Photon Polarization: Explained

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Photon polarization is fundamentally linked to the transverse nature of electromagnetic waves, which allows for two polarization states. Photons are produced when a charged particle moves; linear polarization occurs with up-and-down motion, while circular polarization results from rotational movement, with two possible helicities based on the direction of spin. The concept of gauge invariance explains why photons exhibit only two polarization states instead of three. The term "photon polarization" can be seen as contradictory, as it combines particle and wave characteristics. Understanding these principles is essential for grasping the behavior of photons in various physical contexts.
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can some one explain 2 me the basis of photon polarization ...?
 
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See any freshman physics book, or one more advanced, to learn that electromagnetic waves are transverse, and that does it --transverse => two polarization states. Very basic stuff.
Regards,
Reilly Atkinson
 
preet0283 said:
can some one explain 2 me the basis of photon polarization ...?

Photons are "generated" when a charge moves in a specific way. If
the charge moves up and down, the photons that fly off will be polarized
linearly.

If the charge spins around, they will be polarized circularly and one of two
ways depending on which way it spins, CW or CCW.
 
It's called "helicity operator eigenvalues". Why the photon has two instead of 3, well, it's called "gauge invariance". "Photon polarization" is a bit of an oxymoron, as "polarization" is typical to wavelike phenomena, while "photon" is a particle.

Daniel.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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