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Is the energy difference between two energy levels unique for that particular pair of levels for a hydrogen atom ? If so how can one prove it?

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Good question, I came across this in past but never found an answer. Perhaps it would help others to understand the question if you formulated it as a problem from the theory of natural numbers. –  Ján Lalinský Jan 7 '14 at 23:37
If I understand right: I believe you are asking whether, using the Dirac formula for the energy levels $E_{n\,j} = \mu c^2\left(1+\left[\dfrac{Z\alpha}{n-|k|+\sqrt{k^2-Z^2\alpha^2}}\right]^2\right)^{‌​-1/2}$ and then proving $E_{n\,j}$ is unique for each pair$(n,j)\in \mathbb{Z}_0^+\times \mathbb{Z}_0^+$? If so, you might try Maths SE. This is a harder problem than it looks. –  WetSavannaAnimal aka Rod Vance Jan 8 '14 at 0:23
Why bother with the Dirac formula, it is messy and contains $\alpha$ which is not known to be simple number. More interesting is the Schroedinger non-relativistic case : for given $\Delta E$, find all $n,m$ such that $ \Delta E = \frac{1}{n^2}- \frac{1}{m^2}$. –  Ján Lalinský Jan 8 '14 at 0:33

1 Answer 1

As I am sure you are aware, the electrons in free atoms are in quantized energy states which means that they can only be found on a set of discrete energies. These states can be referred to as energy levels. Since these states are discrete, the difference between any two states must also have a finite value.

To answer your question as to how one could prove this, there is a simple method called atomic emission spectroscopy. You can take a tube of hydrogen gas and excite it through the use of a high voltage transformer. This addition of energy will cause the electrons to move up to higher energy states, and eventually they 'fall' back down to a lower energy level. Now, where dose that energy go? It is emitted as electromagnetic radiation. You can use a detector such as a photomultiplier to obtain rather beautiful data on the emitted spectrum from this tube. An example of such is here: Hydrogen Emission Spectrum

You can see that there are certain distinct lines of emission. These correspond directly to the difference in energy between the differing states. If there were not unique energy levels, the graph would have no distinct peak of intensity. Thus, it can be concluded that the energy states of an hydrogen atom are quantized.

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Richard, nice picture - what element is it? However, it does not answer the OQ - whether the $\omega$ of a transition corresponds to a unique pair of Hamiltonian eigenfunctions (perhaps there are more). –  Ján Lalinský Jan 7 '14 at 23:51
Otherwise, your post assumes that the atom can be only in discrete states with definite energy. This was working assumption years around 1912 and is still taught today (Bohr model, buck-shot model of light), but is not consistent with diff. equations and is not necessary to explain the spectrum today - the Schroedinger equation explains the spectral function with continuously changing $\psi$ functions, most of which do not imply that atom has one of the discrete set of energies. –  Ján Lalinský Jan 7 '14 at 23:52
Thanks for the comment, I misunderstood the OQ. I have yet to reach QM in my studies, and thought this was a question I could tackle, as I find explaining to others is one of the best ways to reinforce the concepts and ideas in my head. –  Richard P Jan 8 '14 at 0:06

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