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  Is it possible to make statements about bosonic/fermionic systems by taking the limit $\theta\to \pi$ or $\theta\to 0$, of an anyonic system?

+ 5 like - 0 dislike

One might naïvely write the (anti-)commutation relations for bosonic/fermionic ladder operators as limits

$$ \delta_{k,\ell} = \bigl[ \hat{b}_{k}, \hat{b}_{\ell}^\dagger \bigr] = \hat{b}_{k} \hat{b}_{\ell}^\dagger - \hat{b}_{\ell}^\dagger \hat{b}_{k} = \lim_{\theta\to\pi} \Bigl( \hat{b}_{k} \hat{b}_{\ell}^\dagger + e^{i\theta}\cdot\hat{b}_{\ell}^\dagger \hat{b}_{k} \Bigr) $$ $$ \delta_{k,\ell} = \bigl\{ \hat{c}_{k}, \hat{c}_{\ell}^\dagger \bigr\} = \hat{c}_{k} \hat{c}_{\ell}^\dagger + \hat{c}_{\ell}^\dagger \hat{c}_{k} = \lim_{\theta\to 0} \Bigl( \hat{c}_{k} \hat{c}_{\ell}^\dagger + e^{i\theta}\cdot\hat{c}_{\ell}^\dagger \hat{c}_{k} \Bigr). $$ I.e. as limits of Abelian anyonic commutation relations. Assuming now that some system could be solved for anyons with $0 < \theta < \pi$, would taking the limits of e.g. the energy eigenstates for $\theta\to \pi$ yield in general the correct eigenstates of the bosonic system (which might be harder to solve directly)?

I'm inclined to think it would work, but after all, the whole Fock space looks different depending on $\theta$, with all kinds of possible topological nontrivialities.

This post imported from StackExchange Physics at 2014-04-05 17:35 (UCT), posted by SE-user leftaroundabout
asked Oct 31, 2012 in Theoretical Physics by leftaroundabout (25 points) [ no revision ]

1 Answer

+ 6 like - 0 dislike

There is no such thing as "Abelian anyonic commutation relations", in the sense that the "Abelian anyonic commutation relations" that you write down does not describe Abelian anyons. So the starting point of the question is not valid.

Also anyons do not have a Fock space description. The standard many-body text books stress on Fock space too much, which lead people to think about many-body systems only in terms of Fock space. Such Foack-space picture can only describe a very small subset of many-body states. Most many-body states (the interesting ones) require a new picture (such as tensor network) to visualize.

This post imported from StackExchange Physics at 2014-04-05 17:36 (UCT), posted by SE-user Xiao-Gang Wen
answered Oct 31, 2012 by Xiao-Gang Wen (3,485 points) [ no revision ]
Could you elaborate on why the Fock space picture fails, and what are the properties of these more exotic beasts (e.g. tensor networks) that allow them to save the day?

This post imported from StackExchange Physics at 2014-04-05 17:36 (UCT), posted by SE-user Doug Packard
Quite honestly, I've never actually dealt with anyons, so I can't much judge what you say here. It would be nice if you could make your points a bit clearer. — Anyway, the main question wasn't really about Fock space / creator/annihilator commutations, but about the $\theta\to k\pi$-limit of observables in anyonic systems. How about that?

This post imported from StackExchange Physics at 2014-04-05 17:36 (UCT), posted by SE-user leftaroundabout
Fock space picture only lead to totally symmetric or total antisymmetric many-body wave-function. So it fails on anyons. Also, $\theta$ is always a rational number for anyons and the properties of an anyon system is not a continuous function of $\theta$. So we cannot take the limit.

This post imported from StackExchange Physics at 2014-04-05 17:36 (UCT), posted by SE-user Xiao-Gang Wen

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