Re: Do free particles have spin?

From: Gregory L. Hansen (glhansen_at_steel.ucs.indiana.edu)
Date: 09/27/04 

Date: Mon, 27 Sep 2004 01:46:33 +0000 (UTC)

In article <hfKdna-aNvKi9MrcRVn-vg@prairiewave.com>,
Old Man <nomail@nomail.net> wrote:
>
>"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.

I don't understand what a "pure" quantum state means in this context.
Whatever quantum state you have, to express it you must first pick your
basis. What is diagonal in one basis won't necessarily be diagonal in
another, but so what? If the spin of a neutron is (|+z>+|-z>))/sqrt(2),
is that a pure state? Doesn't look like it. But switch to a basis S,m_x
and the neutron polarization is |+x>. Now it looks pure. And if your
neutron beam has the polarization |+x> it will pass unimpeded through a
spin filter that passes neutrons in the state |+x> (e.g. a glass cell
filled with polarized 3He pointing in a direction transverse to the beam).
You could still say the spin filter passes neutrons in the state
(|+z>+|-z>)/sqrt(2), but that's clumsy.

>
>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 ?

  |vertical> = (|right> + |left>)/sqrt(2)

(With a possible sign or sqrt(-1) error.)

  P = |<right|vertical>|^2 = 1/2

Half of them will get through. 

-- 
"In any case, don't stress too much--cortisol inhibits muscular
hypertrophy. " -- Eric Dodd