Some hydrogen atom exists in some excited quantum state, and after some time $\Delta t$ it's de-excited, emitting a photon carrying the energy difference.

It is claimed that this photon will carry some uncertainty with respect to its energy (and therefore, continuous energy spectrum), attributed to the uncertainty in the difference between the two hydrogen states due to the uncertainty principle.

How true is this? And how does this square with the fact that the energy values of a hydrogen orbitals are eigenvalues of the Hamiltonian, which, in principle, are completely discrete numbers?

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    $\begingroup$ Related: The proper interpretation of the $\Delta t$ appearing in the "time-energy uncertainty relation" is not as straightforward as many would have you believe. See this old question $\endgroup$
    – ACuriousMind
    Oct 16, 2014 at 19:02
  • $\begingroup$ My OP has no interpretational issues with $\Delta t$. $\Delta E$ is exactly what troubles me. $\endgroup$ Oct 16, 2014 at 19:05
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    $\begingroup$ No need to waste time interpreting $\Delta t$, please. Since even if I am wrong about it, correcting me isn't an answer to my question. Reread to find exactly what troubles me, please. $\endgroup$ Oct 16, 2014 at 19:12
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    $\begingroup$ What troubles you about $\Delta E$? It can be observed in a sufficiently high resolution spectrometer as a widening of the spectral lines. The shorter lived a state is, the wider the emission/absorption line. The only reason we don't work that out in Schroedinger QM is because it can't predict line widths. For that one needs quantum field theory or an ad-hoc assumption like Fermi's Golden Rule. $\endgroup$
    – CuriousOne
    Oct 16, 2014 at 20:02
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    $\begingroup$ @CuriousOne I think it will be nice if you can write a more detailed answer about this problem of the "uncertainty of a single spectrum line". I think this is just what OP asks. I'm very curious about this, too. $\endgroup$ Oct 19, 2014 at 14:51

1 Answer 1


I think what you are missing is that these energies are eigenvalues of the time-independent Hamiltonian. i.e. They correspond to stationary states that do not change in time.

The scenario you describe is not time-independent - therefore the difference between the energy levels will carry some uncertainty corresponding to the lifetime of the excited state.

  • $\begingroup$ Fine. But how does the time dependence come into the picture? is it the mere coupling to the electromagnetic field? $\endgroup$ Oct 16, 2014 at 20:32
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    $\begingroup$ Yes, the Hamiltonian is time-dependent because of the (time-variable) electromagnetic fields. $\endgroup$
    – ProfRob
    Oct 21, 2014 at 9:49

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