How can electron transitions produce EM waves without classical acceleration?

Question

Hi guys,

I've heard about wave-particle duality, $E = hf$, and the photoelectric effect. I never thought quantum mechanics was that weird until now.

What I mean is, our classical theory of electromagnetic waves is much simpler. For instance, if we know the frequency of a radio wave, say X MHz, then we also know the electrons in the antenna are vibrating at X MHz, assuming we ignore the Doppler effect and other complications. This is generally true for radio and microwave frequencies.

But statistically speaking, the majority of visible light around us comes from electron transitions, which are quantum mechanical in nature. We can calculate the frequency of visible light, which falls in the range of 400 THz to 790 THz.

So is anything actually vibrating that fast to produce such waves?

I guess not, because electron transitions don’t seem to involve acceleration in the classical sense.

This makes me wonder, if no physical motion or vibration is happening at that frequency, what exactly does the frequency of visible light represent when it originates from electron transitions?

Answer 1

So is anything actually vibrating that fast to produce such waves?

Yes and indeed Emilio Pisanty's answer to Is there oscillating charge in a hydrogen atom? includes a figure that shows exactly this oscillation.

In a hydrogen atom the $1s$ and $2p$ orbitals are time independent and there is no oscillation with time. However if we mix the two orbitals then the resulting superposition does oscillate, and that's what Emilio's diagram shows. And indeed the frequency of this oscillation is exactly the frequency of the light that is emitted (or absorbed) in the $2p ⟶ 1s$ transition. So basically the oscillating charge in the hydrogen atom emits light in an analogous way to a classical oscillating charge emits light.

We need to be cautious about interpreting this as a simple classical oscillation. For a start the atom emits a single photon not a continuous wave, and the exact physical meaning of a wavefunction in a superposition of states is the subject of some argument. However at least at an arm waving level we simply have an oscillating charge emitting light at the same frequency as the oscillation.

Answer 2

is anything actually vibrating that fast to produce such waves?

Yes, the psi function $\Psi(q,t)$ of the electrons.

To connect this to EM radiation, we can calculate from $\Psi(q,t)$ e.g. the expectation value of electric current density due to an atom, or a molecule, or a piece of solid.

Schroedinger analyzed these vibrations for hydrogen atom based on his equation in his papers from 20's. Similar analysis can be done (in principle) also for larger systems of charged particles. Usually, interaction of an atom with radiation in non-relativistic quantum theory is described in terms of electric dipole moment; when the psi function is a superposition of Hamiltonian eigenfunctions associated with different eigenvalues, expectation value of dipole moment oscillates. I think the conceptually clearest way to formulate it, and to connect to dispersion/absorption theory of material media, is when we talk about electric current density and how we get it from Schroedinger's equation. The two ways (dipole moment/current) are mathematically the same.

It turns out that the expectation value of dipole moment or current density can vibrate at all the transition frequencies $\frac{E_m-E_n}{h}$ where $E_m,E_n$ are the Hamiltonian eigenvalues. Different frequencies have different intensity though, depending on matrix elements of the dipole moment operator.

I guess not, because electron transitions don’t seem to involve acceleration in the classical sense.

Quantum theory doesn't have acceleration of particles in the classical sense, but it can calculate expected value acceleration based on $\Psi$. To connect with EM theory, expected value of electric current density and its time derivative are the quantities of interest.

Answer 3

Yes, something is oscillating; the wave function of the emitting electron cloud has a significant overlap with the oscillatory electric wave that is the emitted photon. The probability of emission of light goes to zero if those functions don't share a wavelength and frequency.

The quantum mechanical balance of the momentum of that outgoing photon is sometimes a many-atom collective recoil, instead of a single atom (this is the Mossbauer effect). That recoil is another bit of the puzzle, and is a classical acceleration.

The light and matter interaction possibilities are many, but every detail of spectrum and intensity seems to be well covered by our theories.

Answer 4

I think introducing the concept of non radiative transitions is useful here, because those very clearly show that the primary quantity at play is the energy of the transition.

A non radiative transition is an electronic transition that will not produce a photon. Instead, the energy difference is deposited somewhere else. The most clear example is Auger electron emission. If there is a hole in one of the inner shells of an atom, which may be caused by ionizing radiation for example, then an electron from a higher energy level may transition into the lower shell. In doing so it can transfer the energy difference not into the electromagnetic field, creating a photon - rather, into another atomic electron.

So, what is the role of the photon here? The frequency of the emitted photon in a radiative electronic transition simply represents the energy difference between the states involved in said transition.