Re: Do free particles have spin?

From: Old Man (nomail_at_nomail.net)
Date: 09/27/04 

Date: Sun, 26 Sep 2004 20:13:32 -0500

"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
news:cj6fbc$984$2@hood.uits.indiana.edu...
> In article <DqadnbKgLpcTiMvcRVn-rA@prairiewave.com>,
> Old Man <nomail@nomail.net> wrote:
>
> >Electromagnetic waves, which consist of photons,
> >can have linear transverse polarization. A photon
> >cannot. A photon has only circular polarization,
> >either in the direction of propagation, or opposed
> >to it, as indicated by the direction of its spin vector.
>
> What if I prepare a photon in the state (|right>+|left>)/sqrt(2)?

After the above reply, Old Man looked in "Quantum
Electrodynamics" by Landau & Lifshitz:

They say that, (|right>+|left>)/sqrt(2) is a legitimate
single photon wave function. However, they also say
that it doesn't represent a "pure" quantum state, as do
the helicity states, and that, under axial symmetry,
only helicity is conserved.

So, Old Man is simply wrong. Evidently a linear
polarizing "filter" or a reflector (which don't possess
axial symmetry) can absorb a circularly polarized
photon and re-emit it in a mixed state, such as
psi = (|right>+|left>)/sqrt(2), as Gregory suggested.

So, what happens if one attempts to detect linearly
polarized photons with a detector possessing axial
symmetry along the photon's propagation vector ?

[Old Man]