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| Kyle... |
 Posted: Thu Nov 05, 2009 8:28 am |
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Guest
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I often see in the literature that circularly polarized light consists
of a large number of photons with a certain propensity for their spin
helicity: either + or - hbar. However, the spin multiplicity for a
photon (spin quantum number of s = 1) is 2s+1 = 3. What polarization
state does a photon with no spin vector component along the direction
of propagation correspond to, if any?
I ask because I have seen many times that a linearly polarized light
beam consists of equal numbers of photons with projections onto the
propagation axis of -hbar and +hbar.
I use the following definitions:
spin quantum number: s = 1
magnitude of spin vector: ms = hbar*sqrt(s(s+1)) = sqrt(2)hbar
component of spin vector along propagation axis (z-direction): sz = -
hbar, 0, or +hbar
spin multiplicity: 2s+1 = 3 (number of possible projections along z-
direction)
Thanks! |
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| Bob May... |
| Posted: Thu Nov 05, 2009 8:18 pm |
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Be it noted that circularly polarized light doesn't reflect specularly
depending upon the angle of the edges of the surface. As such, as the actor
lifts something, that something isn't sending out unique bright reflections
of light into the audience.
--
Bob May
rmay at nethere.com
http: slash /nav.to slash bobmay
http: slash /bobmay dot astronomy.net |
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