Line broadening: What is actually broadened?

Question

Is it just the spectrum of emitted photons that is broadened by e.g. finite lifetime or Doppler effects, or is it actually the electronic energy spectrum?

In my head it would only make sense if it is actually the energy spectrum itself.

Could you please clarify what you mean by 'energy spectrum' and 'emitted spectrum' and 'electronic energy spectrum'? — Nihar Karve, Oct 25 '20 at 14:57
@NiharKarve I wonder whether these effects lead to a uncertainty in the binding energy of the electron or whether the uncertainty arises only when the photon is transmitted. — , Oct 25 '20 at 16:46
Related: What is the meaning of natural line broadening?, Atomic natural line width, and links therein. — Emilio Pisanty, Oct 26 '20 at 10:14
there is a Maxwell velocity relation so it results as a spectral emission line — , Jul 11 '22 at 12:09

Nihar Karve · Answer 1 · 2020-10-26T10:06:28.263

2

The two processes you've mentioned, finite lifetimes and Doppler effects, have different effects:

The finite excited state lifetime results in an uncertainty in the transition energy due to the energy-time uncertainty principle - heuristically, $\Delta E \Delta t \ge \frac{\hbar}2$ means that a short lifetime will increase $\Delta E$, so the emitted photons will have a slight 'range' in frequencies, via $E = h\nu$. You might view this as intrinsic to the emission process.
For Doppler effects, the emitted radiation will be Doppler shifted due to the velocity of the particles - you may view this effect as extrinsic to the actual emission process itself.

Of course, both of these are factors in making the spectroscopic lines not infinitesimally thin.

edited Oct 26 '20 at 10:06

answered Oct 25 '20 at 15:13

Nihar Karve

8,445

Isn't the variance of energy of the excited state 0 as it is an eigenvalue? Why then broadening takes place? – Anton Aug 26 '21 at 17:39
1

@Anton the excited states are eigenvalues of the atomic Hamiltonian, but not of the full Hamiltonian (atomic + external quantized EM field). Rather, it is a resonance (finite lifetime = imaginary mass), which gives the spectral line a Lorentzian profile - hence it's a superposition of energy eigenstates of the full Hamiltonian. – Nihar Karve Aug 27 '21 at 00:40

score 0 · Answer 2 · answered Oct 25 '20 at 15:10

Since $E = h\nu$, broadening the spectral (frequency) width of photons necessarily broadens the energy spectrum.

Doppler broadening means that atoms are moving at different directions and speed from thermal motion. When an atom runs upstream into a photon, the frequency is increased from the Doppler effect. Also the photon "hits harder", even though it moves at the speed of light relative to any atom. See How can a red light photon be different from a blue light photon?

If the time between when photon was emitted and absorbed is short, the uncertainty principle increases the uncertainty of the wavelength, and therefore the energy. $\Delta E \Delta t > \hbar$. See The more general uncertainty principle, beyond quantum, a video from 3blue1brown.

Line broadening: What is actually broadened?

2 Answers2