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In this question about absorption of continuous energies by discrete atom states, one of the reasons given to explain the width of spectral lines is the uncertainty principle (natural broadening): the decay of an electron from an excited state emits a photon with an uncertain energy (until observed) within a certain range.

In order to respect energy conservation, does that mean that the energy levels of ground and/or excited states in an atom also have uncertainty (and thus, a continuous range of possible energies for each level, instead of a unique energy value)? As the electron can decay between two excited states, not necessarily to the ground state, I would expect that to be at least true for excited states.

However, this page gives energy values for the ground and first excited states of hydrogen electron without any range: I'm not sure if that means that there's no range (and thus, no uncertainty), or if there's simply no need to indicate it (not useful, can be calculated from the theory, ...).

Wikipedia states that both ground and excited states are quantum states, but I don't know if that necessarily implies that all associated "properties" are subjected to uncertainty.

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OxTaz
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2 Answers2

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Yes, excited states have a finite lifetime $\tau$ due to spontaneous emission, and therefore an energy uncertainty (linewidth) given roughly by $\hbar$/$\tau$. That is exactly what natural broadening means. The atomic ground state cannot decay to a lower energy state (it has an infinite lifetime) and so there is no energy uncertainty. Presumably the page you linked quotes values for the hydrogen energies that are derived assuming an atom in isolation, i.e. with no electromagnetic field. If you also include the EM field in your model then it is possible to calculate the lifetimes of the excited states and therefore their natural linewidth.

Mark Mitchison
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  • Oh, I see! I somehow misunderstood the lifetime, thinking it was about the emitted photon, and trying to find a consequence on the electron energy while it's the other way around: the uncertainty is initially already there in the excited electron energy, because its state is unstable, and is then "transfered" to the emitted photon when the electron decays. – OxTaz Jan 06 '15 at 19:33
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In order to characterise the energy of a state precisely you need to monitor it for an infinitely long time. You can show this easily for a free particle described by a plane wave, but while I don't know of an easy explanation it's also true of more complicated states like atomic wavefunctions.

So if you excite an atom from the ground state to an excited state you don't know the energy of either state precisely because neither state lasts an infinitely long time. However the ground state usually lasts a lot longer than the excited state, so it's normally safe to take the ground state as exact and only worry about the uncertainty in the excited state. This means the energy of the absorbed photon will have some variability, as will the energy of the emitted photon when the atom relaxes. However the energy of the absorbed and emitted photons will be the same to a very good approximation.

The uncertainty principle is involved because the uncertainty in the energy $\Delta E$ is related to the lifetime of the excited state $\Delta t$ by the time energy form of the uncertainty principle:

$$ \Delta E \Delta t \ge \frac{\hbar}{2} $$

John Rennie
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  • Thanks for your answer! Sadly, I can only accept one answer :) However, the part about the ground state made me curious: I thought the ground state was the "stable" state, with a theoretical infinite lifetime. Is it wrong? – OxTaz Jan 06 '15 at 20:37
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    @OxTaz: the ground state lifetime can usually be safely treated as infinite. It isn't of course, because as soon as you excite an atom the ground state disappears i.e. its life has ended! When the atom relaxes a new ground state is created and the lifetime counter starts again. – John Rennie Jan 07 '15 at 06:10