# Attraction of + and - charge

1. Aug 28, 2014

### Boundless

if a hydrogen atom has only 1 proton and 1 electron what is the reason that the electron does not collapse onto the proton if there is a + and - charge?

2. Aug 28, 2014

### ShayanJ

3. Aug 28, 2014

### Staff: Mentor

I moved this to QM where you should get some good information.

4. Aug 28, 2014

### Matterwave

This was in fact one of the conundrums faced by classical physicists who started out with the model of the atom as a + charge and a - charge separated by a distance.

A first guess might be that one charge is orbiting the other, just as the planets orbit the sun. The charges don't collapse, because they are in orbits.

The problem with this explanation is that classically speaking, a charge in orbit should radiate EM waves (since it's accelerating!). The EM waves would steal energy from your system and the charges would again collapse (in a very short time too).

People saw that actually the charges don't collapse...so...the founding fathers of QM (e.g. Niels Bohr) basically took it by fiat that the charges can not decay to below some ground state. Essentially there is a hierarchy of states that the charges can occupy, and there are no more states below the ground state, so the particles can't decay and collapse into each other. This is more or less taken as fiat, one of the assumptions of the QM theory.

5. Aug 28, 2014

### fluidistic

I don't think it's taken as a fiat nowadays. If we take for granted that the Schrödinger's equation is valid, then you naturally get a ground state for the energy for the electron.
But I agree that you have to take the Schrödinger's equation validity as a fiat. :)

6. Aug 28, 2014

### Matterwave

You are right, I should have specified in my post that I was referring to the semi-classical Bohr atom picture. However, by the time you get to the Schrodinger equation and the wave mechanics formalism, this question, which is a semi-classical question, becomes very odd since it no longer becomes clear what it is meant by the "collapse" of charges onto one another since they are both just probability densities. For example, the ground state of the hydrogen atom actually has the electron's distribution's mode being at r=0, with the probability distribution an exponentially decaying function of r...

7. Aug 29, 2014

### tom.stoer

I don't agree that non-rel. QM with Schrödinger equation provides a full explanation of the stability of atoms. In principle one would have to prove this using QED, but this is out of reach, unfortunately.

The reason I disagree is the following: in classical electrodynamics the electron would radiate away energy as electromagnetic waves; so we have to explain the mechanism which forbids this radiation. In non-rel. QM the electromagnetic field is not a dynamical, quantum mechanical degree of freedom (like the electron) but a fixed, classical field. So the quantum mechanical setup cannot explain why no photons are emitted b/c it does not contain photons by construction! It's an approximation to the full theory obtained by integrating out the photon d.o.f.

So I would say that non-rel. QM provides a strong hint that atoms are stable b/c they have a stable ground state; but this is not a rigorous proof.

8. Aug 29, 2014

### Boundless

Thank you DaleSpam

9. Aug 29, 2014

### Delta²

This will sound crazy but why do you thank only Dalespam and not the other people that answer your question?

xD!???!

10. Aug 29, 2014

### Boundless

I should have expounded....I thanked that person because he or she put my question into the correct category.

11. Aug 29, 2014

### vanhees71

Here, I disagree. Starting with non-relativistic atomic physics is well justified by the fact that it is much simpler than the fully relativistic picture and also a good approximation as long as you study not too heavy atoms.

Particularly the hydrogen atom is a prime example for the fact that the non-relativistic approximation gives an excellent description, at least on a quantitative level. What's going into this standard calculation is the idea that the electron and proton are bound by the instantaneous Coulomb force. This makes it, of course, completely unrelativistic, but it's still a good approximation! Also qualitatively you get an approximately correct answer: The hydrogen atom doesn't radiate in this model, because you are considering a stationary state, i.e., nothing is moving in the atom. It's thus just a statically bound system. Of course, this is not possible to achieve in any classical picture.

The full QED calculation involves a bit more fancy technique. The right hint is that you get an even better description of the hydrogen spectrum by simply using the Dirac equation in a semiclassical way, i.e., putting an external static Coulomb field for the nucleus. This gives you the correct fine structure of the spectrum.

To get this in a full QED treatment the right ansatz is to work in Coulomb gauge, use the soft-photon resummation to justify the Coulomb field as substitute for the proton and then do systematically perturbation theory around this configuration, including the quantized em. field. This leads to a perfect description of the spectrum, including radiative corrections like the Lamb shift. This program has been driven to at least 4 (or even 5?) loops now, and the agreement between theory and experiment is overwhelming.

You find the principle technical details of this calculation in

Weinberg, Quantum Theory of Fields, vol. I

12. Aug 29, 2014

### tom.stoer

Correct - but this is not the point.

No. In this model the atom does not radiate b/c it simply can't radiate. There is no radiation degree of freedom! The fact that there is no radiation is not a result but an input of the model!

(the model gives you a well-defined ground state, that's a success)

Think about a potential which does not provide a well-defined ground state, i.e. V(r) ~ 1/rn. There is no radiation in this model, either!

Correct - but again this is not the point. The Dirac eq. does not contain a dynamical photon field, either.

This is the correct starting point. The remaining problem is that the perturbation series is not well-defined mathematically :-(

But I think you understand what I want to say: to prove the absence of photons you must not start with a model w/o photons.

13. Aug 30, 2014

### Boundless

This is a very interesting article I found :
This is a great question! I will give you a little bit of data, and I think you will have an "aha" moment.

Experiments have shown, to date, that the lifetime of an electron is more than 4.6X10^26 years, or essentially infinite. (http://pdg.lbl.gov/2006/tables/lxxx.pdf) Experiments have also shown that the lifetime of a proton is > 1.9X10^29 years, or also essentially infinite. (http://pdg.lbl.gov/2006/tables/bxxx.pdf)

The lifetime of a free neutron, however, is only 885.7 seconds, or 14 minutes and 52 seconds. (http://pdg.lbl.gov/2006/tables/bxxx.pdf) By free, I mean that the neutron is wandering around by itself and is not part of a nucleus. Have that "aha" moment?

What this means, as I'm sure you figured out, is that the free neutrons decayed away in the very early universe before they could "meet up" with a proton and an electron and make a hydrogen with a neutron (also known as a deuterium). In the early universe, so much hydrogen was made that it began to coalesce into clumps due to gravitational attraction. These clumps grew and grew, and eventually, became stars. Inside stars, hydrogen atoms "fuse" to become helium, and then some of the helium fuses with other hydrogens or other atoms to make some of the heavier atoms, on up to iron. The so-called "binding energy" of each nucleus is less and less as you go up in atomic number, meaning it is more energetically favorable to be in that state, again, until you hit iron. See http://en.wikipedia.org/wiki/Nuclear_fusion for more details on how fusion works.

Could this mean that the proton electron don't really have an affinity for each other until the neutron comes into the picture?

14. Aug 30, 2014

### vanhees71

In this sense I agree, of course :-). Indeed the quantum-field theoretical treatment shows that you always get out photons from an excited state. Except the ground state all "bound states" of the semiclassical theory (i.e., with the quantized radiation field being neglected) have in fact a finite life time due to spontaneous emission, i.e., due to the coupling of the electron to the quantized (!) electromagnetic field, which is a bosonic field.

I disagree with the statement that the perturbative treatment is mathematically not well defined. It's perfectly defined, and it is even done down to the point where you "get the numbers out" which can be compared to experiment, and the agreement is overwhelming.

Of course, you need renormalization theory, and the Dyson series is very likely not to be a convergent series at all (even in usual quantum theory you can show that the perturbation series of simple anharmonic oscillators in one dimension are not convergent; I think there is a nice treatment in Hagen Kleinert's book on path integrals, but I have to check this). It's to be seen as an asymptotic series (in the sense of small coupling constants or small $\hbar$).

15. Aug 30, 2014

### tom.stoer

This is what I mean. Results derived from perturbation theory do not have the status of a mathematical theorem.

16. Aug 30, 2014

### Matterwave

How in the world did you draw your last conclusion from your previous discussion? I do not see how that statement follows at all...

17. Sep 2, 2014

### geoduck

Why doesn't the uncertainty principle provide a ground state for 1/r^n potential? If the particle falls into origin, then it gets localized so it's momentum should get large and be kicked out of origin.

18. Sep 5, 2014

### tom.stoer

As far as I remember the Hamiltonian for 1/r^n with n≥2 is not well-defined (or self-adjoint) w/o additional constraints.

But that's not the point: all what I want to say is that in non-rel. QM problems with potentials w/o stable ground state there are still no photons. You can use other potentials w/o ground state, if you like. The point is that "stable ground state" and "absence of radiation" are two different statements; they are by no means equivalent.

Last edited: Sep 5, 2014