I've been having trouble understanding why transition lines in unperturbed states are not infinitely narrow. I'll run through my reasoning and hopefully that helps find the flaw. I imagine that if we were to remove an hydrogen atom completely from any external interaction (including vacuum fluctuation), then the possible energy levels for this hydrogen atom are infinitely narrow, or if not infinitely narrow, then extremely narrow with the only broadening due to the perturbation of whatever minor interaction arises between the proton and electron that might jiggle around the fine structure (if there is any?).

Whether or not this is true is my main question, but I feel like it follows directly from the understanding of transitions that I'll describe below. So in that previous example the atom was by definition in steady state, so the solutions to the S.E. can be done in the T.I.S.E and we get discrete (which I interpret as infinitely narrow, right?) possible energy levels. However the second we introduce any kind of energy potential we have to treat it as time dependent. (Although I'm not 100% sure why this is the case, but I was told this and I can't think of any example that would contradict it.)

Anyway once you get into the T.D.S.E then the discreteness goes and you have eigenvectors with a time dependence which I guess oscillate about some central frequency and create that broadening.

OK and last one. If all this is the case and we imagine a perfect lattice of these 2 level systems then I understand the transition lines will broaden homogeneously. Why is that? If they all have exactly the same central frequency and the same breadth about it, I can understand something like density of states increasing and having a super high absorption rate or whatever, but I can't think of any reason why the actual frequency range becomes broader.

Am I on track at all?


1 Answer 1


If you had an electron in a Coloumb potential, completely isolated from everything else, then indeed the atom would be in a stationary state and there would be no transitions.

In fact, since, the lifetime of a state is inversely proportional to the width of the transition line, so the fact that the electron is in a stationary state and cannot decay also relates to your statement that the line is infinitely narrow. This is an expression of the energy-time uncertainty principle, $\Delta E \Delta t \gtrsim \hbar$. In a stationary state you know the energy infinitely precisely, so the state cannot be localized in time.


A hydrogen atom is certainly not an electron in a Coloumb potential, completely isolated from everything. For one thing, as you said, there is fine structure, so the energy eigenstates of the Coloumb problem are not the true energy eigenstates of an isolated hydrogen atom. However, in principle even with fine structure one could imagine putting the hydrogen atom into an energy eigenstate, even though it is very difficult to compute those energy eigenstates exactly.

The real problem is that the hydrogen atom also interacts with the electromagnetic field, which is quantized. As a result, the hydrogen atom can never truly be thought of in isolation and in particular we can never truly put it into an energy eigenstate besides its ground state. It is always possible for an excited state of hydrogen to emit a photon and decay into a ground state. As a result, the hydrogen atom by itself is not an energy eigenstate of the whole world, so there is always a finite width.

  • $\begingroup$ Fantastic! Thank you, this is precisely the answer I wanted! $\endgroup$
    – twoodeep
    Commented Oct 27, 2015 at 3:40

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