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  What is the "physical" or "experimental" meaning of fusion rule / multiplicity $N_{ab}^c$?

+ 7 like - 0 dislike

In 2-dimensional topologically ordered system with anyon excitations, we write the fusion of a anyon $a$ with another anyon $b$ outputting several possible outcomes, say anyons $c$. And there is a mysterious integer weight in the front of $c$, called the fusion rule or the fusion multiplicity

$N_{ab}^c$: $$a \times b = \sum_c N_{ab}^c c$$

  • question 1: What are the physical meanings or physical observables for $N_{ab}^c$?
  • question 2: does $N_{ab}^c$ say anything about the probability of having the anyon $c$ by fusing $a$ and $b$? Say, if $a \times b = N_{ab}^{c_1} c_1+ N_{ab}^{c_2} c_2$, then do $N_{ab}^{c_1}$ and $N_{ab}^{c_2}$ represent the probability weight of having a output anyon $c_1$ or $c_2$? Please try to answer this question from a experimental quantum physicist's perspective. Make the answer down to the earth. Thanks. :-)
asked Dec 25, 2014 in Theoretical Physics by RKKY (325 points) [ revision history ]
Kitaev and Preskill have a nice application of these topics in computing the topological entanglement entropy of a ground state whose effective TQFT is known. http://arxiv.org/abs/hep-th/0510092

2 Answers

+ 6 like - 0 dislike

Question 1: Before understanding "What is the physical or experimental meaning of fusion rule/multiplicity $N^c_{ab}$?", one needs to undertstand what is topological excitation and what is the fusion space. The section II of our paper http://arxiv.org/abs/1311.1784 has an explaination of all those notions.

Question 2: The setup that leads to the probability is not specified, and the question is hard to answer. I guess in one setup, the probability should be proportional to $N^c_{ab}d_c$ where $d_c$ is the quantum dimension of $c$. 

answered Dec 25, 2014 by Xiao-Gang Wen (3,485 points) [ revision history ]
edited Aug 8, 2015 by Xiao-Gang Wen
+ 4 like - 0 dislike

For question 2, consider a setup as follows: from the vacuum create a pair of $a$ and $\bar{a}$, and a pair of $b$ and $\bar{b}$. Then fuse $a$ and $b$ and measure the resulting topological charge. The probability of finding $c$ is $\frac{N_{ab}^c d_c}{d_ad_b}$. I think this is the setup Professor Wen had in mind in his answer.

answered Aug 9, 2015 by Meng (550 points) [ no revision ]

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