As I understand it, Schrodinger's wave equation predicts the allowable energy states an electron can have under the electromagnetic forces of a given nucleus (and I assume other 'orbital' electrons). If I understand correctly the change of electron state therefore has a 'quantum' difference of energy that is released (if the electron changes to a lower energy state). Schrodinger's equation simply gives the quantity of the energy difference.

What causes this expressed energy to become/be electromagnetic? Has the electron or its charge been 'jerked' in space in some oscillatory fashion (congruent with Heisenberg's uncertainty principle) by the energy loss, as is the method to create lower frequency electromagnetic emissions in an antenna? Has the electron spin fluctuated to initiate an electromagnetic emission? Is the 'orbital energy state' of the electron somehow a non-propagating (stationary) electromagnetic energy that lets some of its energy 'loose' when the electron changes state to another non-propagating electromagnetic energy state?

What is causing the conversion of the energy to be electromagnetic (a photon)?

To put it in a nutshell, when an electron changes energy states, why does it emit energy in the form of a photon as opposed to anything else?

Thank you for your time to respond and my apologies in advance if there is a commonly understood answer to this ... :-|

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    $\begingroup$ Electrons and the nucleus are charged, which means that they interact strongly trough the electromagnetic force. This has nothing to do with the Schroedinger equation, which can't describe the transitions in a quantum mechanical framework to begin with (it can only describe a classical EM field). For that we need a theory that is a) relativistic and b) contains the proper description of the massless EM field. That's quantum electrodynamics. $\endgroup$ – CuriousOne Jul 24 '16 at 11:28
  • $\begingroup$ Photons can also be emitted from electrons without the involvement of transitions between orbital energy levels physics.stackexchange.com/q/20635 $\endgroup$ – user108787 Jul 24 '16 at 11:40
  • $\begingroup$ The photon carries no electromagnetic charge, so the electron can "afford" to lose energy in this way, whilst still remaining an electron. $\endgroup$ – user108787 Jul 24 '16 at 14:49

The atomic model developed starting from the light spectrum emitted by the hydrogen atom. It was known that hydrogen was one proton and one electron.

The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885.


The visible hydrogen emission spectrum lines in the Balmer series. H-alpha is the red line at the right. The two leftmost lines are considered to be ultraviolet as they have wavelengths less than 400 nm.

Classical physics would predict a continuum emission of light for an electron falling on a proton. The Bohr atom was the first attempt at a quantization model giving the Balmer etc seires and the Schrodinger equation was the final theory, giving the series, using the electric potential, and stable orbitals.

You are putting the carriage in front of the horse when you start by the Schrodinger equation. The theory is driven by the data and not vice verso.

It is not a matter of "why there is radiation", but a matter of "why there is quantized radiation and not a continuum"

What is causing the conversion of the energy to be electromagnetic?

The problem is electromagnetic, both classically and per force quantum mechanically.

  • $\begingroup$ I think the OP struggles specifically with why the Schroedinger equation selects the phenomenon as electromagnetic, which it doesn't, of course. What makes it electromagnetic in the Schroedinger equations is if we plug in an explicit classical, time dependent electric potential or magnetic vector potential, which would give a semi-classical theory of atomic transitions that is quite OK for many purposes, I believe. Maybe you can also mention that one can plug in a gravitational potential, e.g. for a neutron fountain and get correct results, just as a counterexample? $\endgroup$ – CuriousOne Jul 24 '16 at 12:12
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    $\begingroup$ @CuriousOne it is an ###-bacward question. The classical electromagentic expectations came first, and then a model was necessary to explain the observations which ended up being the hydrogen model of the schrodinger equations, is what I am saying. The S equation is not a GOD, it is a tool discovered to explain data and predict new situations. $\endgroup$ – anna v Jul 24 '16 at 13:50

Interesting question.

Electrons have electric charge and are therefore a source of electric field. An atom can be thought as a series of energy levels in which electrons can be in. When the electron transit between these levels, the energy will be emitted as electromagnetic radiation, since the fundamental interaction involved here is the electromagnetic interaction (we are dealing with EM charges) .

This is not because the electromagnetic interaction is 'special'. For example, if you have a massive object falling, such that it is losing gravitational potential energy, you are meant to observe a similar phenomenon (i.e. gravitational radiation emitted). It is just that these radiation cannot be observed using the apparatus we currently have.


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