# What is the meaning of natural line broadening?

I have recently found one exercise in an exercise book:

During the transition from the first excited state of a hydrogen atom into the ground state, photons with a wavelength of 121.5 nm are emitted. The lifetime of the excited state is 10 ns. Calculate the natural line broadening of the excited state using the energy uncertainty.

Could someone please explain to me briefly what the 'natural line broadening of the excited state' is?

• Please take a minute to read our guidelines for homework and exercise questions as well as check-my-work questions. We intend our questions to be potentially useful to a broader set of users than just the one asking, and we prefer conceptual questions over those just asking for a specific computation. – Emilio Pisanty Nov 24 '18 at 21:52
• I did not ask for a solution to this exercise. Instead, I only asked for a specific concept, because the explanations on the Internet made me very confused. I mentioned the exercise in my question for a contextual purpose only. – Caitlyn Nov 24 '18 at 21:56
• I'm not saying you did - your question is within the guidelines. But you should still read the guidelines. – Emilio Pisanty Nov 25 '18 at 8:35

Despite the convenient fiction that transitions between quantum states can only happen if the photon energy exactly equals the energy difference between the two states, $$\hbar \omega = E_2-E_1,$$ if you take that literally then that would imply that even if the photon energy were only off by one part in $$10^{10^{10}}$$, since that's no longer exactly equal to $$E_2-E_1$$, then the transition wouldn't take place $$-$$ but that, of course, is obviously plain nonsense. There's no room in experimental physics for mathematically exact equalities between real numbers, and all physical quantities, including spectral line positions, come with an associated width.

For spectral lines, though, there are two main contributions to that width:

• It can happen, for example, that your sample has a bunch of different atoms, each under similar but slightly different conditions (say, different electric or magnetic fields, or moving at a different thermal velocity with respect to the observer, what have you), which causes all the different atoms to have transitions at slightly different frequencies. This is known as inhomogeneous broadening of the line.
• There is also, however, something called the homogeneous broadening of the line, and this affects the emission of even a single atom. This is caused by the fact that the emission starts at a given time (it hasn't been going on since $$t=-\infty$$) and it's decaying exponentially so at some point it will run out of excited-state population to decay (at a point rather short of $$t=\infty$$), which means that the emission isn't strictly monochromatic, and a Fourier transform of the single-atom emission will find that it has a nonzero width.

You're being tasked with estimating the linewidth caused by the homogeneous broadening, based on the uncertainty-principle requirements caused by the lifetime given in the problem.

In quantum mechanics, electrons can jump from one state to another with radiation absorption/emission. However, the frequency of the radiation doesn't have to be exactly $$\frac{E_2-E_1}{h}$$. One can use a slightly different frequency and still get the energy transition. This means that if you look at the spectrum of light emitted/absorbed from an atom, you son't see a sharp delta function, but a smooth curve (line) that looks like a Lorentzian. This means the absorption line is broader than the case of just a single frequency getting absorbed. This is where the term "line broadening" comes from - it is the phenomenon of an absorption curve getting broader (meaning more and more frequencies are able to get absorbed by an electron or emitted by one). There are a couple of reasons for the absorption line to get broader, such as pressure, Doppler effect, and more - but as said earlier, even a single atom in vacuum has a "natural broadening".

This is due to the finite time an electron spends (on average) in the excited state. This transition is because that the hamiltonian of an atom in a vacuum is perturbed by the electromagnetic field in the vacuum. In perturbation theory, one can see that the more time it takes for a perturbation to induce a transition, the more precise its frequency needs to be, meaning $$\Delta \omega \Delta t \geq 1$$.

To sum up, you should calculate the $$\Delta \omega$$ of the absorption line using this uncertainty principle, but I hope the explanation helped (sorry for English mistakes)

• Please don't post complete answers to homework questions. – Emilio Pisanty Nov 24 '18 at 21:52

Usually when performing calculations, we regard the energy levels of the system as being discrete. But in fact, excited states have a certain probability of decay via the emission of photons, and therefore have a finite lifetime. This implies that the levels become quasidiscrete, with a small but finite width; they can be written in the form $$E-\tfrac{1}{2}i\Gamma$$ where $$\Gamma$$ is the total probability of all possible decay channels (this fact is alternatively stated as the Optical Theorem). Often the width that develops is rather small compared to the gap between the discrete levels, so we can still measure sharp transitions. Someone else comments we shouldn't provide complete answers to homework questions, so instead I'm just going to elaborate on the concept and what it means since that's what you requested. For instance, another answer mentions the energy levels broaden into Lorentzians - it's not too hard to derive this fact.

Let's start from scratch, with the Schrodinger equation $$$$\label{sequation} i\frac{\partial \Psi}{\partial t}=\left(\hat{H}^{(0)}+\hat{V} \right)\Psi$$$$ and expand the solution in terms of the wave functions of the unperturbed states of the system $$$$\label{tdpt} \Psi=\sum_\nu a_\nu (t) \Psi^{(0)}_\nu=\sum_\nu a_\nu (t) e^{-iE_\nu} \psi^{(0)}_\nu$$$$ Inserting this expression either side of the Schrodinger equation and taking inner products with the state $$\nu$$ gives $$$$\label{tdpt2} i\frac{\partial a_\nu}{\partial t}=\sum_\nu\langle\nu|V|\nu '\rangle a_{\nu '} (t) \ e^{i(E_\nu-E_\nu')}$$$$ The usual procedure for time dependent perturbation theory goes as follows. We assume the system is in some initial state with probability unity $$a_1=1$$ and $$a_\nu=0$$ for $$\nu\neq 0$$. Then to leading order we integrate both sides to find $$a_\nu$$ by only keeping terms on the left where $$\nu'=1$$ ie only keeping terms to leading order in $$V$$. The procedure for calculating $$a_\nu$$ more accurately proceeds by iteration. We are particularly interested in the long-time probability of decay $$$$dw=|a_{\omega, 2}(\infty)|^2 \ d\omega$$$$ where $$\nu=2$$ is some excited state and $$\omega$$ denotes the emission of a photon. In the usual case of time-dependent perturbation theory, we are effectively assuming our results apply to time longer than the inverse level spacing but short compared to the decay lifetime of the energy levels. Now let's relax that assumption - when we reach times comparable to the decay life of the state $$1$$ then $$a_1=\exp{\left( -\tfrac{1}{2}\Gamma_1 t \right)}$$ and the equation for $$a_{\omega, 2}$$ becomes $$$$\label{modtdpt} i\frac{d a_{\omega, 2}}{dt}=\langle \omega, 2| V |1\rangle e^{i(\omega-\omega_{12})t-\frac{1}{2}\Gamma_1 t}$$$$ Integrating otherwise as usual, and substituting into the formula for the long-time decay probability gives $$$$\label{bwdist} dw=|\langle\omega, 2| V|1\rangle|^2 \frac{1}{(\omega-\omega_{12})^2+\frac{1}{4}\Gamma_1^2} \ d\omega$$$$ If we assume that the width is small, then this expression is dominated by the frequency range $$\omega\approx \omega_{12}$$, and we recover the usual Fermi Golden Rule.

Now we define the expression $$$$\Gamma_{1\rightarrow 2}=2\pi \sum_{pol,\boldsymbol{k}}|\langle \omega 2| V|1\rangle|^2$$$$ which gives the total probability of emission, after summing over the polarisations and directions of motion of the photon. Then summing our expression for the long-time decay rate similarly over the polarisations and momenta we arrive at the total frequency distribution for emitted radiation $$$$dw=\frac{\Gamma_{1\rightarrow 2}}{2\pi}\frac{1}{(\omega-\omega_{12})^2+\frac{1}{4}\Gamma_1^2} d\omega$$$$ This broadening of the spectral line occurs for an isolated atom at rest, as distinct from broadening caused by the interaction of atoms with other atoms ($$\textit{collision broadening}$$) or by the presence of atoms in the source which move with various velocities ($$\textit{Doppler broadening}$$). It is called the $$\textit{natural shape}$$, and it's clear that it's a Lorentzian peak as claimed by the other answers. (Also: we could continue to add refinements to this calculation by taking into account the lifetime of the level $$2$$).