see the traces of the non-Abelian statistics in the partition function. To be more precise,(and this fact was also written in not so much words in Witten's paper): A non-Abelian Wilson loop can be thought of as a particle moving on a group or a flag manifold stuck to the boundary in the limit where its mass vanishes. Then its dynamics restricts upon quantization to the lowest Landau level producing the correct Wilson loop insertion.

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user David Bar Moshe
Does the Knizhnik-Zamolodchikov equation also hold for other surfaces? Like the torus or the sphere?

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user Hamurabi
@Hamurabi There exist generalizations of the Knizhnik-Zamolodchikov, for example arXiv:arXiv:hep-th/9510143, hep-th/9410091, for elliptic curves and Riemann surfaces.

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user David Bar Moshe
Thank you!!! Well right on the first page of hep-th/9410091v2 it says that the formula you gave above is the one for the spherical case. So where is the difference to the plane?

This post imported from StackExchange Physics at 2014-04-05 03:33 (UCT), posted by SE-user Hamurabi

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