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I am trying to form a clearer picture of how a radio transmitter emits waves from a quantum point of view.

The classical description is quite easy: the electrons oscillate in the antenna and, as accelerating charges, emit electromagnetic radiation. When looking at it from the quantum viewpoint, we obviously cannot directly apply the picture of electronic transitions between discrete states, since we are dealing with free-moving electrons in metal. More likely, we have to describe this in terms of Bremstrahlung. This raises a number of questions:

  • Why the radiation is on a well-defined frequency, rather than in a broad range of frequencies?
  • If the resulting frequency is a result of constructive interference, does it mean that we have significant losses due to the destructive interference at other frequencies?
  • Why constructive interference works for the radiation, even though electrons are involved in incoherent transport (on quantum level)?
  • Is there a quantum equivalent of Larmor formula?

I appreciate your ideas.

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  • $\begingroup$ "we are delaing with free-moving electrons in metal. " the electrons are not free moving, their orbitals cover the whole antenna lattice.. It can be shown mathematically that the classical emerges from quantum electrodynamics , but it needs quantum field theory to show it.motls.blogspot.com/2011/11/… . To show it for an antenna would be another level . Hand waving: the wavefunction of the photon comes from quantizing Maxwell' equation so it is not surprising that the classical emerges from the quantum: arxiv.org/abs/quant-ph/0604169 $\endgroup$
    – anna v
    Commented Sep 30, 2021 at 9:19
  • $\begingroup$ @annav Yes, I agree with all that - including that the "free" electrons are not really free. I am looking for a more technical answer. $\endgroup$
    – Roger V.
    Commented Sep 30, 2021 at 9:22

4 Answers 4

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Bremsstrahlung is not needed for explaining antenna radiation. In fact, no quantum mechanisms at all are needed to do this.

You can think of an antenna as being the primary side of a coupling transformer in which the secondary side is free space with a certain characteristic impedance. When the impedance of the antenna is matched to that of free space at the frequency of operation, the coupling between the two is good and the current flow through the antenna performs work on the free space by creating electromagnetic waves in it.

Those waves are almost monochromatic because the transmitter that is driving the antenna is furnishing a signal which is similarly monochromatic. For a signal which is modulated, the modulation signal shows up as sum-and-difference sidebands that closely accompany the carrier signal.

If the antenna is driven at a frequency for which the antenna has not been tuned (i.e., it is not resonant at), then an impedance mismatch occurs and the incoming energy piles up on the feedline and the antenna in the form of standing waves which are not radiated. These can build in size to the point where the transmitter is damaged by them.

Constructive and destructive interference works for antennas because the wavelengths involved are enormous (tens to thousands of meters) compared to the length scales where quantum effects must be taken into account- that is, it is the waves that interfere, not the wavefunctions of the electrons themselves.

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    $\begingroup$ Thank you. However the question is specifically about how to describe the emission of antenna in quantum language. $\endgroup$
    – Roger V.
    Commented Sep 30, 2021 at 18:19
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Antenna emits a radio-photon when the antenna changes its energy state in a quantum jump. As the antenna works in cyclic way, it reaches the same state many times, which results lots of identical radio-photons to be emitted.

There is lot of uncertainty on the time when some radio-photon was emitted. But more radio-photons are emitted at certain phases of the cycle, compared to some other phases of the cycle.

When a classical observer says that now the acceleration of charges is high, then many radio-photons are emitted. This can be tested by using a photon detector moving very fast towards a radio-antenna. The frequency of detection events will oscillate. The frequency of detection events is proportional to $q^2a^2$, where q is the total charge of accelerating electrons and a is the acceleration. So that would be the quantum equivalent of the Larmor formula.

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In quantum language the current in the antenna can be described by saying each conduction electron is in a Glauber coherent state of motion, oscillating forward and back along the antenna, just like the charge in the classical picture. You have a very large number of electrons in such state, oscillating in phase, and differing in details such as the mean position and momentum. This kind of coherent state can be visualized as a 'wave packet'. The wave packet corresponds roughly to a classical particle, but its position and momentum spreads are limited by Heisenberg uncertainty principle.

All these wavepackets meanwhile also have charge so they act as the source of electric and (when they are moving) magnetic fields. The way they do this, for a radio antenna, is very well modelled by classical electromagnetism. The quantum picture basically agrees except all the charges have a non-zero uncertainty in position and momentum, and the fields have a non-zero uncertainty in amplitude and phase. However the uncertainty is small compared to the size in the regime of operation of a typical antenna.

The state of the emitted radiation is another Glauber coherent state. Its direction in space is the same dipole pattern as classical physics suggests. In terms of photon number it has a Poissonian distribution over photon number states, with the uncertainty $\Delta n = \sqrt{\langle \hat{n} \rangle}$ where $n$ is photon number. The mean photon number $\bar{n} = \langle \hat{n} \rangle$ is very high so this uncertainty is very small compared to $\bar{n}$: $$ \frac{\Delta n}{\bar{n}} = \frac{1}{\sqrt{\bar{n}}} \ll 1 $$ In terms of frequency of the emitted radio waves (and their associated photons), the mean frequency matches that of the electron oscillations as you would expect. The frequency can in principle be arbitrarily well-defined if the oscillation goes on for long enough. It is limited by Fourier analysis: $$ \Delta \omega \Delta t \ge 1 $$ where $\Delta t$ is the time during which continuous oscillation has been going on.

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Naive approach
The source of confusion in this question is thinking in terms of electronic transitions, as we usually do when discussion emission by atoms, molecules and lasers. If one follows this path, one could start with a Hamiltonian like $$ H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\mathbf{A}(\mathbf{x})\right)^2-e\varphi(\mathbf{x})+e\mathbf{E}(t)\mathbf{x},$$ which describes an electron in the conduction band of the antenna, where $$\mathbf{E}(t)=\mathbf{x}_0E_0\cos(\omega_0t)$$ is the driving periodic electric field which accelerates the electrons. One could then express $\mathbf{A}$ and $\varphi(\mathbf{x})$ in terms of the photon creatinga nd annihilation operators and calculate the rate of photon production. There are many levels of complexity that could be added: underlying crystal lattice (necessary in optical problems for obtaining correct matrix elements), Fermi surface, electron-electron and electron-impurity scattering, etc. All this would eventually produce photons generated at the frequency $\omega_0$, since this is the frequency of the driving field, and all that we do is essentially time-dependent perturbation theory...

Succinct answer
It is logical to ask, where the driving field comes from: it is the oscillations of the LC curcuit, which can be seen as collective electron modes. Thus, a more natural description of the emission by an antenna is in terms of these collective modes (with a bit of a stretch one could call them plasma modes/plasmons).

We thus have $$H=\frac{Q^2}{2C}+\frac{LI^2}{2}=\hbar\omega_{LC}\left(b^\dagger b+\frac{1}{2}\right),\\ Q=Q_0(b+b^\dagger),I=iI_0(b-b^\dagger),\\ \frac{2Q_0^2}{C}=2LI_0^2=\hbar\omega_{LC}=\hbar\sqrt{LC} $$ The current density in the electron-photon coupling (which emerges in the very first equation) could then be taken proportional by the current in the LC curctuit, with the effective Hamiltonian looking like this: $$ H_{LC-photons}\propto\sum_\mathbf{k} (b-b^\dagger)\left(a_\mathbf{k}e^{i\mathbf{k}\mathbf{x}}- a_\mathbf{k}^\dagger e^{-i\mathbf{k}\mathbf{x}}\right).$$ (I disregard the photon polarization, which is typically along the antenna)

Summary
Emission of radio waves can be viewed as transitions between discrete states of an LC oscillator, which is why they also have a discrete frequency. The need to invoke Bremsstrahlung arises only insomuch as we need a detailed description of coupling between the photons and the LC circuit.

The circuit diagram of a simple radio receiver (see also here):
enter image description here

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