From: gregegan@netspace.zebra.net.au (Greg Egan) Subject: Re: The belt trick Date: 19 Sep 2000 16:36:57 GMT Newsgroups: sci.physics.research Summary: Applications of the Belt Trick to particle physics In article <39C5E4F3.EDE6A1D0@xs4all.nl>, Gerard Westendorp wrote: > But still, I don't yet see the relationship with electrons, > except that in both cases the group SU(2) turns up. All I can really add is the following stuff about representations, which is a paraphrase (though hopefully not too much of a distortion) of Baez & Muniain's _Gauge Fields, Knots and Gravity_ (Chapter 1 of Part II) and Weinberg's _The Quantum Theory of Fields_ (Chapter 2 of Volume I). If you want lots of gory details, Weinberg writes about this at some length in Section 2.7, and Appendix A of Chapter 2. Suppose you have some quantum system, states of which are described by rays in a Hilbert space, H: that is, equivalence classes of state vectors that differ only by a phase. If you rotate the *spatial* basis that you're using to describe the system, by some rotation R, then the effect of this on the state vector/ray description will be given by a unitary operator U(R) acting on the state vectors in H. At first glance, you might think that for consistency you'd need U to obey the following rule: U(R)U(S) = U(RS) If that's true, then U gives a *representation* of SO(3) on H: that is, it's a homomorphism from SO(3) into the group of linear operators on H. But in fact, it would still make perfect physical sense if instead we had: U(R)U(S) = exp(i phi(R,S)) U(RS) because multiplication by a phase leaves a physical state unchanged. This is known as a *projective representation* of SO(3), as opposed to a *true representation*. OK, if multiplication by a phase is no big deal, why not sweep it into the unitary operators themselves? We can always define: V(R) = exp(i alpha(R)) U(R) for any function alpha we like, and it will still work for the same physical system. So maybe by the right choice of alpha, can we turn any projective representation into a true representation? The answer turns out to be no. Some groups have "intrinsically projective" representations, which can't be turned into true representations, and SO(3) is one of them. However, you can always choose alpha(R) so that V obeys either: V(R)V(S) = V(RS) or V(R)V(S) = (+/-)V(RS) What's more, both cases can be related to the *true* representations of the universal covering group of SO(3), SU(2). You can cook up true unitary representations of SU(2) by the following method: the homogeneous polynomials of degree 2j (where j lies in {0,1/2,1,3/2,2,...}) in two complex variables x and y form a complex vector space H_j of dimension 2j+1, spanned by the functions: x^(2j), x^(2j-1)y, .... y^(2j) If g is some element of SU(2), i.e. a 2x2 unitary matrix with determinant 1, then you can multiply (x,y) by g^{-1} before feeding it to any function f in H_j, and that will give you a new function of x and y, i.e. U_j(g) : H_j -> H_j (U_j(g)f)(x,y) = f(g^{-1}(x,y)) If you check, these U_j give true unitary representations of SU(2) on H_j, for whatever j you choose. Now, the two different homotopy classes of paths between I and R in SO(3) correspond to elements of SU(2) that are opposites, g and -g, as matrices. Pick a function lambda: SO(3)->SU(2), which takes every rotation R to one of the homotopy classes of the paths from I to R. Then define: V_j(R) = U_j(lambda(R)) If j is an integer, U_j(-g)=U_j(g), since taking the opposite of x and y in a polynomial of even degree has no effect. So V_j will be "blind" to which homotopy class lambda assigns to each R, and it turns out that this is good enough to make V_j a true representation of SO(3), however we chose lambda. If j is 1/2,3/2, etc., then U_j(-g)=-U_j(g), so things don't work out so simply. In that case, the best we can guarantee is that V_j satisfies: V_j(R)V_j(S) = (+/-)V_j(RS) Electrons, and all fermions, fall into the latter class. Quantum systems don't need to give representations of SO(3), merely projective representations. But the *projective* representations of SO(3) come from the *true* representations of its universal covering group, SU(2) ... which is the set of homotopy classes of paths in SO(3) ... which is the set of belt configurations that can be deformed into each other. What the belt trick specifically shows is that the fundamental group (the group of homotopy classes of loops) of SO(3) is Z_2, which is why it's the number of rotations modulo 2 that matters for a quantum system, and why there can be such things as fermions that only return to their original state after two complete rotations. -- Greg Egan Email address (remove name of animal and add standard punctuation): gregegan netspace zebra net au