Re: Do free particles have spin?

From: David Rutherford (drutherford_at_softcom.net)
Date: 09/27/04 

Date: Mon, 27 Sep 2004 10:40:42 -0700

Gregory L. Hansen wrote:
>
> It's a vector sum of horizontal and vertical motion. The "direction" is
> given by the axis of rotation, which is mutually perpendicular to the two
> components.
>
> That's what I tried to express above. Circular polarization can be
> expressed as a combination of two transverse polarization modes. The
> field of a circularly polarized light wave doesn't point in the direction
> of propagation, but a vector that describes how the field rotates
> (clockwise or counterclockwise) points along the direction of propagation.

But the polarization is always transverse to the direction of
propagation, not in the direction of propagation like the spin
orientation is, even for circularly polarized light. Just because you
can describe the rotation of the field by a vector perpendicular to the
plane of rotation doesn't mean that the polarization is now in the
direction of that vector. The direction of polarization is changing, but
it's always transverse to the direction of propagation. So even for
circularly polarized light, polarization and spin orientation are not in
the same direction, so they can't be the same thing.

If the spin is described by the same vector that describes the rotation
of the field, then what about linearly polarized light? That vector
would be the zero vector, since the field is not rotating. Does that
mean that in the case of linear polarization the photon has no spin? 

-- 
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
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"Proposed Quantum Mechanical Connection"
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