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#1
Feb1212, 11:45 AM

PF Gold
P: 1,168

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 nonrelativistic 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 nonrelativistic 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
Feb1212, 04:40 PM

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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 e^{2}/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. 


#3
Feb1212, 06:06 PM

PF Gold
P: 1,168

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 wellknown 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: Nonrelativistic Schroedinger's equation is constructed from the classical Hamilton's function which describes conservative system that moves according to prerelativistic notion of electric force (radial, timeindependent, 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 nonrelativistic quantum theory as much as it is not taken into account in the nonrelativistic 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 nonrelativistic 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? 


#4
Feb1212, 06:20 PM

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P: 4,160

Stability of the atom in QFT
Sorry I couldn't be of help.



#5
Feb1212, 06:27 PM

PF Gold
P: 1,168

That's OK, thanks anyway.



#6
Feb1212, 07:42 PM

P: 149

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? 


#7
Feb1212, 09:45 PM

P: 8




#8
Feb1312, 05:55 AM

PF Gold
P: 1,168

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 electronproton when they are separated by a distance [itex]r[/itex] is
the work done by a nonelectromagnetic force to bring the electron from an infinite distance into the distance [itex]r[/itex]: [tex] U(\mathbf r) = \int_{\infty}^{\mathbf r} Kq_e (\mathbf E_p(\mathbf r')) \cdot d \mathbf r', [/tex] 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]: [tex] \mathbf F(\mathbf r_e) = q_e \nabla U(\mathbf r_e). [/tex] 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). 


#9
Feb1312, 11:11 AM

P: 8

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 nonrelativistic Schroedinger's equation



#10
Feb1312, 12:54 PM

PF Gold
P: 1,168

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 wavefunction. Schroedinger's wavefunction is not the electron. It is a function on the configuration space of a nonrelativistic 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. 


#11
Feb1312, 02:48 PM

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



#12
Feb1312, 04:14 PM

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PF Gold
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An alternate approach, closer to what you've described above is the absorber theory http://en.wikipedia.org/wiki/Wheeler...bsorber_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 twobody problem, I don't believe any exact solutions are known, so perturbation theory is used to evaluate corrections to the familiar nonrelativistic solution. Firstorder 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. 


#13
Feb1312, 07:27 PM

P: 189

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] and [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 ? 


#14
Feb1312, 07:58 PM

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PF Gold
P: 2,602

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. 


#15
Feb1412, 04:51 AM

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P: 303

Stability of hydrogen in QED is way beyond current technology. However "Dirac Hydrogen", so coulomb term + dirac operator is known to be selfadjoint 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.



#16
Feb1412, 05:44 PM

PF Gold
P: 1,168

FZero, thanks. I'd like to comment:
This interesting comment on retardation on Wikipedia seems relevant: "If higherorder 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 timedependent 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?) DarMM, which atoms are unstable without QED? Why? 


#17
Feb1412, 06:08 PM

PF Gold
P: 1,168

ytuab,
It was a good success, so he got the Nobel prize. 


#18
Feb1512, 01:47 AM

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