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We need non-abelian fractional hall states because of the ground state degeneracy http://rmp.aps.org/abstract/RMP/v80/i3/p1083_1 (arXiv version for free). But we can also have degeneracy even in case of abelian fractional quantum hall states on topologcally non-trivial surfaces http://prb.aps.org/abstract/PRB/v41/i13/p9377_1. Could anyone tell me the reason (conceptual) for not using abelian state for quantum computation?

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  • $\begingroup$ We won't have universal computation. $\endgroup$ – Ali Jul 19 '13 at 9:25
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    $\begingroup$ Welcome to SE.physics. When it's possible, please prefer to give the arXiv, free version of the paper you cite. I edited your question. It's about abelian vs. non-abelian anyons. $\endgroup$ – FraSchelle Jul 19 '13 at 9:26
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You need some criteria to perform computation at the quantum level. One of them is the ability to perform any manipulation of your qubit. Abelian anyons does not provide full possibility to manipulate the qubit state (in particular you can not rotate the phase the way you want, only by $\pi/4$ for instance for Majorana modes in superconducting wires, and the braiding is then called "abelian braiding").

So for short you can use abelian anyons to make simple quantum computation. One nevertheless prefers to use the name "quantum memory" since the simple thing you can do is to encode and store a quantum state into an abelian anyon. But you will not be able to use them to make a universal quantum computer. For a universal quantum computer you need non-abelian anyons.

Please comment to obtain more details.

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  • $\begingroup$ I agree,but couldn't the degenerate states can be used for the same logic in the sense of protecting the information even in the case of abelian hall states ? $\endgroup$ – Krishna Tripathi Jul 19 '13 at 9:31
  • $\begingroup$ @KrishnaTripathi Yes, that's what I said: you can use abelian anyons to make (possibly good) quantum memory having a long coherence time, but you can not use them to make universal quantum computers since they lack full manipulation of the quantum state (rotation essentially). $\endgroup$ – FraSchelle Jul 19 '13 at 9:38

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