Spectral line broadening [duplicate]

Question

From Wikipedia article Spectral line:

Spectral lines are the result of interaction between a quantum system (usually atoms, but sometimes molecules or atomic nuclei) and a single photon. When a photon has about the right amount of energy (which is connected to its frequency) to allow a change in the energy state of the system (in the case of an atom this is usually an electron changing orbitals), the photon is absorbed. Then it will be spontaneously re-emitted, either in the same frequency as the original or in a cascade, where the sum of the energies of the photons emitted will be equal to the energy of the one absorbed (assuming the system returns to its original state).

A spectral line extends over a range of frequencies, not a single frequency (i.e., it has a nonzero linewidth).

The lifetime of excited states results in natural broadening, also known as lifetime broadening. The uncertainty principle relates the lifetime of an excited state (...) with the uncertainty of its energy.

Can you provide proof for the last statement? (Maybe without using QED, if possible)

Answer 1

Well mathematically you could ask what is the frequency behaviour of a damped harmonic oscillation?

If the excited energy state is represented by a quantum harmonic oscillation centred around some frequency $\omega_0$, but the oscillation amplitude decays exponentially with timescale $(\gamma/2)^{-1}$, then $$a(t) \propto \exp(i\omega_0t - (\gamma/2)t)\, \ \ \ \ \ t>0. $$ The Fourier transform of this is $$A(\omega) \propto \frac{1}{(\omega - \omega_0) + i\gamma/2}\ ,$$ with $$ A^* A \propto \frac{1}{(\omega -\omega_0)^2 - (\gamma/2)^2}\ .$$

This is a Lorentzian profile with FWHM of $\gamma$. Thus if $\gamma$ is large, the lifetime of the state is short and the probability distribution for $\omega$ (where the energy of the state would be characterized by $\hbar \omega$) would be broader.

If the energy uncertainty is characterized as $\Delta E \simeq \hbar \gamma$ and the lifetime of the state is $1/\gamma$ (in terms of a squared amplitude), then $$\Delta E \Delta t \simeq \hbar $$