Stability of the atom in QFT

  1. Jano L.

    Jano L. 1,211
    Gold Member

    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 [itex]H = \frac{p^2}{2m} - \frac{Ke^2}{r}[/itex].

    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?
  2. jcsd
  3. Bill_K

    Bill_K 4,160
    Science Advisor

    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.
  4. Jano L.

    Jano L. 1,211
    Gold Member

    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?
  5. Bill_K

    Bill_K 4,160
    Science Advisor

    Sorry I couldn't be of help.
  6. Jano L.

    Jano L. 1,211
    Gold Member

    That's OK, thanks anyway.
  7. 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?
  8. There is no electric force, only potential energy, in non-relativistic Schroedinger's equation
  9. Jano L.

    Jano L. 1,211
    Gold Member

    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 [itex]-Ke^2/r[/itex]; 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 [itex]r[/itex] is

    the work done by a non-electromagnetic force to bring the electron from an infinite distance into the distance [itex]r[/itex]:

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

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

    \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).
  10. 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
  11. Jano L.

    Jano L. 1,211
    Gold Member

    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.
  12. atyy

    atyy 9,758
    Science Advisor

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

    fzero 2,601
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    This interaction describes both interaction with an external field and single photon processes. In QFT it is natural to split A up in to a background term describing an external field and a quantum term generating single photons. This is known as the background field method.

    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 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–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 while QED corrections lead to things like the 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.
  14. 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,
    [tex]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][/tex]
    [tex]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][/tex]
    where r (= distance) is close to infinity, the second term (= E(2) ) is left, which means the accelerated term.
    And the emitted energy is
    [tex]E= \frac{1}{2}\epsilon E^2 + \frac{1}{2\mu}B^2 [/tex]
    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 ?
  15. fzero

    fzero 2,601
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    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.
  16. DarMM

    DarMM 331
    Science Advisor

    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.
  17. Jano L.

    Jano L. 1,211
    Gold Member

    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.

    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."

    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?)

    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.

    which atoms are unstable without QED? Why?
  18. Jano L.

    Jano L. 1,211
    Gold Member


    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.

    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.

    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.
  19. DarMM

    DarMM 331
    Science Advisor

    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.
  20. Jano L.

    Jano L. 1,211
    Gold Member

    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?
  21. DrDu

    DrDu 4,088
    Science Advisor

    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.
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