Natural Science Forum / Physics / Research / May 2005

I prefer to think about it as a passive rotation, because I don't know
what it means to actively rotate a point particle.

It's just a preference.

Hello Jeff:

Here's how I operationally picture 4 pi rotation.  I put a beer mug in
my hand and rotate it.  After 2 pi's worth of rotation, my arm is in an
awkward spot.  Now do 2 pi more, and you can toast 1/2 integral spin.
This is clearly a classical system with 4 pi rotational symmetry.

If I had just sat on the bar stool and done a spin, in 2 pi my mug gets
back to where it started.  Not too shocking.

This is odd though: if you really understand the math behind those two
classical systems with different rotational symmetry, then you have the
means to understand the difference between spin 1/2 and spin 1
particles.  Take the classical measurements and turn them into
operators.  If you are unfamiliar with this road, please study chapter
3 of "Modern Quantum Mechanics" by J. J. Sakurai.  He starts with
classical 3D rotations, moves to infinitesimal rotations in quantum
mechanics, then finite rotations.  This was the coolest chapter I ever
read on the topic, enjoy it.

Newsgroups are not good for doing too much math, so you'll have to look
to the book for details.  For a 3D infinitesimal classical rotation, he
considered rotations around the z axis to order epsilon^2.  He showed
that small rotations around z (R_z) were algebraically the same as a
commutator of rotations around x and y:

   R_x(e) R_y(e) - R_y(e) R_x(e) = R_z(e^2) - R_anyaxis (0)
   
 = [R_x(e), R_y(e)]

This is how the classical barstool, 2 pi rotation can be explained.  The
riddle, not explained in Sakurai, is how to understand the
infinitesimal rotations of the classical beer mug in hand.

Here's how I do it.  I tack on a time dimension to my rotation matrix
(it does not a thing but waste more ink and electrons, and let in sail
into a relativistic formalism without skipping a beat).  Define the
conjugate of a rotation matrix as flipping the signs of the x, y, and z
parts, leaving the time part alone, not that it does anything.  It
should be no surprise that this changes nothing:

   [R_x(e)*, R_y(e)*] = R_z(e^2) - R_anyaxis (0)

Now we introduce a new type of conjugate which does something similar to
the old one: it flips the sign of three out of the four.  The first
conjugate keeps the sign of x the same, flipping the sign of time, y
and z.  Operationally, the conjugate is (1 R 1)* while the first
conjugate is (i R i)*.  Calculate the commutator when you take the
first conjugates (symbolized here with a *1)

   [R_x(e)*1, R_y(e)*1] = -R_z(e^2) + R_anyaxis (0)

Basically there is a factor of i^2 in between the two rotation matrices
R_x and R_y that generates the extra minus sign in the infinitesimal
rotation calculation.  In 2 pi, you get around to minus where you
started, and another 2 pi will flip the sign again back to where it
should be.

A nice feature about using conjugates to differentiate between classical
2 pi and 4 pi rotations was the same dimensional (4x4 real) rotation
matrix was used in both cases.  It had to be 4D so that the conjugate
would always flip three out of four signs.

Practically speaking, if you treat the three directions of space the
same, you will get a integral spin symmetry, whether a bar stool or a
photon.  If you treat two of the three directions of space differently,
you may get the half spin situtation.  With the mug in hand, the
shoulder effectively constrains the rotation to two dimensions of
space.

That's how I view the story.
doug

Epilog
"Look ma, no group theory!"  
Mother was not at all pleased.  She sent the lad to his room to write
out the multiplication tables for over a hundred groups, telling him
not to utter the word "first conjugate" again, lest the neighbors
thought there was a heretic in this family group.

Take a point-like particle and have it trace out a path around the
surface of a cylinder like so.  Start at one end of the cylinder and go
around the surface to the same spot on the other end.  You have gone
around 360 degrees.  Now go back to the starting spot on the original
end.  You have gone around 720 degrees and you are back at the starting
point.  This is the easiest way I know of to picture a spin 1/2
point-like particle in 3D.  In reality for fermions, maybe the length of
the cylinder is not one of our 3D dimensions?  Or maybe the radius of
the cylinder is not?  Hmm...  Maybe the circumference is a compact
dimension?  Now if you imagine that the point-like particle is also
relativistic, then this two turn "loop" is like a string that has an
"internal" circulation.  Does string theory consider internal
circulation within a string?

FrediFizzx

http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps