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?

  • 2
    $\begingroup$ 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 '14 at 19:02
  • $\begingroup$ My OP has no interpretational issues with $\Delta t$. $\Delta E$ is exactly what troubles me. $\endgroup$ – kalkanistovinko Oct 16 '14 at 19:05
  • 1
    $\begingroup$ 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. $\endgroup$ – kalkanistovinko Oct 16 '14 at 19:12
  • 1
    $\begingroup$ 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 '14 at 20:02
  • 1
    $\begingroup$ @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. $\endgroup$ – 喵喵是我的猫猫 Oct 19 '14 at 14:51

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? $\endgroup$ – kalkanistovinko Oct 16 '14 at 20:32
  • $\begingroup$ Yes, the Hamiltonian is time-dependent because of the (time-variable) electromagnetic fields. $\endgroup$ – Rob Jeffries Oct 21 '14 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.