Stability of the atom in QFT

[Jano L.]

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?

 

[Bill_K]

Jano, Quantum field theory is not the issue. I'm afraid to say you have multiple misunderstandings about regular Schrodinger quantum mechanics, and I probably can't clear them all up at once. The idea that the electron can "fall onto the proton because it's accelerating", even stated hypothetically, is pretty far off base.

You need to understand at least three separate things:
- that the hydrogen atom has a ground state.
- that the lowest Bohr level is the ground state.
- that perturbations don't produce new states, they can only connect an existing state with another existing state.

When you get that far, consider this: interaction with the electromagnetic field, even nonrelativistically, is more than just the e²/r term. There's also a p·A. This term is what produces matrix elements between the bound states of hydrogen and causes transitions between them.

Relativity is a separate issue. The Dirac equation for the hydrogen atom has an exact solution, very similar to the Schrodinger solution. There are twice as many bound states of course, because of the two spin states. Also the degeneracy between energy levels that you see in the Schrodinger solution are split.

But the answer to your question won't be found in quantum field theory, you need to go back and understand it on the Schrodinger level first.

 

[Jano L.]

Hello Bill_K,

I am sorry to read your statement that I do not understand Schroedinger's equation. Which part do you think I do not understand?

To save us from silly accusations, I declare honestly that I understand and accept the existence of the lowest discrete proper value of the standard hydrogen atom Hamiltonian. There is no need to discuss this.

I do not understand why do you think quantum field theory is not the issue. I asked what is the status of hydrogen atom in quantum field theory. That means I am interested in how _quantum field theory_ defines and analyzes the bound system electron+proton. I think this is pretty clear question about quantum field theory.

How can you say the acceleration of the electron is far off base? It is well-known that Maxwell's equations connect radiation with accelerated motion of charge.

The p.A term in the Hamiltonian is irrelevant - I want to discuss spontaneous emission, not interaction with the external field.

Perhaps it is the motivation which I stated in a cumbersome way. So I try again:

Non-relativistic Schroedinger's equation is constructed from the classical Hamilton's function which describes conservative system that moves according to pre-relativistic notion of electric force (radial, time-independent, no retardation).

However, relativity shows us that the changes in the field of the sources propagate with finite velocity. This effect is not contained in the standard Hamilton's function.

It follows that this effect is not taken into account in the non-relativistic quantum theory as much as it is not taken into account in the non-relativistic classical theory(Kepler's problem).

If one introduces the retardation in the classical model, the system loses energy and the two particles approach (Synge has shown that the motion of proton+electron with retarded fields is unstable, I can give a reference if needed).

My question is, what would happen in a fully relativistic quantum theory which takes into account of the retardation. Is the result the same as in the classical model, or does the quantum theory make the system stable despite the retardation effects?

Dirac's equation for hydrogen atom won't do, because it has essentially the same drawback as the non-relativistic eq.: it does not take into account the retardation. The potential that is used in the calculation is supposed to be radial and time independent.

That is why I ask about the quantum field theory. In this case, quantum electrodynamics, since I am said this is the most accurate theory of electromagnetic interaction.

So the question is: is there a possibility in QED to address the hydrogen atom in an exact way? If yes, what are the results?

 

[Bill_K]

Sorry I couldn't be of help.

 

[Jano L.]

That's OK, thanks anyway.

 

[DeShark]

As far as I can find from a quick google search (and I don't know QED/QFT very well at all!), QED has only been used to add corrective terms, i.e. pertubations to the energy levels of the states found by the schroedinger equation.

I don't know whether QED is capable of describing the interaction between the proton and the electron as you desire (i.e. from the ground up), but I'm certainly interested if there's a link out there to a paper (or just a summary of how one would go about it).

I presume there's a Lagrangian that can be used, which has certain solutions that can be found from QED?

 

[ardenmann0]

Non-relativistic Schroedinger's equation is constructed from the classical Hamilton's function which describes conservative system that moves according to pre-relativistic notion of electric force (radial, time-independent, no retardation).

There is no electric force, only potential energy, in non-relativistic Schroedinger's equation

 

[Jano L.]

That is true, ardenmann0. But this potential energy can be introduced into analytical mechanics by studying the concept of radial electrostatic force. Schroedinger did not invent potential energy $-Ke^2/r$; he has taken it from the classical theory. In classical theory, the definition of the potential energy of the system electron-proton when they are separated by a distance $r$ is

the work done by a non-electromagnetic force to bring the electron from an infinite distance into the distance $r$:

$$U(\mathbf r) = \int_{\infty}^{\mathbf r} Kq_e (-\mathbf E_p(\mathbf r')) \cdot d \mathbf r',$$

where $q_e E_p$ is the electric force on the electron. This potential energy can be used to express the force on the electron in $\mathbf r_e$:

$$\mathbf F(\mathbf r_e) = -q_e \nabla U(\mathbf r_e).$$

This definition requires that the electric field is static.

But in relativity when the proton moves, the force on the electron cannot not be static like that, but should take into account the movement of the proton. With the electric force, there should be the magnetic force as well. This cannot be described by the potential energy. It is much less general concept than the force (field).

 

[ardenmann0]

The electrostatic force can act on the electron as in the Bohr model, but not on the wavefunction which is not a point particle. That's why there is no force in the non-relativistic Schroedinger's equation

 

[Jano L.]

Schroedinger's equation does contain mathematical equivalent of the physical radial electrostatic force - the potential energy. Of course, force is not something acting on the wave-function. Schroedinger's wave-function is not the electron. It is a function on the configuration space of a non-relativistic system of particles.

I do not say we should put Newton's F into Schroedinger's equation. I'm saying the description based on potential energy has the basic flaw that it neglects relativity (field, retardation, ...). In relativity, the potential energy does not exist. If we want to take the results of relativity into account, we have to formulate the scheme by means of fields. That is why many people are working on the theory of field: to account for relativity.

Since radiation is a relativistic effect, to address the radiation and stability of the atom, we need to take the relativistic field (which is not radial electrostatic!) into account.

 

[atyy]

I guess you've already tried googling "bound state QED"? I tried but didn't see anything directly related to your question.

 

[fzero]

Hello Bill_K,

I am sorry to read your statement that I do not understand Schroedinger's equation. Which part do you think I do not understand?

To save us from silly accusations, I declare honestly that I understand and accept the existence of the lowest discrete proper value of the standard hydrogen atom Hamiltonian. There is no need to discuss this.

I do not understand why do you think quantum field theory is not the issue. I asked what is the status of hydrogen atom in quantum field theory. That means I am interested in how _quantum field theory_ defines and analyzes the bound system electron+proton. I think this is pretty clear question about quantum field theory.

How can you say the acceleration of the electron is far off base? It is well-known that Maxwell's equations connect radiation with accelerated motion of charge.

The p.A term in the Hamiltonian is irrelevant - I want to discuss spontaneous emission, not interaction with the external field.

This interaction describes both interaction with an external field and single photon processes. In QFT it is natural to split A up into a background term describing an external field and a quantum term generating single photons. This is known as the background field method.

My question is, what would happen in a fully relativistic quantum theory which takes into account of the retardation. Is the result the same as in the classical model, or does the quantum theory make the system stable despite the retardation effects?

Dirac's equation for hydrogen atom won't do, because it has essentially the same drawback as the non-relativistic eq.: it does not take into account the retardation. The potential that is used in the calculation is supposed to be radial and time independent.

That is why I ask about the quantum field theory. In this case, quantum electrodynamics, since I am said this is the most accurate theory of electromagnetic interaction.

So the question is: is there a possibility in QED to address the hydrogen atom in an exact way? If yes, what are the results?

Dirac's equation is fine. The effect of retardation is accounted for in QFT in a subtle way through the time-evolution operator. In perturbation theory, the Dyson series http://en.wikipedia.org/wiki/Dyson_series is used to evaluate this. The time-ordering there correctly accounts for causality. It is not necessary to modify the equations of motion.

An alternate approach, closer to what you've described above is the absorber theory http://en.wikipedia.org/wiki/Wheeler–Feynman_absorber_theory. This theory does not have a method of dealing with divergences, so was abandoned in favor of QFT.

Bill_K was leading you down the right path. The Dirac equation + interactions is the correct QFT description of the hydrogen atom. As a quantum two-body problem, I don't believe any exact solutions are known, so perturbation theory is used to evaluate corrections to the familiar non-relativistic solution. First-order relativistic corrections lead to fine structure http://en.wikipedia.org/wiki/Fine_structure while QED corrections lead to things like the Lamb shift http://en.wikipedia.org/wiki/Lamb_shift.

There is no instability since the quantum system still has a ground state (with an energy slightly lower than that determined by NR QM, see the figure on the Fine structure page linked above). Transitions between excited states are computed in QED with an astonishing degree of agreement with experimental results.

 

[ytuab]

I heard classical Maxwell equation explains why the accelerated charge radiates energy, which removed Bohr classical orbit.
But Bohr got Nobel prize for this Bohr orbit, so members of a selection committee in Nobel prize didn't understand this reason ?

When I recheck classical electromagnetism book again, the electric fields generated by a changing electric dipole (= p ) are,
$$E^{(1)} (x, t) = \frac{1}{4\pi\epsilon} \left[ -\frac{\dot{p}(t)}{c r^2} + \frac{3x(x\cdot\dot{p}(t))}{cr^4} \right]$$
and
$$E^{(2)} = \frac{1}{4\pi\epsilon}\left[ -\frac{\ddot{p}(t)}{c^2 r} + \frac{x(x\cdot \ddot{p}(t))}{c^2 r^3} \right]$$
where r (= distance) is close to infinity, the second term (= E(2) ) is left, which means the accelerated term.
And the emitted energy is
$$E= \frac{1}{2}\epsilon E^2 + \frac{1}{2\mu}B^2$$
But as shown in the first term of the electric field (= E(1) ), even when the electric charge is not accelerated, the electric field is generated around them.
So why we pick up only the accelerated term ? Eventually, a charged particle which is moving at the constant velocity radiates energy ?
In Borh's orbit, one electron is going around the proton. And when we give attention on each short part of the orbit, the charged particle is moving like the free particle.
Of course, if we observe in the places where r is infinity, the hydrogen atom is neutral, so the electric dipole itself is neglected.

In textbooks, when the accelerated charge emits the electromagnetic wave, many charged particles are oscillating at the same place, which can be expressed as AC current.

For example, one charged particle is supposed to be moving in one direction at the constant velocity v.
And at some interval, the second charged particle follws the first one at the same velocity v in the same direction.
And the third one, and the forth one... in the same way.
In this case, the electric fields E around them are changing periodically. But they don't emit energy ?

 

[fzero]

I heard classical Maxwell equation explains why the accelerated charge radiates energy, which removed Bohr classical orbit.
But Bohr got Nobel prize for this Bohr orbit, so members of a selection committee in Nobel prize didn't understand this reason ?

...

In this case, the electric fields E around them are changing periodically. But they don't emit energy ?

The reason that the electron in the hydrogen atom doesn't radiate is well understood in quantum mechanics. The ground state is the lowest energy state available for the system (also lower energy than the unbound proton-electron system). If the electron were to radiate away some energy in the form of a photon, the system would have to move into a state with a lower energy than the ground state. But there is none, so the electron in the ground state does not radiate.

Even though the Bohr model is qualitatively incorrect, viz a viz the description in terms of orbits, the notion there of discrete energy levels, including a definite ground state energy is still correct.

 

[DarMM]

Stability of hydrogen in QED is way beyond current technology. However "Dirac Hydrogen", so coulomb term + dirac operator is known to be self-adjoint and stable. Also Dirac Hydrogen + One loop corrections from QED, which basically involve adding QED effects like the anomalous magnetic moment directly to the Hamiltonian is also known to be stable. Interestingly enough, some atoms only become stable when you add the QED effects.

 

[Jano L.]

FZero, thanks. I'd like to comment:

Dirac's equation is fine. The effect of retardation is accounted for in QFT in a subtle way through the time-evolution operator. In perturbation theory, the Dyson series http://en.wikipedia.org/wiki/Dyson_series is used to evaluate this. The time-ordering there correctly accounts for causality. It is not necessary to modify the equations of motion.

Do you mean Dirac's equation with Coulomb potential + rel. corrections? Because I think this model lacks the retardation right from the beginning - it is bit like Sommerfeld's enhancement of Bohr's model. It introduces some corrections, but still neglects the rest - retardation and radiation. But these make the atom unstable classically.

This interesting comment on retardation on Wikipedia seems relevant:

"If higher-order terms in v/c are retained then the field degrees of freedom must be taken into account and the interaction can no longer be taken to be instantaneous between the particles. In that case retardation effects must be accounted for."

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

You say the retardation is considered by Dyson's series.

I know Dyson's series from spectroscopy; there we have the perturbation series in the interaction Hamiltonian containing the time-dependent external field. In our problem, one could imagine the external field would be replaced by the internal field due to the proton. But the time evolution of this field is not trivial due to the retardation. How would you write the interaction Hamiltonian for such a thing? (In case it is complicated, can you post a reference?)

... since the quantum system still has a ground state.

Are you sure about this? Without Hamiltonian and exact solutions... I think, if I has some Hamiltonian, say 10 lines long, I guess I could accept that maybe its spectrum has lower bound. But is there an exact Hamiltonian in the first place? If no, I think we cannot claim it has a ground-state.

DarMM,
which atoms are unstable without QED? Why?

 

[Jano L.]

ytuab,

I heard classical Maxwell equation explains why the accelerated charge radiates energy, which removed Bohr classical orbit.
But Bohr got Nobel prize for this Bohr orbit, so members of a selection committee in Nobel prize didn't understand this reason ?

In fact Bohr made a rather strong departure from electrodynamics when he said there is no radiation in stationary states. It allowed him to develop his theory and partially succeed in explanation of the line spectra.
It was a good success, so he got the Nobel prize.

So why we pick up only the accelerated term ? Eventually, a charged particle which is moving at the constant velocity radiates energy ?

The reason why only the acceleration term is connected with the radiation is that this term is important far away from the atom (it falls off only as 1/r). The other term falls off as 1/r^2 and so is negligible.

In this case, the electric fields E around them are changing periodically. But they don't emit energy ?

I like to think that they do emit radiation, but there is also an incoming radiation (thermal, random,...), so the atom can be in sort of dynamical equilibrium.

 

[DarMM]

DarMM,
which atoms are unstable without QED? Why?

Heavy element atoms. The mechanism is that the more you squeeze an electron in an atom the more you lower its energy due to moving deeper into the coulomb potential, however through the uncertainty principle you're also increasing its kinetic energy by localising it. At some point these effects balance, which is the ground state. Without QED corrections the potential energy effects always outweigh the kinetic energy effects, so the electron can localised arbitrarily far and basically can be collapsed right on to the nucleus.

 

[Jano L.]

That's interesting. Why does the potential energy win? I've read I think in Landau that steep potential (like 1/r^3) can eat away the wave-function. But 1/r should not be able to do this. Can you please post a short explanation for the effect, or a link?

 

[DrDu]

In principle, QED of the Hydrogen atom can be solved using the Bethe-Salpeter equations.
However, without further approximations like the ladder approximation, these are arbitraryly complicated.

 

[Irigi]

Without QED corrections the potential energy effects always outweigh the kinetic energy effects, so the electron can localised arbitrarily far and basically can be collapsed right on to the nucleus.

This doesn't seem correct. If we squeeze the atom to size x, we also know that Δx ≈ x and from uncertainty principle Δp ≈ p ≈ ℏ/x. Since E_kin ~ p² ~ x^-2, it grows faster that Coulombic potential -1/x around x = 0, so the kinetic term ougweights the potential one. That's why there should always be ground state in standard quantum mechanics, even without QED. Am I not right?

 

[ytuab]

fzero, I want to know why we denied the Bohr's orbit in the point of radiating energy now.
As you know, Bohr's orbit uses the "quantization of de Broglie's wavelength", which gives the same results as Schrodinger equation.

Jano L., thanks for your reply.
But there are some points I want to ask in your answer.
Actually, we can not go to infinity. And when we observe moving atoms near them, the radiation energy is stronger in the "constant-velocity" charged particle (considerring 1/r and c ) than "accelerated" one ?
And if you think about an incoming radiation, too, Bohr's orbit doesn't radiate ?
By the way, radiation energy by Maxwell law originates in the electric vacuum energy (u) of
$$u(x) = \frac{1}{2}\epsilon_0 E^2 (x)$$
Using pointing vector S, the total energy change is
$$- \frac{d}{dt} \left[ \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2\mu_0} B^2 \right] = \frac{1}{\mu_0} div (E \times B)$$

The above vauum electric energy u(x) is gotten from a spherical conductor with a radius of a.
when the total charge of the spherical conductor is q, the potential energy on the surface area is
$$\phi(q,a) = \frac{q}{4\pi\epsilon_0 a}$$
So the energy needed for giving the charge Q to the spherical conductor is
$$U = \int_0^Q \phi dq = \int_0^Q \frac{q}{4\pi\epsilon_0 a} dq = \frac{1}{2} \frac{Q^2}{4\pi\epsilon_0 a}$$
Using Gauss law, we can get
$$U = \frac{\epsilon_0}{2} \int_a^{\infty} d^3 x E^2 (x)$$

But as you know, a single electron (-e) can NOT be divided.
So the above calculation method can not be used in the case of a single electron.
(Single electron is not a spherical conductor. )

So the idea of the radiation from an accelerated charge applies only to a classical conductor which includes many charged particles ?
(In this case, we can treat the charge e as infinitesimal one in comparison with the total big charge Q. )

 

[DarMM]

This doesn't seem correct. If we squeeze the atom to size x, we also know that Δx ≈ x and from uncertainty principle Δp ≈ p ≈ ℏ/x. Since E_kin ~ p² ~ x^-2, it grows faster that Coulombic potential -1/x around x = 0, so the kinetic term ougweights the potential one. That's why there should always be ground state in standard quantum mechanics, even without QED. Am I not right?

That's not relativistic though. Relativistically Energy is proportional to momentum not its square.

 

[Irigi]

That's not relativistic though. Relativistically Energy is proportional to momentum not its square.

OK, you are right then, there might be a problem. Can you give me some link on this topic, please? Is the problem already in Dirac's equation without radiative corrections? How does QED fix it? Thanks.

 

[Jano L.]

I want to know why we denied the Bohr's orbit in the point of radiating energy now.
As you know, Bohr's orbit uses the "quantization of de Broglie's wavelength", which gives the same results as Schroedinger's equation.

You are right. Bohr's model gives some results (energy values) which Schroedinger's equation gives too. And nobody would deny Bohr's model as entirely invalid. It is just much less successful than Schroedinger's equation; extending it for many electron atoms was not successful. It also left the question of mechanism untouched; Schroedinger's theory is much more elaborated - for example, it can describe the excitation of the atom as a process in time, which is impossible in Bohr's theory.

We did not deny Bohr's model due to neglection of radiation. Quantum theory does the same thing, as this thread shows. Bohr and physicists of his period were aware of the result of electrodynamics that the atom should radiate until it collapsed. He just decided to go on and develop his ideas anyway; and the partial success is the reason.

However, the problem is still there; with only mutual retarded em forces in classical model, the atom should collapse. Even if we did not accept Poynting's expressions and Larmor's formula as valid for point-like particles (as you correctly point out, they cannot be derived from the equations of motion for point-like particles), a system of such point-like particles would still lose energy on account of the retardation. This was shown by Synge, J. H. in "On the electromagnetic two-body problem", Proc. Roy. Soc. London, A177, 118 (1940). He shows that if the proton and electron acted on each other by mutual retarded forces and no self-force was present, they would still lose energy, albeit 925x more slowly than according to Larmor's formula.

This is easy to understand, as the finite velocity of interaction gives the field non-trivial degrees of freedom. Generally one expects there will be energy associated with these new degrees of freedom and will be transported away.

Personally I like to think that the atom is (at least quasi-) stable on account of some non-electromagnetic forces or chaotic background radiation - search for random electrodynamics, especially the paper by Daniel C. Cole , Yi Zou, "Quantum mechanical ground state of hydrogen obtained from classical electrodynamics", Physics Letters A 317 (2003) 14–20.

 

[Jano L.]

I came across an article on positronium. This is a bound state of electron and positron. It is claimed to be unstable with very low lifetime.

Can this collapse and the lifetime be calculated in QED? Why would positronium collapse if hydrogen atom is stable?

 

[fzero]

FZero, thanks. I'd like to comment:



Do you mean Dirac's equation with Coulomb potential + rel. corrections? Because I think this model lacks the retardation right from the beginning - it is bit like Sommerfeld's enhancement of Bohr's model. It introduces some corrections, but still neglects the rest - retardation and radiation. But these make the atom unstable classically.

No, you don't put the Coulomb potential in by hand. The tree level diagram for single photon exchange already leads to the Coulomb potential. As I said, it's possible to get excellent quantitative results in perturbation theory, so you use the nonrelativistic states as your 0th order starting point. You can verify that all perturbations are small.

As DrDu mentioned, there is the Bethe-Saltpeter equation around if one wants a more rigorous approach to bound states, but it's still necessary to introduce approximations there.

You say the retardation is considered by Dyson's series.

I know Dyson's series from spectroscopy; there we have the perturbation series in the interaction Hamiltonian containing the time-dependent external field. In our problem, one could imagine the external field would be replaced by the internal field due to the proton. But the time evolution of this field is not trivial due to the retardation. How would you write the interaction Hamiltonian for such a thing? (In case it is complicated, can you post a reference?)

Well, as I said above, you don't introduce an external potential anywhere.

fzero, I want to know why we denied the Bohr's orbit in the point of radiating energy now.
As you know, Bohr's orbit uses the "quantization of de Broglie's wavelength", which gives the same results as Schrodinger equation.

This is explained in any modern QM textbook. Basically, Bohr's treatment of the H atom turns out to be correct in the case of the energy spectrum. But the picture fails because the electron is not confined to circular orbits. The correct QM treatment is in terms of wavefunctions and probability distributions.

Personally I like to think that the atom is (at least quasi-) stable on account of some non-electromagnetic forces or chaotic background radiation - search for random electrodynamics, especially the paper by Daniel C. Cole , Yi Zou, "Quantum mechanical ground state of hydrogen obtained from classical electrodynamics", Physics Letters A 317 (2003) 14–20.

As I said several posts ago, the atom is stable because the ground state has the lowest energy of any other configuration. I took a look at the paper you refer to (on the arxiv at http://arxiv.org/abs/quant-ph/0307154). It's not particularly convincing of anything, since they fail to discuss how their results depend on the box size and minimum frequency. They also confine the orbit to a plane by hand, which changes the problem significantly from realistic physics.

I came across an article on positronium. This is a bound state of electron and positron. It is claimed to be unstable with very low lifetime.

Can this collapse and the lifetime be calculated in QED? Why would positronium collapse if hydrogen atom is stable?

Positronium is unstable because eventually the electron and positron coincide and annihilate. The decay rate is proportional to $|psi(0)|^2$. It has nothing to do with any instability of the H-atom.

 

[ytuab]

You are right. Bohr's model gives some results (energy values) which Schroedinger's equation gives too. And nobody would deny Bohr's model as entirely invalid. It is just much less successful than Schroedinger's equation; extending it for many electron atoms was not successful. It also left the question of mechanism untouched; Schroedinger's theory is much more elaborated - for example, it can describe the excitation of the atom as a process in time, which is impossible in Bohr's theory.

We did not deny Bohr's model due to neglection of radiation. Quantum theory does the same thing, as this thread shows. Bohr and physicists of his period were aware of the result of electrodynamics that the atom should radiate until it collapsed. He just decided to go on and develop his ideas anyway; and the partial success is the reason.

However, the problem is still there; with only mutual retarded em forces in classical model, the atom should collapse. Even if we did not accept Poynting's expressions and Larmor's formula as valid for point-like particles (as you correctly point out, they cannot be derived from the equations of motion for point-like particles), a system of such point-like particles would still lose energy on account of the retardation. This was shown by Synge, J. H. in "On the electromagnetic two-body problem", Proc. Roy. Soc. London, A177, 118 (1940). He shows that if the proton and electron acted on each other by mutual retarded forces and no self-force was present, they would still lose energy, albeit 925x more slowly than according to Larmor's formula.

This is easy to understand, as the finite velocity of interaction gives the field non-trivial degrees of freedom. Generally one expects there will be energy associated with these new degrees of freedom and will be transported away.

Personally I like to think that the atom is (at least quasi-) stable on account of some non-electromagnetic forces or chaotic background radiation - search for random electrodynamics, especially the paper by Daniel C. Cole , Yi Zou, "Quantum mechanical ground state of hydrogen obtained from classical electrodynamics", Physics Letters A 317 (2003) 14–20.

Jno L. Thanks for your interesting comments.

This paper explains Synge's paper in 1940 about two body problem.
But as I see this paper, this method is not classical. and he used some approximations.
If we think about the effect of retardation effect correctly, we have to use a computer.
Of course in 1940, it was difficult to calculate it correctly.

Even in 1983 (Bohr orbit theorey revisited I. Ground state energies for the helium isoelectnic sequence, R D Harcourt), he didn't use a computer to get the helium three-body orbit.
Since about 1990, they tend to use computers to get "classical helium three-body atom".
In this case, due to the repulsive force between two electrons, it causes "autoionization", or very unnatural stable orbits (one electron is very close to nucleus, and another is far away).
Of course, these classical helium atoms didn't consider de Broglie waves (and two electron's interference) at all.
(If we don't consider de Broglie wave, the hydrogen ground state energy can be much lower than the original value.)

Even in the recent paper (Semiclassical quantization of Bohr orbits in the heluum atom, V.V. Belov, 2007), they didn't use the classical idea based on the computational method, instead they took WKB-Maslov method.
(This result is a little different from the experimental value.)

Suppose, a positive nucleus exists at the origin.
Even when an electron enters any point, which is far away from the origin, the electron can always feel the attractive Coulomb force from the nucleus.
This means the nucleus continues to emit "infinite virtual photons" (which act as "sensors" ) in all space ?
If the nucleus is emitting very few sensor photons, it would be very difficult to find the electron, which accidentally enters one place in large space.
So is it natural that we think all space itself is affected and distorted by Coulomb force of the point-like nucleus ?
In this case, when the electron is at the distance of r from the origin, it feels the constant force from near its field (local action), which has already been distorted.

I'm glad you agree with my idea about Poynting's expressions and Larmor's formula in a point-like electron.
This classical idea of Larmor formula (radiating energy) reminds me about what we call "classical electron radius", I think.

 

[Jano L.]

Thanks to everybody for the contributions. I have discussed this a bit with my friends too and my understanding of this question so far is this:

Stability is hard to prove in QFT because the theory is not well-fitted to deal with stationary states. It deals mainly with asymptotically free particles and scattering.

We have only results based on perturbation theory.

These indicate that in case electron-positron, the system is unstable. Electron and positron can transform into two or more photons.

The system electron - proton is believed to be stable. But because of the previous statement, electromagnetic interaction is not sufficient for this. In order to prevent the annihilation of electron and proton, terms in Lagrangian describing the proton have to be not purely electromagnetic. There has to be something that reflects the law of conservation of baryon number.

So, the stability is not some new non-trivial result of the theory, but is believed to be true in real world and corresponding conservation laws are observed when constructing the Lagrangian.

Whether the system in the resulting theory is indeed stable or not, seems to be too difficult to ascertain.

Still, if something of this seems wrong to you, please tell me.

 

[fzero]

The system electron - proton is believed to be stable. But because of the previous statement, electromagnetic interaction is not sufficient for this. In order to prevent the annihilation of electron and proton, terms in Lagrangian describing the proton have to be not purely electromagnetic. There has to be something that reflects the law of conservation of baryon number.

You've actually got this backwards. 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. However, the ground state of hydrogen is forbidden to do this by energy conservation.

 

[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?