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A radio generator feeds the antenna rod with alternating current. This accelerates the electrons in the rod towards the end of the rod, then decelerates them, accelerates them towards the generator and decelerates them again. How many photons are emitted during one period of the radio generator?

One more side question, which can be removed if the "Only one question" policy should apply.
In a synchrotron, the motion of an electron can be decomposed into an X-sinusoid and a Y-sinusoid. For each of the two components, we get the same acceleration sequences as in the antenna rod. For a synchrotron, what does the number of emitted photons depend on and how many of them get emitted?

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  • $\begingroup$ Doesn't this depend on the power of your generator and the size of your antenna? $\endgroup$ Commented Mar 25, 2021 at 21:12
  • $\begingroup$ I guess you have to derive the number from the emitted energy. Divide the energy through the energy of one photon with the emitting frequency, and subsequently through the inverse of oscillation time associated with the frequency. $\endgroup$ Commented Mar 25, 2021 at 21:17
  • $\begingroup$ @DescheleSchilder I fully agree with you. The question aims at whether it is one photon per period, 2, 4 or a much larger number of photons emitted by each of the surface electrons. Thank you for reflecting on this question. $\endgroup$ Commented Mar 25, 2021 at 21:20
  • $\begingroup$ Ah! This should depend on the number you get from what I wrote in the comment. Or not? $\endgroup$ Commented Mar 25, 2021 at 21:23

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The question as written is unanswerable, because the photon number operator does not commute with the electric/magnetic field operators, and so a state with definite electric and magnetic fields (which is what we usually consider the wave emitted by an antenna to be) has an indefinite number of photons. For instance, a state of definite photon number always has constant vanishing expectation value for the electric field and so cannot in any way correspond to the classical idea of an electromagnetic wave.

Instead, monochromatic classical light corresponds to something like coherent states for $\alpha \in\mathbb{C}$ which have a mean photon number $\lvert \alpha\rvert^2$, an uncertainty in the mean photon number of $\lvert \alpha\rvert$ and an uncertainty in the electric field strength that goes as $\propto\frac{1}{\lvert\alpha\rvert}$, while the expectation for the electric field amplitude goes as $\propto\lvert \alpha\rvert$, with the time-dependence being the periodic function we actually expect of an EM wave. For large $\lvert \alpha\rvert$, $E$ becomes more and more well-defined, and so this state resembles a classical wave with sharply defined $E$ more and more.

So, while the average photon number corresponds to the naive computation you get when you compute how many photons you need to emit the total energy of the wave, it is wrong to say that this is "the number of photons emitted" - that number is indefinite for a state that resembles a classical wave.

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You have not specified a lot of the details. The number of photons per electron over one cycle would be given by $$\frac{e \sqrt{p z}}{f h}$$ where $p$ is the power, $z$ is the antenna impedance, and $f$ is the frequency, $e$ is the charge on an electron, and $h$ is Planck’s constant.

For a 100 W ham radio with a dipole antenna operating at 145 MHz that works out to about $1.5 \ 10^8$ photons per cycle for each electron with a quantum uncertainty of $1.2 \ 10^4$ (which is much less than than the uncertainty/rounding in the other estimated quantities)

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  • $\begingroup$ Number of photons per electron over one cycle is 1,5x10^8? $\endgroup$ Commented Mar 27, 2021 at 7:59
  • $\begingroup$ For the conditions specified, yes $\endgroup$
    – Dale
    Commented Mar 27, 2021 at 12:24

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