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1

In fact you can think of freely propagating electromagnetic field as an infinite set of harmonic oscillators. To see this, note that (with appropariate choice of unit system) energy density of electromagnetic field is proportional to $\vec E ^2 + \vec B ^2$. This expression is sum of two quadratic terms, which closely resembles Hamiltonian of harmonic ...


0

In quantum electrodynamics, the photon propagating is modeled as an excitation propagating along coupled harmonic oscillators. From this point of view, the photon is the excitation of the vacuum state. Including the other answers above one can see that this relationship comes up over and over again.


2

A photon is produced by a transition between two levels and by definition of "photon" its energy is $h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave that will emerge from a great number of the same energy photons. So it is a matter of coincidence only because Maxwell's equations have sinusoidal solutions for the electromagnetic ...


3

It's not a coincidence! You can see the reason even in classical mechanics: if you take a charge and shake it sinusoidally at frequency $\omega_q$, it makes light with equal frequency $\omega_{\gamma} = \omega_q$. If you quantize light wave emission into individual photons, so that $E = \hbar \omega_{\gamma}$, the spacing between harmonic oscillator energy ...


0

All radiation whether it's blackbody or as you say "non blackbody" is still made up of photons. It's the photons that are oscillating and have a frequency.


8

There doesn't exist any procedure to uniquely associate a Hermitian operator $L$ to a function of the phase space $f(x,p)$. Quantum mechanics is a theory that exists independently of classical physics. Quantum mechanics is not just a cherry on a classical pie that needs the classical theory to exist at every moment. If we want to define a quantum theory, we ...



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