If atomic energy levels are theoretically discrete, what mechanisms lead to the broadening of spectral lines?

Question

From what I understand, one key contributor to spectral line broadening is the finite lifetime of excited states, which results in natural linewidth broadening due to the uncertainty principle. I have issue with this approach because it doesn't directly answer the problem at hand but diverts it. Since these decaying states acquire an imaginary component, their energies are no longer purely real. This raises a question: are such decaying states truly part of the Hilbert space? Or, because they are non-Hermitian, do they exist only as approximations, influenced by their coupling to the environment?

Additionally, I’ve read that an excited atom can interact with virtual photons, transitioning briefly to a virtual state before returning to the ground state. This interaction also broadens spectral lines by perturbing the energy levels dynamically, effectively reducing the state’s lifetime. Are such transient processes another way of blurring the distinction between discrete and continuum states?

However, I think I might be misunderstanding something fundamental. If the state of the atom is an energy eigenstate, it should be stable, and there would be no broadening of the energy level. But if the state is not an energy eigenstate, then I have no reason to expect its energy to be strictly discrete. Instead, it should be allowed to have a distribution centered around the old discrete energy value.

How do these factors—finite lifetime, virtual photon interactions, and the nature of eigenstates versus non-eigenstates—collectively explain the transition from seemingly discrete energy levels to the observed spectral line broadening? I am trying to understand the problem from the perspective of the energy levels. I’m thinking that in the presence of a perturbing external time-dependent potential, the energy eigenstates are not exactly discrete but form a distribution centered around the unperturbed Coulomb Hamiltonian. I would like further clarification on the subject.

Answer 1

Approximations.

When we solve the Coulomb hydrogen atom, for instance, we get discrete energy levels, $E_{n(lm)}$, that only depend on principle quantum number $n$ (because symmetries make $l$, $m$ degenerate).

Note that eigenstates are stationary states, that is:

$$\psi_{nlm}(\vec r, t) = \psi_{nlm}(\vec r)e^{iE_n/\hbar t} $$

and $|\psi_{nlm}|^2$ is constant.

With that, there are no transitions. An excited state remain so, forever.

This is obviously not what happens.

There are corrections to $\psi_{nlm}$ that fall under "fine structure" and "hyper-fine structure"...but we don't need to consider those for this question.

What affect us here is that those solutions are for:

$$ V(\vec r) = \frac 1 {4\pi\epsilon} \frac{Ze} r $$

which is a static electric field. IRL, the EM field is a dynamic part of the system, and a full treatment requires QED (quantum electrodynamics), where Schrödinger equations and stationary states aren't a thing.

Luckily, the fine-structure constant:

$$ \alpha \approx \frac 1{137} << 1 $$

is small, so we can retain $psi_{nlm}$ as approximate descriptions of atomic states, and then use Fermi's Golden Rule to calculate transition rates between various states.

In first order perturbation theory, we get an interaction operator, e.g, a dipole operator:

$$ \vec d \propto {\bf z} $$

which then links orthogonal states (here with the ground state):

$$ \langle 100|{\bf z}|nlm\rangle \propto {\mathfrak M}_{fi} $$

Fermi's Golden Rule (https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) then gives the initial state a lifetime, $\tau$, and we can now consider a correction to the excited state energy as:

$$ E_n \rightarrow E_n + i\frac{\hbar}{\tau} $$ $$ E_n \rightarrow E_n + i\Gamma $$

where $\Gamma$ is the width.

(Note this can be contrasted with the quark wave functions of $x$ in a proton: there are none. The color field is so strongly coupled that we don't have approximate wave functions for quarks, we just have parton distribution functions, which are functions $x$, but it's Feynman-$x$, which has nothing to do with position).

Answer 2

From what I understand, one key contributor to spectral line broadening is the finite lifetime of excited states, which results in natural linewidth broadening due to the uncertainty principle.

The evolution of $\psi$ during de-excitation towards the lower state is enough. There is no need to additionally refer to some time-energy uncertainty principle, that is a rather confusing element that has to be carefully formulated first, and when that is done, it turns out it's about the fact that superposition of many close but orthogonal energy eigenfunctions is behaving as exponential decay with rate constant proportional to width of the eigenvalue distribution. Real decay of excited state with sharp energy is not due to such close eigenvalues of the atom, but due to interaction with EM field.

Since due to interaction with EM field, psi function of the atom evolves towards lower state, electric current oscillation component at the resonance frequency $\omega_{21} = \frac{E_2-E_1}{\hbar}$ changes intensity and asymptotically approaches zero. Just as in classical theory, such oscillation produces oscillatory change in the state of EM field of similarly decreasing strength in time. From the Fourier theory, we known such oscillation in the field, with fixed frequency but with decreasing amplitude, actually has Fourier transform that is a peak at $\omega_{12}$ with non-zero width, proportional to rate constant at which the amplitude decays. This is a mathematical fact which applies both to classical and quantum theory of radiation.

Since these decaying states acquire an imaginary component, their energies are no longer purely real.

That is just simplified effective mathematical description of the decay of the amplitude, it's cheating. Actually, the eigenvalues $E_n$ of isolated atom Hamiltonian $H_0$ are real numbers, and the decay of the excited state component of psi is due to interaction with EM field, not due to $\hat{H}_0$ having eigenvalues with non-zero imaginary component. But mathematically, such decay can be approximately described using a factor $e^{-iEt/\hbar}$ in psi function, where $E$ has imaginary component. It's an effective description, not an explanation. Different more detailed descriptions show the decay of excited state isn't actually exponential in time, but inverse power of time.

are such decaying states truly part of the Hilbert space? Or, because they are non-Hermitian, do they exist only as approximations, influenced by their coupling to the environment?

Yes, states decaying in time can be in the Hilbert space. States are not "non-Hermitian", only operators can be non-Hermitian. Maybe what you mean is the effective time evolution is not purely due to time-independent atom Hamiltonian (as such Hamiltonian can't de-excite the state). But it can be due to combined atom+EM field Hamiltonian.

Additionally, I’ve read that an excited atom can interact with virtual photons, transitioning briefly to a virtual state before returning to the ground state. This interaction also broadens spectral lines by perturbing the energy levels dynamically, effectively reducing the state’s lifetime. Are such transient processes another way of blurring the distinction between discrete and continuum states?

This seem to be misleading words, some people like to use similar ones to describe state transitions, but they are not implied by mathematics. Mathematically, for spontaneous emission of single isolated atom, there is no such thing as broadened eigenvalue of Hamiltonian, there are only many different sharp eigenvalues. Sometimes they can be very close to some accumulation point or region, but they are still sharp. IF the Hamiltonian depends on some parameter of environment, time dependence of the parameter will cause time dependence of the eigenvalues, but this another topic, stochastic models. Even isolated system with fixed Hamiltonian and fixed discrete spectrum will produce radiation of non-zero width (natural line width). This aspect of mathematics is the same as in classical theory: an oscillator that has sharp constant natural frequency $\omega_{12}$ (sharp levels), when its oscillation is damped by some effect (coupling to EM field) produces radiation whose spectrum is not delta sharp, but has non-zero width proportional to damping constant.

However, I think I might be misunderstanding something fundamental. If the state of the atom is an energy eigenstate, it should be stable, and there would be no broadening of the energy level.

Correct.

But if the state is not an energy eigenstate, then I have no reason to expect its energy to be strictly discrete. Instead, it should be allowed to have a distribution centered around the old discrete energy value.

A quantum state that is not a Hamiltonian eigenstate does not have single energy. One can assign to it probability distribution of energies, but this probability distribution is immaterial to de-excitation of single atom, since we work directly with the quantum state.

How do these factors—finite lifetime, virtual photon interactions, and the nature of eigenstates versus non-eigenstates—collectively explain the transition from seemingly discrete energy levels to the observed spectral line broadening? I am trying to understand the problem from the perspective of the energy levels. I’m thinking that in the presence of a perturbing external time-dependent potential, the energy eigenstates are not exactly discrete but form a distribution centered around the unperturbed Coulomb Hamiltonian.

No, all those except finite lifetime are red-herrings. Again, the natural line broadening is due to decay of excited state component due to interaction with EM field, and only effective approximate description of this in terms of psi function of the atom introduces stuff such as modified levels, imaginary eigenvalues, etc.

Answer 3

If atomic energy levels are theoretically discrete, what mechanisms lead to the broadening of spectral lines?

The discrete energy levels arise only under specific conditions, notably:

• We are solving for an isolated atom (atom alone in the Universe - no other particles and no EM field, even vacuum EM field.)
• We are solving for the atom in its center-of-mass system (the transformation that is typically done in the very beginning, and quickly forgotten - yet, all the positions are typically calculated in respect to this COM, which nearly coincides with the nucleus, due to the nucleus being much heavier than electrons.)

Now, if we lift the second of this conditions, we immediately get Doppler broadening, so successfully used in laser cooling. The spectra of the atoms atoms moving to us shifts to higher wavelength, and the spectra of the atoms moving away shifts towards longer wavelengths. Take the Maxwell-Boltzmann distribution of velocities, and we get a continuously broadened line (inhomogeneous broadening - often approximated by the Gaussian shape.)

If we now take into account that the atom is coupled to electromagnetic field, then it always can decay to a lower energy state via emission of photons. Describing it as a complex energy component is indeed an approximation - for an exact solution we would need to solve for joint eigenstates of an atom coupled to the electromagnetic field (having an infinite number of modes) - see, e.g., the Hamiltonian in this thread. The resulting solutions will not be atomic levels, but polaritons - atomic states dressed by a photon cloud, with continuous spectra centered on the atomic level. Natural linewidth is an estimate of the width of these polaritonic bands.