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According to Quantum Mechanics, in order for an atom to absorb a photon the energy of the photon must be precisely that of a "jump" between energy states of the atom.

How precise must it be?

If I create a photon with an energy within an error of 0.0001% of that of an energy state, will it be absorbed by my atom?

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    $\begingroup$ That depends on the width of the spectral line, which depends on the strength of the coupling between the atomic states and the vacuum (natural line width), radiation environment (thermal radiation leads to stimulated emission), collision broadening and thermal Doppler broadening due to the movement of the atoms. Natural line widths are usually very small and only observable in cold atomic gases in very low pressure vacuum, otherwise the other effects will dominate. $\endgroup$
    – CuriousOne
    Commented Feb 23, 2016 at 12:36
  • $\begingroup$ So, translating these effects of the Lorentz atom model to the "energy state"-idea, would mean that an energy state isn't precise? That it is some distribution around each level? $\endgroup$
    – Tobias Tovedal
    Commented Feb 23, 2016 at 12:39
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    $\begingroup$ It means that the atom isn't the only thing there is. It is always coupled to the environment. When we are solving the Schroedinger equation, that environment is being described by a classical potential, which can be used to model some of these effects (e.g. a classical electromagnetic wave by adding a time dependent electric field). This is useful to describe stimulated emission and the broadening due to a thermal radiation background. If we want to go beyond that, we may have to use quantum electrodynamics and/or average over a thermal velocity distribution or inter-atomic interactions. $\endgroup$
    – CuriousOne
    Commented Feb 23, 2016 at 13:01
  • $\begingroup$ Well, even in the conceptually perfect case it will depend on the collision of the photon with the electron. If 2eV is required for the transition, a photon carrying $E > 2eV$ of energy could induce the transition and a scattered photon carrying $E-2eV$ would depart in the elastic case. $\endgroup$
    – Greg Petersen
    Commented Feb 23, 2016 at 14:49
  • $\begingroup$ @CuriousOne Someone ought to write down the example of a damped harmonic oscillator absorbing "photons" to illustrate how the $Q$ of the oscillator determines the line width and its ability to absorb slightly off-resonant energy. This is an exactly solvable problem which illustrates all the important features. $\endgroup$
    – DanielSank
    Commented Feb 23, 2016 at 17:52

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In atoms the energy levels do not have a precise energy. When you solve Schrodinger's equation for an atom the results are the energy eigenfunctions. However these are functions that are time independent, and they have an exact energy only because they are time independent.

At the risk of oversimplifying, you can regard this as an example of the energy time form of the Heisenberg uncertainty principle:

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

If $\Delta t$ is the lifetime of a state then $\Delta E$ is the uncertainty in the energy of that state. For the energy eigenfunctions $\Delta t = \infty$ so $\Delta E = 0$ and the energy is precisely defined.

The point of all this is that in an atom an excited state has a finite lifetime and therefore it has a finite energy uncertainty, and this produces an effect called lifetime broadening. This means transitions to and from the state can occur for photons with a range of energies. The range of energies allowed depends on the energy uncertainty of the state, which in turn depends on its lifetime.

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  • $\begingroup$ Well the solutions are time-dependent, it's just that the dependence doesn't matter much since it doesn't change relative phases nor magnitudes of eigenfunction values. $\endgroup$
    – Ruslan
    Commented Feb 23, 2016 at 20:34
  • $\begingroup$ Maybe you could explain how the higher energy state lasts long enough to have a high certainty in energy because the fine structure constant is so small. $\endgroup$
    – Timothy
    Commented Aug 9, 2018 at 23:15
  • $\begingroup$ @John Rennie When we solve the time-dependent Schrödinger equation we get time dependent solutions of the form $\psi(x)\cdot e^{-iE_n t / \hbar} $. Energy is still time independent (we can show it by calculating its expectation value). What then causes the energy of the wavefunction (its expectation value) to be time dependent? A time-dependent Hamiltonian, a transition between electronic states? $\endgroup$
    – Antonios Sarikas
    Commented Sep 1, 2022 at 11:43
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Agree with the above, but also if the atom, or collection of atoms, are in thermal equilibrium, then there is another broadening mechanism, besides lifetime broadening, called Doppler broadening that accounts for the motion of the atom(s). This has the effect to substantially widen the effective line width depending upon the temperature.

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The linewidth broadening everyone is talking about is actually a very classical effect that comes straight from antenna theory and depends only on the size of the antenna as compared to the wavelength of light. It is well known in classical antenna theory that the bandwidth of a lossless short antenna goes as the cube of the electrical length (the physical length divided by the wavelength). For the s-p transition of the hydrogen atom, this parameter is close to the fine structure constant, 1/137. The cube of this number gives the (dimensionless) bandwidth of around 10^7.

Since the frequency of the transition is around 10^16, this gives a transition time of around 10^-9 seconds. I think this is about right for the hydrogen atom. You just treat the atom as a classical antenna and everything comes out.

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    $\begingroup$ Even if treating the atom as a classical antenna gives the right answer, it's still horribly wrong. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 23, 2016 at 17:44
  • $\begingroup$ So you're agreeing that I get the right anwr. $\endgroup$
    – Marty Green
    Commented Feb 23, 2016 at 17:44
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    $\begingroup$ No, because I have not done the calculation. Getting the right answer doesn't make the derivation correct - you can also compute the Schwarzschild radius of a black hole by an essentially Newtonian computation, but that doesn't make applying Newtonian mechanics to a black hole make any more sense. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 23, 2016 at 17:46

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