Submitted by Grand-Tension8668 t3_1125ccr in askscience
wooly_boy t1_j8llybf wrote
Reply to comment by Movpasd in When measuring the wavelength of EM radiation... what's actually being measured? by Grand-Tension8668
Ok, so now assuming I understand that radiation is a traveling variation in EM field, how do photons come into play?
1ndiana_Pwns t1_j8ltc70 wrote
It takes energy to create those little EM waves, right? A photon is a little packet of that energy. The exact amount of energy a photon has is determined by the wavelength of the light, with the relationship being a result of all the math behind light propagation and electromagnetism (thank Maxwell and his equations)
Movpasd t1_j8qxs6w wrote
It's quite complicated, but I'll try to give a brief answer --
Maxwellian electrodynamics is a classical field theory. When you canonically quantise such a theory, you find that a (sort of*) conserved, discrete quantity pops out, which can be interpreted as "particle number".
This is in line with observations from a century back around black-body radiation that appeared to show quantisation of energy levels.
It's also a satisfying interpretation because certain calculations in QFT have combinatorial properties which allow them to be represented using Feynman diagrams (which you've probably heard of). Together with the path integral formulation, you get a really useful picture of the physics of scattering. But this is mired in complications (renormalisation, confinement (not a problem for photons but it is for quarks), divergences around every corner, ...).
As a side note, instead of starting from the field theory, you can build up a quantum field theory by starting from a particle-based theory; a common approach for effective theories in condensed matter theory, because it's much more tractable mathematically. The fact that QFT unifies classical many-particle and field theories is an advanced form of the "wave-particle duality" you might have heard of.
*: conserved in the free theory
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