Frequency of EM waves in classical and quantum physics

Question

In classical physics, when a charged particle oscillates, it emits an electromagnetic wave, and the frequency of the wave depends on the frequency with which the particle oscillates.

But in quantum physics, when an excited atom emits a photon, the energy of the photon depends on the magnitude of the quantum leaps that the emitting electron makes (if it jumps one level, the photon will have a certain energy; if it jumps two, a greater energy, and so on). So the frequency of the electromagnetic wave that corresponds to the photon will depend on the amplitude of the quantum leaps made by the electron. I don't understand why these two cases are so different. In analogy with the classical case, shouldn't the frequency of the wave emitted by the atom depend on the frequency with which the electron makes quantum jumps? Or is there a quantum explanation of the classical case that I don't understand? I know that classical physics cannot be used to explain quantum phenomena, but it seems strange to me that there is this asymmetry in the two cases. Sorry in advance if the question is dumb; I am approaching quantum physics because it is a topic that I am very passionate about, but I have a totally different background ... thanks if you can clarify me in a simple way!

Answer 1

To add to the answer of anna v:

In your position it may give you better intuition by looking at the problem of a driven atom (a two-level system, to be precise) in the semi-classical picture (i.e. quantized electron states, classical field).

You find that the expectation value of the dipole operator oscillates with the frequency of the driving field. You can think of this dipole as the "antenna" that emmits electromagnetic radation. The classical analogon is just two particles with opposite charge and oscillating distance between them. They will emit radiation at the frequency of oscillation.

This oscillation comes from the electron being in a weighted superposition of the ground- and excited state, where this weighting oscillates at a frequency that is equal to that of the driving field. The amplitude of oscillation gets smaller, the further detuned the driving field is from the atom. The detuning is defined as the difference between the frequency that is assigned to the transition between the excited and ground state and the frequency of the driving field. For furhter information you can look up "optical bloch equations".

If you understand this, you can go to the "fully" quantized picture, where you also quantize the modes of the electromagnetic field. What you find is that every mode is a harmonic oscillator which you can quantize in the standard way. A mode is occupied by $n$ photons, if the corresponding oscillator state is $|n\rangle$. This $n$ you can think of as the quantized amplitude of the harmonic oscillator. You see that the further we go into the quantum mechanical picture, the harder it gets to give you intuition about what is going on. The truth is that you'll only gain some intuition by working with those problem on an intimate level (i.e. not only reading answers on stack exchange, but also diving into the math and solving problems).

Hope this helps!

Answer 2

Classical electromagnetism is completely described by the solutions of the Maxwell equation. Atomic spectra were the observation that could not be modeled with classical physics, in contrast to the continuous distributions from antennas, and were one of the reasons for introducing quantum mechanics.

The maxwellian solution would be having the atoms as bound states of the electron and the nucleus, similar to the planetary model of the planets bound around the sun, and this cannot work because rotating electrons would radiate, according to the maxwell equations, in a continuous spectrum from the rotations losing energy and falling on the nucleus. This led to the quantum mechanical equations and solutions for bound states.

In analogy with the classical case, shouldn't the frequency of the wave emitted by the atom depend on the frequency with which the electron makes quantum jumps?

The classical case for the atom is invalid, the quantum jumps have nothing to do with oscillations or rotations , When light comes from atomic transitions , here for helium,

it is the addition of a large number of individual photons from individual atoms from the same energy levels that make up the lines, not of anything oscillating. The photon is described by the energy $E=hν$ where $ν$ is the frequency of the collection of many many photons. Maybe this single photon at a time experiment will help you understand the difference between electromagnetic radiation and its constituent photons.

Quantum electrodynamics, QED , is a mathematical theory based on quantum theory, which can show how the classical electromagnetic fields emerge from the underlying quantum fields, but it needs some years of study to be really able to understand the mathematics.