High School Photon Energy & Wave Amplitude

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A photon's energy is defined by the equation E=hv, where frequency is a key wave property, while amplitude relates to the expectation value of photon number. Amplitude does not affect the energy of individual photons but indicates the quantity of photons present. This distinction clarifies common misconceptions about treating photons with a wave function similar to non-relativistic quantum mechanics. The electromagnetic field must be quantized since photon number is not conserved, allowing for the creation and destruction of photons in interactions. Ultimately, a single photon in a specific frequency mode consistently possesses energy equal to ##\hbar \omega##, with detection occurring as a whole or not at all.
jeremyfiennes
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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?
 
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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.
 
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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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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