Stability of the atom in QFT

[fzero]

Fzero, thank you for your effort to discuss this. I am glad someone is interested in these things too. For the sake of clarity, I think it is important to say once again that I am concerned with the question:

Is a system electron+proton stable provided their interaction is described by purely electromagnetic terms?

I see your point with the annihilation. The interaction between the proton and the electron is not described in the same way as that between the positron and the electron, so there is no problem with the annihilation in the standard sense. In this point I was wrong; hydrogen does not suffer from the kind of annihilation the positronium does.

How would you write purely electromagnetic Lagrangian for electron-positron+em-field+proton-antiproton? Is such thing even known in the theory? Most sources I have, give only the electron-positron+em/field terms or full Standard Model Lagrangian, which however contains non-electromagnetic terms, so it does not help in answering the question.

You can write a Lagrangian with two Dirac fields (proton and electron) coupled to a U(1) gauge field. At the energy scale associated with the H-atom, the quark substructure of the proton and weak and strong forces are very small corrections.

What I had on mind: it could be that there is no stable solution of the equations derived from a purely electromagnetic Lagrangian electron+field+proton. If so, other interactions (new terms in Lagrangian) are necessary to make it stable. Please let me know if this possibility can be immediately rules out. That would answer my question.

No non-EM terms are needed. I'll address this in a bit more detail below.

Of course, conservation of baryon and lepton numbers without conservation of energy is not enough for the process to go on. This process is supposed to take place in heavy atoms, where some additional energy can be gained from other particles.

But I did not mean this process. I meant a process where proton+electron transform into _radiation_. Any positive energy is enough for this process, so energy conservation does allow it. I meant that non-electromagnetic terms in Lagrangian imply conservation of lepton and baryon numbers and this does not allow lepton + baryon transforming into radiation.

No terms in the SM Lagrangian allow an electron and proton to transform into a photon. That's usually what you mean by radiation. Non-EM terms do not "prevent" this, since it isn't allowed in the first place.

Of course, we can be more general and consider the inverse beta decay process, which is mediated by the weak interaction. Energy conservation prevents it from happening to an isolated H-atom, which is what we have to consider in the definition of "stable." If we put the H-atom in a heat bath of some sort, then it is possible for it to happen eventually. This is not what we mean by unstable. A heat bath is not needed to show that the classical system is unstable.

You also say

The non-EM terms are miniscule compared to even the relativistic corrections and play no role whatsoever in the stability of hydrogen.

How do you support the second part of the claim? Can you give me a reference or some equations?

It's just a matter of comparing the coupling constants (at a few eV) to that of the EM coupling.

Smallness of some quantity does not always imply it has negligible role in stability. Other properties of the quantity are also important. Radiation damping in classical equation of motion is also minuscule compared to Coulomb's force; on the time scale of one revolution of the electron around the proton, it is completely negligible. Yet it makes the atom unstable after millions of revolutions, which still is something like $10^{-10}$ s or so. Some equivalent of this damping has to be in quantum theory too and something has to prevent the atom from collapse. I see two possibilities:

- either full relativistic quantum theory incorporates the charges and em. radiation in such a way that stable charge distribution can exist without radiating energy away

or

- non-electromagnetic forces accomplish this, counteracting the Coulomb attraction and radiation damping during accelerated motions.

You assume the first position; can you give me some reference supporting it so I can study this in detail?

As I've said a number of times already, what prevents these terms from making the H-atom unstable is the nonexistence of any quantum states with a lower energy than the bound ground state. The ground state cannot radiate energy away. The bound state is stable because the state consisting of a proton and electron separated asymptotically has a higher energy.

Adding additional small terms from the SM Lagrangian can only modify the ground state energy slightly. For instance, weak interactions at an energy scale $E$ are suppressed relative to EM by a factor which includes $(E/m_W)^2$, where $m_W\sim 80~\mathrm{GeV}$ is the W mass. For the H-atom, $E\sim 10~\mathrm{eV}$, so this is a correction of 1 part in $10^{-20}$. This cannot affect stability.

 

[A. Neumaier]

You can write a Lagrangian with two Dirac fields (proton and electron) coupled to a U(1) gauge field. At the energy scale associated with the H-atom, the quark substructure of the proton and weak and strong forces are very small corrections.

But this would not resemble the real proton. its electromagnetic form factor is far from that of an electron with the mass of a proton.

Thus you need a non-renormalizable interaction with the e/m field.

 

[Jano L.]

Hello Arnold,

welcome to the discussion. What do you mean by em form-factor? That proton cannot be described the same way as the electron, by Dirac field? Or that some non-trivial charge density of proton should be introduced into the Lagrangian?

I presume the simplest Lagrangian suggested by fzero would be


$$\mathscr{L} = \overline \psi_e [ i\gamma^\mu(\partial_\mu + ieA_\mu)-m_e]\psi_e + \overline \psi_p [ i\gamma^\mu(\partial_\mu - ieA_\mu)-m_p]\psi_p - \frac{1}{4}F^{\mu\nu}F_{\mu\nu}.$$


fzero, please can you direct me to some paper which would show nonexistence of any quantum states with a lower energy than the bound ground state from this or similar Lagrangian?

 

[fzero]

But this would not resemble the real proton. its electromagnetic form factor is far from that of an electron with the mass of a proton.

Thus you need a non-renormalizable interaction with the e/m field.

The ratio of the charge radius of the proton to the Bohr radius is $1.5\cdot 10^{-5}$. You do not need to get so elaborate to determine that the H-atom is stable. If you are trying to measure the Lamb shift to high precision, yes you need to consider form factors and other corrections.

 

[M Quack]

The only way the Hydrogen atom could possibly be unstable would be through decay of the proton, e.g. into a pion and positron. The positron would then annihilate with the former Hydrogen's electron (creating two photons), and the pion would decay into two photons.

At present there is no experimental evidence that this happens.

http://en.wikipedia.org/wiki/Proton_decay

The Wiki entry has a list of references.

As far as I can tell, none of the proposed hypothetical mechanisms are electromagnetic. They all require features beyond the Standard Model.

That the free neutron decays into a proton, an electron, an antineutrino and some spare kinetic energy is a dead giveaway that the Hydorgen atom will not collapse into a neutron.

 

[fzero]

Hello Arnold,

welcome to the discussion. What do you mean by em form-factor? That proton cannot be described the same way as the electron, by Dirac field? Or that some non-trivial charge density of proton should be introduced into the Lagrangian?

He means that the proton is not a point particle, so beyond a certain precision, you cannot neglect that it has a finite charge radius.

I presume the simplest Lagrangian suggested by fzero would be


$$\mathscr{L} = \overline \psi_e [ i\gamma^\mu(\partial_\mu + ieA_\mu)-m_e]\psi_e + \overline \psi_p [ i\gamma^\mu(\partial_\mu - ieA_\mu)-m_p]\psi_p - \frac{1}{4}F^{\mu\nu}F_{\mu\nu}.$$


fzero, please can you direct me to some paper which would show nonexistence of any quantum states with a lower energy than the bound ground state from this or similar Lagrangian?

I don't know a great reference that starts from this Lagrangian and goes into exhaustive detail. Most of what I've been saying is obvious after having worked through enough examples in QM to know how perturbation theory works. It's intuitively obvious that the corrections are much smaller than 13.6 eV, so they are not going to raise the ground state energy above that of the unbound state.

The book Quantum Electrodynamics, by Berestetskii, Pitaevskii, Lifgarbagez discuss fine and hyperfine structure, Lamb shift and relativistic corrections to hydrogenic systems. For a brief review of the sorts of corrections you need to consider for precision measurements of the Lamb shift, you can look at appendix A in the thesis http://edoc.ub.uni-muenchen.de/5044/1/Antognini_Aldo.pdf by Antognini, or in the review he cites: M. I. Eides, H. Grotch, and V. A. Shelyuto. Theory of light hydrogenlike atoms. Phys. Rep. 342 (2001), 63–261. I haven't looked at the latter myself, but it no doubt exhausts the subject.

 

[A. Neumaier]

What do you mean by em form-factor? That proton cannot be described the same way as the electron, by Dirac field?

A proton in an external e/m field satisfies a Dirac equation, but one with correction terms.
The latter are called form factors.

But as fzero pointed out, this is not relevant for stability considerations. In fact, on the e/m level alone, any new Fermion species introduced has its own conserved number operator, hence annihilation is impossible. Neutrons and other particles would all be additional species from the point of view of QED. Therefore all these are conserved in the absence of non-QED interactions. Thus even if the neutron were much lighter than the proton so that fzero's argument would break down, it would still require a non-QED interaction to cause the required transition.