Hydrogen atom - complete orthornormal set

In summary: I tried to check his 50 pages long <readable account> of the H-atom and I was dissapointed. Formal mathematics with an abuse of bracket notation, no functional analysis, 90% of his statements unproven, Well, as the title says this is a book for physicists, not for mathematicians. So he takes all the liberties a physicist needs to get to the results quickly. Mathematicians can add the required rigor themselves.For a more rigorous treatment, try the book by Barut and...
  • #1
QuantumCosmo
29
0
Hi,
I was wondering if the bound solutions to the radial part of the hydrogen atom form a complete set for the functions in L^2(0,\infty). I know that the laguerre polynomials are complete and that they only differ from the radial solutions by factors of x^l * exp, so I thought that they would have to be complete too (apparently, it is not so simple as to say: "well, you are missing the scattering states, so of course they are not complete").
But I heard that it followed from Sturm-Liouville-Theory that they are not complete, which doesent seem to make any sense to me.
Does anyone have an idea?
Thank you,
QuantumCosmo
 
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  • #2
As far as I know, the bound states of the H-atom form a complete orthonormal set in a Hilbert space containing a sum of spaces one of them being [itex] L^{2} \left(\left(0,\infty\right), \, r^2 dr\right) [/itex], the other components of the sum being related to the spherical angles.
 
  • #3
So that would mean the solutions to the hydrogen atom do form a complete set. Its what I thought but somehow not what my professor seems to think...
Thanks for the answer :)
 
  • #4
Can't you make a wave packet out of scattering eigenstates to get a perfectly normalizable state that has no projection onto any bound state? That would imply that the bound states do not form a complete basis, yes?
 
  • #5
You have to include the positive energy (scattering) states.
 
  • #6
The_Duck said:
Can't you make a wave packet out of scattering eigenstates to get a perfectly normalizable state that has no projection onto any bound state? That would imply that the bound states do not form a complete basis, yes?

dextercioby said:
As far as I know, the bound states of the H-atom form a complete orthonormal set in a Hilbert space containing a sum of spaces one of them being [itex] L^{2} \left(\left(0,\infty\right), \, r^2 dr\right) [/itex], the other components of the sum being related to the spherical angles.

I based my reply above on the purely Hilbert space analysis of the H-atom (Gerald Teschl's notes) which is obviously incomplete, due to its failure to account for the scattering states (<=>continuous spectrum of the Hamiltonian). Indeed, if one goes to an extension of L^2(R^3), called rigged Hilbert space (RHS), the bound states no longer form a complete set, because, as you say, one can form wavepackets out of scattering states which can be orthogonal to the bound states. One, of course, has to add the generalized eigenvectors corresponding to the continuous values of the energy, which is basically what the RHS achieves.
 
  • #7
dextercioby said:
I based my reply above on the purely Hilbert space analysis of the H-atom (Gerald Teschl's notes) which is obviously incomplete, due to its failure to account for the scattering states (<=>continuous spectrum of the Hamiltonian). Indeed, if one goes to an extension of L^2(R^3), called rigged Hilbert space (RHS), the bound states no longer form a complete set, because, as you say, one can form wavepackets out of scattering states which can be orthogonal to the bound states. One, of course, has to add the generalized eigenvectors corresponding to the continuous values of the energy, which is basically what the RHS achieves.

Even in the Hilbert space, the bound states do not form a complete set, as you can see by looking at the spectral resolution. The space spanned by the bound states contains not a single scattering state. (The full spectrum can be seen by means of the dynamical group SO(2,4) rather than O(4), which only gives the discrete spectrum. Wybourne's book ''Classical groups for physicists'' gives a readable account.
 
  • #8
A. Neumaier said:
(The full spectrum can be seen by means of the dynamical group SO(2,4) rather than O(4), which only gives the discrete spectrum. Wybourne's book ''Classical groups for physicists'' gives a readable account.

I tried to check his 50 pages long <readable account> of the H-atom and I was dissapointed. Formal mathematics with an abuse of bracket notation, no functional analysis, 90% of his statements unproven, 50 articles in his generous bibliography (to which I don't have access anyway) probably containing the missing proofs and making your head spin between the last pages containing the bibliography and the pages containing the links to it.

Oh, I may have only skimmed through, but a mention of the continuous portion of the spectrum appears on 2 pages at most, so I guess I'll give that part a second chance.
 
  • #9
dextercioby said:
I tried to check his 50 pages long <readable account> of the H-atom and I was dissapointed. Formal mathematics with an abuse of bracket notation, no functional analysis, 90% of his statements unproven,
Well, as the title says this is a book for physicists, not for mathematicians. So he takes all the liberties a physicist needs to get to the results quickly. Mathematicians can add the required rigor themselves.

For a more rigorous treatment, try the book by Barut and Raczka. But I am not sure they are rigorous enough, and I am also not sure whether they exhibit the continuous portion of the hydrogen spectrum in detail enough to be useful.
 
  • #10
A. Neumaier said:
[...]
For a more rigorous treatment, try the book by Barut and Raczka. But I am not sure they are rigorous enough, and I am also not sure whether they exhibit the continuous portion of the hydrogen spectrum in detail enough to be useful.

I have checked it and there's no detailed account on the unitary irreds of SO(4,2) or their influence in the quantum mechanical description of the H-atom.
 
  • #11
dextercioby said:
I have checked it and there's no detailed account on the unitary irreds of SO(4,2) or their influence in the quantum mechanical description of the H-atom.
I haven't the book anymore, but there was a long sequence of exercises on the hydrogen atom in the the chapter on dynamical groups.
 
  • #12
A. Neumaier said:
I haven't the book anymore, but there was a long sequence of exercises on the hydrogen atom in the the chapter on dynamical groups.
Actually, I found the book, and the exercises are _not_ there. This surprises me - I am sure that I had seen them somewhere...

Another possible source is: B. Cordani, The Kepler Problem, Birkh"auser 2003.
It treats the classical and the quantum Kepler problem along similar lines.

see also
http://slac.stanford.edu/pubs/slacpubs/0000/slac-pub-0141.pdf
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085769/pdf/pnas01823-0045.pdf
 
Last edited:
  • #13
QuantumCosmo said:
Hi,
I was wondering if the bound solutions to the radial part of the hydrogen atom form a complete set for the functions in L^2(0,\infty). I know that the laguerre polynomials are complete and that they only differ from the radial solutions by factors of x^l * exp, so I thought that they would have to be complete too (apparently, it is not so simple as to say: "well, you are missing the scattering states, so of course they are not complete").
But I heard that it followed from Sturm-Liouville-Theory that they are not complete, which doesent seem to make any sense to me.
Does anyone have an idea?
Thank you,
QuantumCosmo
Although I don't remember the details, there is a difference of how the Laguerre polynomials enter in their definition as a complete set of orthogonal polynomials and how they enter into the radial solution of the Schroedinger equation. Namely in the latter, the radial dependence is scaled by a factor containing the main quantum number n. This makes the decisive difference.
 
  • #14
  • #15
dextercioby said:
I thank you very much for the <stanford> article.

Some more free related articles on the topic:
http://arxiv.org/pdf/quant-ph/0409209
http://www.ifi.unicamp.br/~aguiar/Cursos/FI264/a90110.pdf
 
Last edited by a moderator:
  • #16
A. Neumaier said:
Actually, I found the book, and the exercises are _not_ there. This surprises me - I am sure that I had seen them somewhere...
[...]

Hi, Arnold, I found them at the end of Chapter 12, p. 378 pp 392. :D
 

1. What is a complete orthornormal set in a hydrogen atom?

A complete orthornormal set in a hydrogen atom refers to a set of quantum states that are both orthogonal (perpendicular) and normalized (unit length) to each other. In other words, these states are independent of each other and provide a complete description of the possible energy levels of a hydrogen atom.

2. How many states are included in a complete orthornormal set for a hydrogen atom?

A complete orthornormal set in a hydrogen atom includes an infinite number of states. This is because the energy levels of a hydrogen atom are quantized, meaning they can take on an infinite number of discrete values.

3. What is the significance of a complete orthornormal set in studying a hydrogen atom?

A complete orthornormal set is essential in understanding the energy levels and behavior of a hydrogen atom. It allows for accurate calculations of the probabilities of finding the atom in a particular state, as well as the transitions between energy levels.

4. How are the states in a complete orthornormal set labeled?

The states in a complete orthornormal set for a hydrogen atom are labeled using quantum numbers, which describe the energy level, orbital angular momentum, and magnetic moment of the electron in the atom.

5. How does the concept of a complete orthornormal set relate to the uncertainty principle?

The concept of a complete orthornormal set is related to the uncertainty principle in that it describes the set of states in which the position and momentum of an electron in a hydrogen atom can be known with certainty. This set of states is known as the stationary states, and they obey the uncertainty principle, which states that the more precisely we know the position of an electron, the less precisely we can know its momentum, and vice versa.

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