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Despite Planck's constant being in $E=hf$, it would appear to me that energy is still not discrete, as frequency can be an fraction of a Hertz that one wants. How does this imply that electromagnetic radiation is quantized?

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For light of a given frequency the energy comes in discrete units of $hf$, i.e. you can have $0\ hf,\ 1\ hf,\ 2\ hf,\cdots$. –  Michael Brown Aug 7 '13 at 16:12
$E/f$ comes in steps of size $h$. –  Johannes Aug 7 '13 at 16:28

2 Answers 2

It's because there are quantifiable "packages" of light, called photons. There can be an infinity of frequencies (energies) that photons can have, but one photon can have only one quantifiable energy.

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In my books, the energy content of light is not quantized generally in the same sense that the energy of an electron is not quantized generally. The electron can have a continuous energy spectrum as all other free particles.

It is only particles bound in potentials that get quantized energies, and then the photons that mediate transitions between these energy states are quantized. Thus we get the emission and absorption spectra of atoms.

emissionof iron

emission spectrum of iron

In this sense these particular photons are quantized in energy.

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I don't agree. The bosonic free field are quantized. They are harmonic oscillator $\phi(\vec k,t)$ –  Trimok Aug 7 '13 at 17:07
@Trimok why would they be quantized if they move infinitely? Or do you use some unusual evolution equation for them (maybe in barycentric coordinates?)? –  Ruslan Aug 7 '13 at 20:47
@Trimok one can always state that "everything is quantized" since there is really no space where there does not exist some potential . One can always approximate a symmetric potential by its tailor expansion where the first term is the harmonic oscillator. In theory. In real life there exist continua because 1) either one cannot measure any difference or 2) one has hit the heisenberg uncertainty limit which is an inherent mixer. –  anna v Aug 8 '13 at 3:18
@Ruslan : In Quantum Field Theory, Only free fields are quantized. This means that we could associate creation and anihilation operators, states, etc... For these free fields, one may speak of particles. When one turns to perturbations of fields, or "virtual particles" which appear in Feynman diagrams, representing interactions, you don't have creation/anihilation operators, states, and in fact you don't have "particles". –  Trimok Aug 8 '13 at 6:55

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