Intuitive explanation for the de Broglie / Planck relations A friend asked me to explain "why" a particle's energy is proportional to it's frequency, i.e: $$E=h\nu$$
The reason this result is so un-intuitive, is that in the macroscopic world, A wave's energy (e.g electric power dissipated through a resistor) is not frequency dependent but rather amplitude dependent. Of course a "pure" photon doesn't really have an amplitude due to it's description as a plane wave, but that's beside the point.
So in short: Is there an easy explanation of "why" the energy is only dependent on frequency? can this be "derived" from other principles (say, the particle-wave duality)?

N.B. I'm aware there are no real "reasons" for any law of physics other then the fact that they agree with experiment, but laying those laws on a smaller number of basic principles is always nice when possible.
 A: As other have said, the answer to "why ?" is basically "because !". It is the definition of quantum physics. 
However, one could have an answer linked to the anthropic principle, which is basically :

If the particle energy would not be increasing with its frequency, the blackbody radiation would have an infinite power, because of the ultraviolet catastrophe.

Let's suppose the energy of a photon to be $\varepsilon(\nu)$. In quantum physics, $\varepsilon(\nu)=h\nu$, but we are precisely exploring other possibilities. 
If $\varepsilon$ were constant or decreasing, when $k_BT$ is at least of the order of $\varepsilon$, each mode of the electromagnetic field would carry an energy of the order of $k_BT$. Since $\nu$ is a priori unbounded, there are infinitely many modes, and the sum is infinite. This problem, is the ultraviolet catastrophe and was solved in 1900 by Max Planck when he established Planck's law. (Edit: The previous sentence is historically inaccurate: Planck did not want to solve this problem, which was only stated in in 1905.) 
If $\varepsilon$ increases with $\nu$, then the average population of the high frequency modes decrease exponentially, allowing to sum over all the frequencies and have a finite integral.
More quantitatively, the population of a mode with frequency $\nu$ is $P(\nu)=\frac1{e^{{\varepsilon}/{k_BT}}-1}$ and Planck's law is
$$ I(\nu,T) =\frac{ 2 \varepsilon\nu^{2}}{c^2}\frac{1}{ e^{{\varepsilon}/{k_BT}}-1}.$$
If one wants the total radiated power $\int_0^{+\infty}d\nu I(\nu,T)$, one needs $\varepsilon(\nu)$ to increase fast enough ($\varepsilon\propto \nu^l$ seems enough $\forall l>0$, but $\varepsilon\propto\log \nu$ is not fast enough.) 
A: If you already know that why-questions are always a little problematic and you have an understanding regarding amplitude assumption involving a quantization of the field into photons, then a short argument why the frequency-energy relation pops up in quantum mechanics is that the axiom
$$\left(i\hbar\frac{\partial}{\partial t}\right)\Psi=H\Psi,$$
already contains the relation, the Hamiltonian giving the energy. Check the units - it basically says that energy is one over time. Consider a decomposition of the wave into plane waves 
$$\psi\sim \text e^{-i\frac{t}{\hbar} \omega}\ \ \ \Longrightarrow\ \ \ \left(i\hbar\frac{\partial}{\partial t}\right)\psi=\omega\ \psi$$ 
then the energy is a decomposition into frequencies $\omega$.
A: I suspect this argument just exchanges one non-intuitive bit of physics for another, but you could try it on your friend and see.
The key point is that photons aren't (just) a convenient way of chopping up a light beam into bits. The photon is the unit of interaction between light and matter i.e. when matter interacts with light it does so by absorbing or emitting a photon.
Also, light doesn't interact with all matter, it specificially interacts with electric dipoles. Absorbing light makes a dipole oscillate and an oscillating dipole emits light.
Let's take the quantum version of the oscillating dipole to be a simple harmonic oscillator. I think the argument is simplest if you consider the oscillator emitting light (absorbing light is the same process but time reversed). If you solve the Schrodinger equation for a simple harmonic oscillator you find that if the frequency is $\nu$ the energy levels have a spacing $h\nu$. The oscillator emits a photon when it drops down an energy level, and the frequency of the emitted photon will be the same as the frequency of the oscillator. Since the energy of the photon must match the energy spacing in the oscillator, the energy of a photon of frequency $\nu$ is $h\nu$.
