I'm aware that there're some questions posted here with respect to this subject on this site, but I still want to make sure, is frequency quantized? Do very fine discontinuities exist in a continuous spectrum like the black body spectrum?

The quantization of photon energies


Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e.

$$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$

for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator,

$$E_n=\hbar \omega \left( n+\frac{1}{2}\right)\quad n=0,1,2,\dots$$

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    $\begingroup$ I initially wanted to leave a comment on JamaIS answer but couldn't due to low points. Planck's postulate was: Any physical entity with single degree of freedom whose "coordinate" is a sinusoidal function of time(i.e. executes simple harmonic oscillation) can only possess a total Energy $E$ of: $E = nhv$ where $n=0,1,2,3...$ Source: Quantum Physics of Atoms, Molecules, Solids, nuclei and particles. Robert Eisenberg & Rober Resnick. Frequency available is continuous (spectrum), Photon Energy is quantized as a packet of energy -with a given frequency- i.e. a blue light with wavelength 450nm will $\endgroup$ – user -1 Feb 18 '15 at 4:29

"Yes", but the quantisation depends on the size of the box. In practice the 'box' is large and of variable shape, so all sizes are available, so all frequencies are available.

Ultimately, it is somewhat of a philosophical question who's answer depends on which axioms and base concepts you (they) are using at the various stages of reasoning. Consider, does time pass for a non-interacting particle? Can a non-interacting particle be kept in a box? etc.

Try for a quirky view on the problem. Link Between the P≠NP Problem and the Quantum Nature of Universe

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  • $\begingroup$ I have not looked at the paper the linked article alludes to, but the article itself misses the key piece of the argument. Namely why would the fact that it is hard to evaluate the model have any effect on whether the modelled system can behave in certain way. Especially since it is already known that quantum calculations allow doing certain operations faster (but with a chance of failure) than calculation on traditional computer. $\endgroup$ – Jan Hudec Apr 29 '14 at 20:33
  • $\begingroup$ @JanHudec, As I understand the paper, it considers that our universe is the computational device!, and that it can't compute the solution for macroscopic items (that is, it isn't a quantum super-position solution). It just has to wait until we observe an item to force a state onto it, which would then create a cause-effect cascade (the last bit's my sudden realisation). . . Going off at a tangent, it's like Fast Multi-pole Methods which can 'ignore' effects at a distance by rolling them all up into a single item (e.g. distributed mass -> mass at a C of G) $\endgroup$ – Philip Oakley Apr 30 '14 at 8:53
  • $\begingroup$ @JanHudec An experiment acting like a model means the experimenter can describe what the model would predict the thing to behave like. An experiment (or observation of) a macroscopic system acting quantum implicitly assumes we are are able to describe what a macroscopic system acting quantum looks like: which implies being able to solve the equation. The paper states that this is impossible beyond toy systems. But there is no need to presume that our universe is a computational device; rather, the act of theoretical physics and experimentation is a computational device. I am not convinced. $\endgroup$ – Yakk Jul 30 '16 at 2:41

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