# Is there a harmonic potential explaining why photons have energies $E=n\hbar\omega$?

Is there a harmonic potential (maybe vacuum fluctuations) that is responsible for the energy of the photons being $E=n\hbar\omega$?

In classical and semi classical physics we always consider $E=\hbar\omega$, does that come from the bose statistics which tends to keep the particles to their lowest levels or is this related to the fact that we do not consider the electromagnetic field to be quantized?

I know this question has partly been answered with references to Maxwell's equations but it really makes it look like a coincidence to me where I suspect it is not. I suppose Maxwell's equations can be derived from more fundamental quantum electrodynamics...

Yes, there is a harmonic potential at the heart of quantum electrodynamics. In the electromagnetic Lagrangian density you have $$\mathcal{L}= \frac{\epsilon_0}{2} \mathbf{E}^2 - \frac{1}{2\mu_0} \mathbf{B}^2.$$ The magnetic field plays the part of the potential energy, with the coordinate being the solenoidal part of the vector potential, $\mathbf{A}$, and the momentum being (related to) the solenoidal part of the electric field, $\mathbf{E}$.

The divergent part of the electric field also contributes to the kinetic energy, but it doesn't have an associated potential energy term. Thus, the quantum states are states of definite divergent part of the electric field and a continuum of energy states associated with it are allowed.

$$a^{\dagger}|0\rangle= |k\rangle,$$
and using the dispersion relation $\omega=k c$, so the energy of the one particle state is just $\hbar\omega$ if you keep track of factor of Planck's constant. If a beam or states contains many photons then the energy is just the sum of individual ones. References are