QuantumCuriosity42 said:
At a simpler level, without delving too deep into advanced theory, I'm trying to understand why, in the equation E=hf (for an individual photon's energy), the energy is dependent on the harmonic frequency of the wave (I don't think my question is ambigous?) That is precisely Planck's relation, and I've struggled to find a satisfactory explanation online.
Any free electromagnetic field can be decomposed into plane-wave modes, and so you can for the field operators of the quantized theory.
The corresponding mode functions are all ##u(t,\vec{x})=\propto \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x})##. Also they must fulfill the wave equation,
$$\frac{1}{c^2} \partial_t^2 u - \Delta u=0,$$
from which you get
$$\omega=c |\vec{k}|.$$
The free electromagnetic field is then represented by an infinite set of harmonic oscillators, labelled by ##\vec{k}## and the helicity ##\lambda \in \{1,-1\}## to describe the polarization states too. ##\lambda=1## corresponds to right-circular and ##\lambda=-1## to left-circular polarized em. waves.
Each harmonic oscillator then is described by creation and annihilation operators ##\hat{a}^{\dagger}(\vec{k},\lambda)## and ##\hat{a}(\vec{k},\lambda)##. They obey bosonic commutation relations,
$$[\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=(2 \pi)^3 \delta^{(3)}(\vec{k}-\vec{k}') \delta_{\lambda \lambda'}.$$
A basis is then formed by the occupation-number (Fock) basis, starting from the ground state, the "vacuum", ##|\Omega \rangle##, for which
$$\hat{a}(\vec{k},\lambda) |\Omega \rangle=0$$
for all ##(\vec{k},\lambda)## and the states ##|N(\vec{k},\lambda) \rangle##, which are simultaneous eigenvectors of the number-density operators
$$\hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The energy and momentum can then be derived from the operators given as in classical electrodynamics by the corresponding energy-momentum tensor of the electromagnetic field. In terms of the number operators it results
$$\hat{H}=\sum_{\lambda} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \omega \hat{N}(\vec{k},\lambda),$$
$$\hat{\vec{P}}=\sum_{\lambda} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \vec{k} \hat{N}(\vec{k},\lambda).$$
This can be interpreted that the mode ##(\vec{k},\lambda)## comes in discrete "quanta" of energy and momentum, ##E=\omega## and ##\vec{p}=\vec{k}## (note that I use natural units, where ##\hbar=1##.