Is the Hydrogen Atom Stable in Relativistic Quantum Theory?

[Jano L.]

The electromagnetic interaction doesn't allow the electron and proton to convert to anything else. If we were to include weak interactions, the proton and electron could combine to form a neutron + neutrino.

Hello fzero,

if I understood your argument, it goes like this: the neutron is known to be heavier than the hydrogen atom. Therefore there is not enough energy so that hydrogen atom can form a neutron. You conclude that em interaction doesn't allow the hydrogen to change to anything else.

I do not think this is correct. The electromagnetic forces alone should allow the electron and the proton to absorb each other and produce radiation, irrespective of the properties of the neutron. It should be similar as with the positronium, only proton being 1836x heavier than the positron, or similar to the classical model of damped orbits of electron revolving around the proton.

In your example, you use the fact that the neutron is heavier than the hydrogen atom. This has no electromagnetic explanation and requires other forces. In Standard Model, these forces are described by terms corresponding to strong and weak interactions. Once the Lagrangian contains these non-electromagnetic terms, laws of conservation follow: there are conserved quantum numbers (baryon, lepton numbers) that do not allow the hydrogen to simply transform into radiation (as positronium does).

This does not show the stability yet (electron and proton could be approaching indefinitely), but it shows in what respect the hydrogen is essentially different from the positronium: the Lagrangian is more complicated and contains non-electromagnetic terms.

 

[ardenmann0]

If I may, I would like to give this question another try, especially if guys with some cabala in QFT can address it.

Is it possible to show in the relativistic quantum theory, that the hydrogen atom is stable? (Electron will not fall onto the proton)?

I explain.

In the non-relativistic quantum theory the system is described by an Hamiltonian derived from the classical Hamilton's function for Kepler's problem H = \frac{p^2}{2m} - \frac{Ke^2}{r}.

But this immediately implies the system is conservative. There is no radiation, no spontaneous emission in the model. No surprise that both in classical and quantum theory the hydrogen atom is stable.

But such an Hamiltonian is correct only in a non-relativistic theory (if the Galilei invariance of laws was accepted). In the relativistic theory, the retardation of the EM field and thus of radiation should be taken into account. I suspect that this is not possible to do within the standard Hamiltonian.

The outgoing radiation can however draw energy away if the particles accelerate and the relativistic quantum model of hydrogen atom can still be unstable for the same reason as in the classical model of the hydrogen atom which takes into account the radiation: the accelerated particles will radiate and lose energy.

What do you think? Is there a way in QFT to describe the hydrogen atom exactly (at least in principle) and show it has a stable ground state?

what acceleration is taking place in the hydrogen atom using QM?

 

[Jano L.]

Acceleration of electron is not stressed in textbooks on quantum theory. Wave function in hydrogen s-state is not accelerating, of course. But the electron is not a wave function. Wave function is a mathematical requisite of the quantum theory.

But electron and proton are point-like particles in quantum theory. If the electron stays near the proton, it has to be either in rest, or accelerate. Otherwise rectilinear motion would lead to ionization of the atom.

Moreover, whenever you have psi function which is a sum of two or more eigenstates of the standard Hamiltonian, the expectation value of the radius vector of the electron is oscillating sinusoidally back and forth. This implies electron has to move with non-zero acceleration.

 

[fzero]

Hello fzero,

if I understood your argument, it goes like this: the neutron is known to be heavier than the hydrogen atom. Therefore there is not enough energy so that hydrogen atom can form a neutron. You conclude that em interaction doesn't allow the hydrogen to change to anything else.

Exactly.

I do not think this is correct. The electromagnetic forces alone should allow the electron and the proton to absorb each other and produce radiation, irrespective of the properties of the neutron. It should be similar as with the positronium, only proton being 1836x heavier than the positron, or similar to the classical model of damped orbits of electron revolving around the proton.

The difference between the H-atom and positronium is that, in the latter case, the positron is the antiparticle of the electron. Particle-antiparticle pairs are always special, since the EM gauge interaction couples them to one another through the term \bar{\psi}\gamma^\mu A_\mu \psi. There is no such coupling between the proton and electron, so there is no annihilation.

In your example, you use the fact that the neutron is heavier than the hydrogen atom. This has no electromagnetic explanation and requires other forces. In Standard Model, these forces are described by terms corresponding to strong and weak interactions. Once the Lagrangian contains these non-electromagnetic terms, laws of conservation follow: there are conserved quantum numbers (baryon, lepton numbers) that do not allow the hydrogen to simply transform into radiation (as positronium does).

The process

p + e^- \rightarrow n + \nu_e

conserves baryon and lepton number. That the mass of the neutron is larger than that of the proton is precisely the reason that the proton is stable and the neutron is not. Where this mass difference comes from has nothing to do with the energy spectrum of hydrogen, since the corresponding physics occurs on vastly different scales. m_n-m_p\sim 1.3~\mathrm{MeV}, while the H-atom binding energy is \sim 13.6~\mathrm{eV}.

This does not show the stability yet (electron and proton could be approaching indefinitely), but it shows in what respect the hydrogen is essentially different from the positronium: the Lagrangian is more complicated and contains non-electromagnetic terms.

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

 

[Jano L.]

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.

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.

The process

p+e−→n+νe

conserves baryon and lepton number.

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.

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?

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?

 

[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


<br /> \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}.<br />


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


<br /> \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}.<br />


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.