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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?

Qmechanic
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kalkanistovinko
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    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 – ACuriousMind Oct 16 '14 at 19:02
  • My OP has no interpretational issues with $\Delta t$. $\Delta E$ is exactly what troubles me. – kalkanistovinko Oct 16 '14 at 19:05
  • No, you are troubled by $\Delta t$, since you claim it is the time between excitation and emission. This is not supported in every view (and, in fact, $\Delta t$ is rather the half-life than the actual lifetime in those views that support this kind of interpretation.). – ACuriousMind Oct 16 '14 at 19:07
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    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. – kalkanistovinko Oct 16 '14 at 19:12
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    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. – CuriousOne Oct 16 '14 at 20:02
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    @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. – an offer can't refuse Oct 19 '14 at 14:51
  • If you realize that the actual Hamiltonian is a time-dependent the whole confusion would evaporate. – A.khalaf Oct 19 '14 at 15:07

1 Answers1

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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.

ProfRob
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