Photon Energy & Wave Amplitude

  • Context: High School 
  • Thread starter Thread starter jeremyfiennes
  • Start date Start date
  • Tags Tags
    Amplitude Photon Wave
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
jeremyfiennes
Messages
323
Reaction score
17
TL;DR
Where does a photon's wave amplitude enter into its energy equation?
A photon's energy is E=hv where v, the frequency, is a wave property. Particles don't have frequencies. But a wave's energy also depends on its amplitude. Where does this come into the energy relation?
 
Physics news on Phys.org
jeremyfiennes said:
a wave's energy also depends on its amplitude. Where does this come into the energy relation?

Wave amplitude corresponds to the expectation value of photon number (more precisely, to the square root of it). So in the photon model it has nothing to do with the properties of individual photons; it's just how many photons there are.
 
  • Like
Likes   Reactions: BvU
This seems to be the typical misconception arising from the misunderstanding that one could treat photons with a wave function as in non-relativistic Schrödinger quantum mechanics. That's not the case. The electromagnetic field is what has to be quantized, because photon number is not conserved but photons can easily be created and destroyed in interactions of charged particles.

The free field, describing asymptotic free states is normalized such as to obey the canonical commutation relations in, say, the Coulomb gauge (or for free fields the "radiation gauge" following from it). Then you can define momentum-eigenmodes of single free photons and build the entire Fock space from the photon-number eigenstates. The total energy density is then given as an expectation value with respect to the states. A single photon in a single-frequency mode always has the energy ##\hbar \omega##, no more no less, and all you can say about this photon is the probability that it's detected by some detector. It's always detected as a whole or nothing.