From: Gerard Westendorp Subject: Re: The belt trick Date: 18 Sep 2000 12:37:40 GMT Newsgroups: sci.physics.research Summary: [missing] Greg Egan wrote: [..] > > There's a Java animation of this at: > > > > This shows both the physical motion of the belt, and what's happening in > terms of the group of rotations, SO(3). The idea is that a loop that > crosses SO(3) twice (when drawn as a sphere of rotations) can be > contracted down to a point, because you have the freedom to move the two > separate pairs of antipodal points where you hit rotations of pi. A loop > that crosses only once has its endpoints constrained to stay on opposite > sides of the sphere, so it's non-contractible. > This is quite cool. I read the "mathematical details" link as well. Now I understand what a homotopy class is, and a covering group. I've seen these words before, but the belt makes it come to life. One thing I didn't see at first is that the belt is actually a world sheet of the buckle, you could imagine the buckle traveling along the belt as it twists and turns. The sequence of rotations form a path through SO(3). As you wobble the belt, the path is deformed. But still, I don't yet see the relationship with electrons, except that in both cases the group SU(2) turns up. Gerard ============================================================================== From: gregegan@netspace.zebra.net.au (Greg Egan) Subject: Re: The belt trick Date: 20 Sep 2000 07:47:34 GMT Newsgroups: sci.physics.research Previously, I wrote: > 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 ... On the relationship between belts and electrons, maybe I should add some comments that make the connection a bit more explicit. First, every C^2 spinor is an angular momentum eigenstate for some spin axis. To be specific, if (z,w) is any normalised vector in C^2, then it's an eigenvector with eigenvalue 1 of the Hermitian matrix: sigma(z,w) = n_x sigma_x + n_y sigma_y + n_z sigma_z where n_x = (zw* + z*w) n_y = i(zw* - z*w) n_z = (zz* - ww*) sigma_x, sigma_y, sigma_z are the usual Pauli spin matrices, and * means complex conjugate. All three coefficients (n_x,n_y,n_z) are real, and the sum of their squares is 1. The matrix sigma(z,w) also has a second eigenvector, (w*,-z*), with eigenvalue -1. It's easy to check that this works for some well-known eigenstates: Eigenstate (z,w) sigma(z,w) ---------- ------------- ---------- spin-x +ve (1,1)/sqrt(2) sigma_x spin-y +ve (1,i)/sqrt(2) sigma_y spin-z +ve (1,0) sigma_z What happens to sigma(z,w) if you multiply (z,w) by a phase? Every one of the n_i is a sum of terms with one C^2 component multiplied by one conjugate, so overall there's no effect. Using this correspondence, you can describe (the spin part of) an electron's state vector at any moment by a rectangle lying in the plane orthogonal to the spin axis (n_x,n_y,n_z), along with a phase angle, phi. Now, we have an explicit formula for the spin axis in terms of (z,w), but what about phi? There's a lot of freedom in assigning this -- we could make (1,0) the "standard spin-up eigenvector" with phi=0 ... or we could choose (i,0), or (-1,0), etc. And having picked, say, (1,0) for spin up, what about all the other directions? This leads back to the problem of intrinsically projective representations of SO(3); whatever nice recipe we use for assigning a "standard eigenvector" to every direction, if we follow it around a 360-degree circle, we won't be able to avoid a change of sign. So we just accept the fact that the representation of SO(3) on C^2 will be intrinsically projective, and we use the spin-(1/2) true representation of SU(2). Then if you rotate the electron by an angle theta around some axis (p_x,p_y,p_z), the element of SU(2) that acts on its state vector is: U(theta,p) = cos(theta/2) 1 - i sin(theta/2) (p_x sigma_x + p_y sigma_y + p_z sigma_z) The effect of this is to rotate (n_x,n_y,n_z) as a vector in R^3, just as you'd expect. However, any 360-degree rotation gives a change in phi of -1. It takes a 720-degree rotation to restore both (n_x,n_y,n_z) and phi to their original values. So a belt's configuration really does make a pretty good model for keeping track of an electron's state: the normal vector to the buckle is (n_x,n_y,n_z), and a twist you can't smooth out tells you when the phase has reversed, compared to a configuration where the buckle orientation is the same but the belt *can* be smoothed out. -- Greg Egan Email address (remove name of animal and add standard punctuation): gregegan netspace zebra net au ============================================================================== From: gregegan@netspace.zebra.net.au (Greg Egan) Subject: Re: The belt trick Date: Wed, 20 Sep 2000 21:27:17 +0800 Newsgroups: sci.physics.research I wrote: > If (z,w) is any normalised vector in C^2, then it's > an eigenvector with eigenvalue 1 of the Hermitian matrix: > > sigma(z,w) = n_x sigma_x + n_y sigma_y + n_z sigma_z > > where n_x = (zw* + z*w) > n_y = i(zw* - z*w) > n_z = (zz* - ww*) There's one more detail that I didn't mention before because I didn't quite understand it, but now with the aid of Misner, Thorne & Wheeler's Chapter 41, I think I've got this straight. In addition to the spin axis n, you can associate a second vector with the spinor (z,w): f = ((w*^2­z^2+w^2­z*^2)/2, i(w*^2­z^2­w^2+z*^2)/2, zw+z*w*) This is a unit vector orthogonal to the spin axis. If you multiply (z,w) by a phase, f rotates around the spin axis by twice as much as the phase angle. So f almost keeps track of the phase, but it can't discriminate between (z,w) and (-z,-w). To recover that sign, you need to know the homotopy class of the path through SO(3) you've followed, which is recorded by the belt. -- Greg Egan Email address (remove name of animal and add standard punctuation): gregegan netspace zebra net au