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Irrelevance of parastatistics for space dimension > 2

+ 4 like - 0 dislike

Consider a system of $n$ undistinguishable particles moving in $d$-dimensional Euclidean space $E^d$. The configuration space is $M=((E^d)^n \setminus \Delta)/S_n$ where $\Delta$ is the diagonal (subspace where at least 2 particles have coincidental positions) and $S_n$ is the group permuting the particles

Quantization of this system yields superselection sectors corresponding to unitary irreducible representations of $\pi_1(M)$: $S_n$ for $d > 2$, $B_n$ for $d = 2$. The trivial representation yields bosonic statistics, the sign representations yield fermionic statistics. For $d > 2$ there are no other 1-dimensional repsentations. For $d = 2$ there are other 1-dimensional representations in which switching two particles generates an arbitrary phase. These yield anyonic statistics.

What about higher dimensional irreducible representations? These correspond to "parastatistics". It is said that for $d > 2$ we can safely ignore them because in some sense they are equivalent to ordinary bosons/fermisons. However for $d = 2$ this is not the case. Why?

Why is parastatistics redundant for $d > 2$ but not for $d = 2$?

This post has been migrated from (A51.SE)
asked Feb 17, 2012 in Theoretical Physics by Squark (1,705 points) [ no revision ]

2 Answers

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Let me quote Phys. Rev. B 83, 115132 (2011)

The one-dimensional representations of $S_n$ correspond to bosons and fermions. One might have hoped that higher-dimensional representations of $S_n$ would give rise to interesting 3D analogues of non-Abelian anyons. However, this is not the case, as shown in Ref. 18,19: any higher dimensional representation of $S_n$ which is compatible with locality can be decomposed into the tensor product of local Hilbert spaces associated with each particle. For instance, suppose we had $2n$ spin-1/2 particles but ignored their spin values. Then we would have $2^{2n}$ states which would transform into each other under permutations. Clearly, if we discovered such a system, we would simply conclude that we were missing some quantum number and set about trying to measure it. This would simply lead us back to bosons and fermions with additional quantum numbers. (The color quantum number of quarks was conjectured by essentially this kind of reasoning.)

The papers they are referring to are this and this.

This post has been migrated from (A51.SE)
answered Feb 17, 2012 by Heidar (855 points) [ no revision ]
Thx I'll look into these papers

This post has been migrated from (A51.SE)
+ 0 like - 0 dislike

Inequivalent quantizations are not in correspondence to the unitary representations of the fundamental group $\pi_1(M)$, but rather to the group of its character representations $\mathrm{Map}(\pi_1(M), U(1))$, please see, for example, Doebner and Tolar contribution to Symmetries in science XI. In our case, this group is $Z_2$ whose two points are gived by the trivial and alternating characters corresponding to bosons and fermions respectively. In the language of geometric quantization, every character leads to a different connection on the prequantization line bundle appearing in the prequantization formula, which gives a correspondence between functions on the phase space and operators on the prequantization Hilbert space. Thus, other statistics are not possible because this correspondence cannot be constructed.

This post has been migrated from (A51.SE)
answered Feb 18, 2012 by David Bar Moshe (3,595 points) [ no revision ]
this is because you're doing geometric quantization. In deformation quantization there is nothing wrong with higher dimensional representations. Given any vector bundle on $M$ equipped with a flat connection you get an action of differential operators on $M$ on the sections. The algebra of differential operators is the quantization of the algebra of functions on the cotangent space polynomial in the momenta

This post has been migrated from (A51.SE)

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