In the various presentations I've seen so far in atomic physics of series such as the Balmer series, the wavelength of each spectral line is definite - but in QM, free particles have no definite energy if I understand correctly: none of the photons in a beam of photons that would interact with the electron of an hydrogen atom would carry an energy corresponding exactly to a wavelength of 656.3 nm for example (I'm saying "none of the photons" because if we have a distribution of energies for the photons in the beam, in the limit, there is no particle at a single discrete value of energy). What happens exactly? Should Heisenberg uncertainty be taken into account? Is there a quantum mechanical account of photon absorption by atoms?
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1$\begingroup$ Related/perhaps duplicates: physics.stackexchange.com/q/142373/50583, physics.stackexchange.com/q/11147/50583, physics.stackexchange.com/q/141685/50583 The answer essentially is: The atomic energy levels are not energy eigenstates anymore as soon as the atom "couples to the EM field" (i.e. you write the EM-matter interaction term into the Hamiltonian), so their energies get smeared out a bit, yielding the (tiny) natural line width of atomic transitions rather than sharp peaks. $\endgroup$– ACuriousMind ♦Commented Mar 5, 2015 at 22:53
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$\begingroup$ ACuriousMind has it right. Note that if they really were energy eigenstates, they wouldn't ever decay by spontaneous emission, since they'd be constant in time. $\endgroup$– zeldredgeCommented Mar 5, 2015 at 22:54
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$\begingroup$ I love it - ACuriousMind: you can see how this is a continuation from my line of questioning yesterday about energy eigenstates of free particles and Hamiltonian spectra... I haven't looked at the links yet, but can you tell me what the EM-matter interaction term looks like? Also, just adding that term to the Hamiltonian and using Schroedinger's equation is enough, or do we need to be in QED or QFT territory (which I don't know much about)? $\endgroup$– FrankCommented Mar 5, 2015 at 22:58
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$\begingroup$ You can do it without QFT. As for the term you add, it depends. If you want the interaction of the atom with an external light source that looks like $gE(| 1 \rangle \langle 2 | + g^* E^* | 2 \rangle \langle 1 |$, where g is some coupling constant and E is the electric field. This is what gets you "Rabi flopping" when the atom is illuminated by the laser. It's mildly more complicated if you want to look at spontaneous emission because you have all these continuum states; the key term to Google is "Wigner-Weisskopf" $\endgroup$– zeldredgeCommented Mar 5, 2015 at 23:07
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$\begingroup$ Thanks a ton - I'm going to read up following the links you mention. $|1\rangle$ and $\langle2|$ are spin states of the electron right? Is Dirac's equation required? Or would it yield more precise results than Schroedinger's equation? $\endgroup$– FrankCommented Mar 5, 2015 at 23:41
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Of course the photon will have an amount of energy and that is radiation energy which if defined by Planck's equation. Radiation energy is also related directly to wavelength,frequency and wavelength number.The higher the energy, the less is the wavelength ( all this from Planck's equation). What Heisenberg equation is about, does not include the energy.It includes only the position and the rate of electrons ( you can't know both of them at the same time ). The absorbtion of radiaton by atoms is very well defined by absorbance formula A=abc.I hope I understood the question correct.