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