Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

Question

Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

Answer 1

We have to make clear that a photon is an elementary particle, the quantum of the electromagnetic field, it has a zero mass and a spin of 1, and its energy is given by $E=h\nu$.

An electromagnetic wave is composed of photons, a photon is not a chopped up electromagnetic wave. The electromagnetic field emerges from a huge ensemble of photons in a smooth way that goes from the microworld of quantum mechanics to the macroworld of classical physics. The how is described in this link. In a nutshell, it is the electromagnetic potential ${\mathbf A}$ that enters in the quantum mechanical wave function of the photon, which in tandem with zillions of photons constructs the macroscopic with frequency $\nu$ wave and builds the electric and magnetic fields of the classical wave.

Answer 2

If you ask whether there is a phase difference of 90° between the electric field and the magnetic field, the answer is yes. The electric field and the magnetic field oscillate in quadrature. You can see that from the conservation of the total electromagnetic energy during one oscillation cycle, or as you mentioned it, to conserve the energy of a "single photon" during a cycle.

Plug quickly $$E = E_0 cos(\omega t)$$ and $$B = B_0 sin(\omega t)$$ in the instantaneous electromagnetic energy density), remembering that $$cos(\omega t)^2 + sin(\omega t)^2 = 1 $$ and $$ B_0 = E_0/c $$ in the present case of transverse electromagnetic waves propagating in free-space.