Stability of atoms in QM / QED

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  • #1
tom.stoer
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It is often stated that quantum mechanics is able to explain the stability of atoms.

I think most explanations are cheating b/c the compare apples and oranges.

There are two reasons in classical theory which indicate that atoms should be unstable:
A) there is no minimum for the orbit; the radius r can become arbitrary small, therefore the potential energy V(r) is not bounded from below
B) due to emission of electromagnetic waves the electron will lose energy and will therefore fall into this unbounded potential V(r)

QM solves (A); the eigenvalue problem is well-defined; the spectrum of H is bounded from below; there exists a stable ground state; therefore the potential energy <V> is bounded from below.

However QM does not solve (B), simply b/c there is no dynamical el.-mag. field; QM is not even able to state the problem (B).

Of course there are attempts to address this scenario in QED; the most famous one is the (perturbative!) Lamb shift calculation.

So here's my question: is there a non-perturbative proof (in the physical sense ;-) which answers (B) in the affirmative?
 

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DarMM
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So here's my question: is there a non-perturbative proof (in the physical sense ;-) which answers (B) in the affirmative?
No, there isn't unfortunately.

As you are probably aware, the non-relativistic limit of tree level QED gives an effective quantum mechanical model with the coloumb potential, the Darwin term, Spin-Orbit coupling and a modified Kinetic term.
This Hamiltonian possess a lowest eigenvalue and so is stable.

At one loop the non-relativistic limit gives you an effective qm model which is the same as the tree level one plus the Uehling potential and the Tomonoga Spin-Orbit term. I've never seen a paper with the proof that the spectrum of this is bounded below, but it's not that hard to prove yourself if you read papers dealing with the tree-level hamiltonian.

So you could say the atom is stable to one loop in QED.

Of course this involves checking the stability of a non-relativistic Hamiltonian whose predictions match those of the non-relativistic limit QED to a certain loop order, rather than a direct proof in QED.

Since QED on its own probably doesn't exist non-perturbatively, you'll never get such a proof. However QED + QCD almost certainly does exist. So this is the theory you'll have to deal, i.e. to check the stability of hydrogen you can't get away with treating the proton as a fundamental object. However non-perturbative control over QCD for something like the stability of hydrogen is way beyond current mathematical technology.
 
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tom.stoer
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As you are probably aware, the non-relativistic limit of tree level QED gives an effective quantum mechanical model with the coloumb potential, the Darwin term, Spin-Orbit coupling and a modified Kinetic term.
This Hamiltonian possess a lowest eigenvalue and so is stable.
I know

At one loop the non-relativistic limit gives you an effective qm model which is the same as the tree level one plus the Uehling potential and the Tomonoga Spin-Orbit term. I've never seen a paper with the proof that the spectrum of this is bounded below, but it's not that hard to prove yourself if you read papers dealing with the tree-level hamiltonian.

So you could say the atom is stable to one loop in QED.
I know

Since QED on its own probably doesn't exist non-perturbatively, you'll never get such a proof ...
I could see it coming ;-)

... However non-perturbative control over QCD for something like the stability of hydrogen is way beyond current mathematical technology.
It's a one million dollar problem.

OK, that was my impression from the very beginning. But thanks a lot for the response!
 
  • #4
DarMM
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OK, that was my impression from the very beginning. But thanks a lot for the response!
:smile:

You might be interested to know that with the Dirac Hamiltonian it is not known if Helium is stable, see:
Open problems about many-body Dirac operators, by Jan Derezinski.
 
  • #5
tom.stoer
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But in reality they are stable; that saves my day ;-)
 
  • #6
Jano L.
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It is often stated that quantum mechanics is able to explain the stability of atoms.
I think most explanations are cheating b/c the compare apples and oranges.
Yes, exactly! The boundedness of spectrum of basic Schr. equation is as convincing explanation of the stability of atom as is the stability of Kepler's orbits. The very problem of instability was introduced because of radiation, so we can only really answer the question if we do not neglect relativity.

I asked similar question some time ago here:

https://www.physicsforums.com/showthread.php?t=576848&highlight=stability

with similar answers.

Funny, after all these years, we still do not know whether the theory can describe stable atom :-)
 
  • #7
tom.stoer
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Thanks for the link
 

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