Scattering of an unpolarized photon off an electron

AI Thread Summary
The discussion centers on the scattering of unpolarized photons off electrons, specifically addressing the polarization of the outgoing photons. It is noted that while classical electrodynamics describes the scattering process, the interaction with unpolarized light can still result in polarized emissions due to the electron's spin. The concept of an unpolarized beam is clarified as a random selection of electromagnetic waves, which can scatter in a way that aligns with the dipole moment of the electron. The participants express confusion over how unpolarized light can lead to polarized scattering and the role of the electron's spin in this process. Ultimately, the conversation highlights the complexities of photon-electron interactions and the implications for understanding phenomena like cosmic microwave background polarization.
fairy._.queen
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Hi all!

I have been told that if an unpolarized photon hits an electron in a Thomson scattering the outcoming photon will be polarized because of the electron's spin. I didn't understand what it means, nor do I get how an electron reacts to an unpolarized photon: in Classical Electrodynamics, if an electron is hit by an electromagnetic wave, it starts to vibrate in the direction of the electric field, i.e. of polarization. If the photon is unpolarized, there is no preferred direction for the electron to vibrate in, so I do not understand how it can even emit an outcoming electromagnetic wave.

Thanks in advance!
 
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The expression "unpolarised photon" gives me a problem. What does it mean? Polarisation is a field / wave description, surely.
 
I meant an unpolarized beam of photons, in any case, an unpolarized EM wave.
 
OK
A "beam of photons" is best described as a wave until you are dealing with the actual interaction between the wave and a charge system (e.g. an atom), at which time there will or will not be absorption of one photon's worth of energy and momentum. The interaction can only be in energy steps of one quantum. By "unpolarised beam" you must mean the result of a random set of emissions at the source - in the form of Photons, if you like - as in the emission from an ionised gas. These photons then have to lose their individual identity as the wave travels from A to B until a photon reacts (or not) with your receiving atom. Whether or not there is an interaction will be a statistical thing.
In your implied model, if an atom had to wait for a 'whole photon' of precisely the right 'orientation' before it could interact then only an infinitessimal proportion of photons could be absorbed - rather than the 1/√2 of the total that would occur in practice.
 
Thanks for replying!
What I don't understand is:
1) How can you describe, using Classical Physics, the reason why an unpolarized EM wave is scattered at all? Perhaps your post suggests that Classical Physics doesn't actually describe the phenomenon, but I want to be sure.
2) How can an interaction between an unpolarized beam of photons and an electron result in a polarized beam of photons?

Thanks again!
 
fairy._.queen said:
Thanks for replying!
What I don't understand is:
1) How can you describe, using Classical Physics, the reason why an unpolarized EM wave is scattered at all? Perhaps your post suggests that Classical Physics doesn't actually describe the phenomenon, but I want to be sure.
2) How can an interaction between an unpolarized beam of photons and an electron result in a polarized beam of photons?

Thanks again!

OK. Bog standard classical approach:
If you take a dipole in the presence of a single polarised EM wave then you will get scattering, according to the relative angle of the dipole to the E plane of the wave. If you hit it with a random selection of different EM waves (radio transmissions, for instance), around the same frequency and with different polarisations, from the same direction, you will find that the resulting scattered waves will have the same polarisation as the dipole and that the unscattered (majority) power will still be largely unpolarised. An 'unpolarised' beam of light is equivalent to this random selection of EM waves - with a narrow range of frequencies and not coherent, or the resultant would be necessarily polarised because it would always have a maximum E in an particular plane.

This translates fairly easily to a single atom, I think - as long as you don't need to insist that the photon 'exists' before the interaction occurs - which is the way I find it easiest to look at things.

PS My "1√2", in the previous post, should really be 1/2, I think.
 
Thanks again for replying. What puzzles me is that, if you have a free electron and not an atom, then you have no dipole. I've been told that the outcoming EM wave is polarized due to the electron spin, but I don't exactly understand why.
Is it perhaps that the magnetic field of the wave will be aligned to the spin direction and so the wave will be partly polarized?
 
The only effect like this that I have heard of is what you get in the ionosphere, where the electrons are caused to move along the E field of a radio wave and, due to the Earth's magnetic field, they follow a curved path. This causes a cross-polar component. (Google Appleton - Hartree equation)
 
Thanks for replying.
The contest is that of the CMB polarization, created due to a quadrupolar variation in the temperature. The Thomson scattering cross section depends on the polarization of the incoming and outcoming photons, so that a net polarization is created if a quadrupolar variation in the temperature is present. However, I have been told polarization is created even if the incoming wave is not polarized, due to the electron spin.
 
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