Quantum Spin and Occam's Razor
Monday, April 29, 2013 at 08:22AM
Doug

It’s interesting to read online discussions of quantum spin. Bruce Schumm admits physicists “don’t have a clue as to what the physical origin of spin really is,” but you wouldn’t know that from reading these forum explanations.

For instance, the discussion on one threaded forum goes like this (edited for clarity):

Commentor A:

[The question is] why a particle’s spin doesn’t remain invariant when you rotate it by 2pi, using the corresponding rotation operator. The easy answer is that spin doesn’t live in normal 3D space so it doesn’t transform with the usual rotation matrices from classical mechanics. Classical objects rotate using a representation of the group SO(3), whereas spin vectors rotate with a representation of the group SU(2).

Commentor B: “

A 2pi rotation changes the sign of the state vector, |\sigma\rangle\mapsto -|\sigma\rangle, but any state vector of the form C|\sigma\rangle where C is a complex number represents the same state as |\sigma\rangle.
I don’t think this has anything to do with the fact that we’re using SU(2) instead of SO(3). SU(2) is used instead of SO(3) to make sure that we always have U(R’R)=U(R’)U(R) where U(R) is the rotation operator (acting on state vectors) that corresponds to the rotation matrix R (acting on vectors in \mathbb R^3). If we use SO(3), we’d sometimes get U(R’R)=-U(R’)U(R), depending on what paths in SO(3) that we use to get from the identity to R, R’ and RR’.
I think that any representation of SO(3) on a 2-dimensional complex vector space will always have the property that a 2pi rotation changes the sign of the vector on which it acts. One could ask why we would choose to use a 2-dimensional representation. The answer to that is that we didn’t choose. The number of dimensions needed to represent the spin state of a spin S particle is always 2S+1.

Commentor C:

Why is it 4pi (it’s not 2pi) in spin space? Please give a picturesque explanation.

Commentor B:

There’s no way to really draw a picture of this. 2 complex dimensions are equivalent to 4 real dimensions and we can’t draw more than 3. Roger Penrose suggests the following picture, but I don’t think it’s helpful at all: Put one end of a long belt on the table, under something heavy, and put the other end between two pages in a book. Now rotate this book by 4 pi around the axis defined by the belt. Without rotating the book any further, you can “undo” the 4 pi rotation by looping the belt around the book. This doesn’t work with a 2 pi rotation.

Commentor D:

I read also that a waitress with a tray works nicely. If she rotates the tray once, her arm is twisted. To regain the original state, she has to rotate the tray by 720 degrees. I think you have to try it to see what I mean. I believe it was Feynman who gave me this picture, to give credit where it’s due.

Commentor E:

It is cute, but not perfect. Like a recently deceased kitten. Cheers.

Commentor F:

It’s nice to see a 4pi rotation is significant somehow, but this doesn’t really make it clear what this has to do with spin 1/2 particles. What part of an electron is analagous to the belt?

Here’s one way to understand it. Let the belt represent the path of the electron through spacetime. The electron may twist and turn as it travels, but let’s say that at certain points, it has the same orientation it started with. When this happens, the electron must be in a state related to the original one by some constant factor.

What can this constant be? Well, we need to assign it in such a way that the state of the electron varies continuously along its path, ie, no sudden jumps. Let’s say the electron has done two complete rotations since it started. Can the constant we assign at this point be something besides one?

The belt trick says no. To see why, let’s suppose we could make it something else, say A. Then we’ve assigned a state to the electron at every point along the path in such a way that it starts at some state |\psi> and ends at some state A|\psi>. Now, using the belt trick, we can continuously deform this path to one where the electron doesn’t rotate at all. But we’re holding the ends of the belt fixed, so it still has to start at |\psi> and end at A|\psi>. But this should be true no matter how short the path is, and when we take the path to be very small, this means the state must change discontinuously, unless A=1. On the other hand, a single rotation cannot be so deformed, so the constant doesn’t have to be 1. However, it’s restricted by the fact that doing this twice must return you to the original state, so it must be either 1 or -1. For an electron, it turns out the constant is -1.

Here’s a more technical explanation. The belt is supposed to represent is a path through SO(3), the group of rotations in 3 dimensions. Namely, each point along the belt corresponds to a rotation: the rotation that brings that slice of the belt into its current orientation. So we have a one parameter family of elements in SO(3), ie, a path in SO(3). Then by deforming the belt while holding its two ends fixed, we get a deformation of the path, ie, a homotopy of paths. The fact that a twice twisted belt can be deformed to a straight belt (while a once twisted belt can not) is then just the demonstration that a path in SO(3) corresponding to a 4pi rotation is homotopic to the constant path (while a 2pi rotation is not), ie, that the fundamental group of SO(3) is Z2. Thus the rotation group is not simply connected, and so it has representations that are not single valued. For one reason or another, nature chooses to use one of these representations for certain particles, such as the electron.

Commentor C:

Thank you! It is a visualized explanation. But I think it is similar to a picture of Berry phase, isn’t it?

Commentor G:

Another model is when two complete revolutions are only projections of single complete revolutions which are structured from two nested rotations:

You can imagine it like path in 3D space for pendulum with same inner and outer frequencies (axes of rotations are perpendicular). Resulting trajectory is on half-sphere (Viviani window curvature). One of the trajectory projection is circle with half radius and this is what we are able to measure like magnetic moment for particle with ½ spin. When structured rotors complete 2pi revolutions their composite projection complete 4pi.

That was the end of the discussion. I don’t know why the last comment remained unanswered. It’s not clear to me what this person is saying, but it seemed to be out of the ordinary context of rotation groups, which is what legacy physics (LST) is based on today.

Even though Schumm says physicists haven’t a clue as to the physical origin of quantum spin, this undefined concept is an indispensable part of legacy physics, as can be seen in this online discussion of spin 0 particles:

http://physics.stackexchange.com/questions/31119/what-does-spin-0-mean-exactly

Alas, however, you will not find a definition of “particle” in any of these discussions of quantum spin, since that concept itself is undefined. If you dig deep enough, you will find that the word “particle” is qualified, as “point-like.” That is, it’s not strictly a point, since a point is defined as something with no physical extent!

Consequently, the theoretical concepts of legacy physics are necessarily a mess, but they work well, to a certain extent, for calculations of observed physical properties. This places the new system of physical theory at a disadvantage, but then we don’t have the research resources the LST has.

Still, it’s nice to have a simple, straight forward explanation of the physical origin of quantum spin, as shown in the previous entry. The real satisfaction, though, comes from the fundamental explanation of magnitude, dimension and “direction,” which underlies the RST concept.

Article originally appeared on LRC (http://www.lrcphysics.com/).
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