Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So far in condensed matter physics, I only know anyons(abelian or nonabelian) can emerge as quasiparticles in 2D real-space.

But is there any possibility to construct anyons in momentum-space ? And what about the braiding, fusion rules in momentum-space ? I mean do anyons always live in real-space rather than in momentum-space ?

Maybe this is a trivial question, thank you very much.

share|improve this question

1 Answer 1

I was wondering something similar few month ago. Then I concluded that most of the topological staffs appear at the boundary between two different topological sector. A sector being characterised by a Chern number, or if you prefer a topological charge, one needs a boundary / an interface between two systems characterised by different topological charge.

A $k$-space (or momentum, or reciprocal, or Fourier, ...) is well defined only for periodic boundary conditions. The fact that the $x \leftrightarrow k$ is a Fourier transform imposes a periodicity in $x$ or in $k$. That's the stringent condition under which $k$ is a good quantum number. Note that we can still define some quasi-$k$ for disordered media. So we could not in principle define a $k$-space when a system has boundary. Note that infinite system are usually closed by periodic boundary condition, also called Born-von-Karman conditions.

I'm not aware so much about anyons (I'm still learning about that) but I believe they (almost all of them ? all of them ? I don't know) appear due to boundary conditions in condensed matter, for the reason I gave about the topological charge transition. So I believe it should be impossible to define anyons in $k$-space, for the simple reason that the $k$-space is not a correct description of the matter when anyons exist.

I would really appreciate comments/critics about what I said, especially if it's (partially) wrong.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.