Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus.

Now Schrödinger describes the electron as a wave function. His theory is able to describe all atoms (in contrast to Bohr's model), but how is the radiation problem solved? I understand that the wave has not an exact position in the time anymore. But the electron still "moves", so it has an acceleration anyway (because of vibrations or so).

Why is in this theory the electron not radiating anymore? And if it is, why doesn't the atom collapse?

share|improve this question
add comment

8 Answers

up vote 7 down vote accepted

I think the problem here is with E&M, and it is in the assumptions implicit to the question. Here is the assumption:

the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus.

This statement can be demolished in short order, considering the topic of a non-radiating condition. The definition of the condition is:

the conditions according to classical electromagnetism under which a distribution of accelerating charges will not emit electromagnetic radiation.

The statement that acceleration leads to radiation and therefore collapse must be completely abandoned. We still have room for skepticism - formally we are arguing over whether the Bohr model of an electron fits the radiating condition.

I will argue that there is not a simple answer "yes" or "no". The reason is that we could currently re-interpret it as a model where the answer is that it does not fit a radiating condition. The picture of the Bohr atom I am most familiar with looks like this:

Bohr atom

This is, an electron moving around a nucleus doing two things, a) exhibiting orbital behavior and b) fitting the orbital circumference to a mode of its de Broglie wavelength. But is the electron a particle or a wave in this interpretation? Following from that, does the electron have a defined location? If it's a defined particle, then it must radiate. But then why the requirement that the wavelength fit in the first place?

Let's turn to a classical analog. Imagine a ring of wire with no resistance and current flowing. This is an ideal electromagnet. It has a static electric field and a static magnetic field, but nothing is changing. It will sit there and maintain the current forever. (img source)


What's more, we could paint a picture of a simple electron orbital that fits this. Mentally combine the above two images. You'll probably be confused about the peaks and troughs of the wave, since they are not smooth with respect to position along the orbit. But this is an electron wave, not a EM wave. You can not distinguish the peaks and troughs in reality because the probability density is the imaginary wave function squared.

This doesn't describe all orbitals perfectly. For one, the electron in this model has a net angular momentum. But still, this gets surprisingly close to a passable model of nature. I think this is the correct stepping stone to move between the Bohr atom and real QM. The only question not sufficiently answered is why the electron arranges itself in this torus-like state, and the answer is that it doesn't because the actual dynamics are explained by QM, which not only accurately describes the emission lines (better than the orbital calculations confined to wavelength intervals), but also correctly predicts chemical bonds and the entire world around us.

What Schrödinger (really QM in general) did to advance beyond the Bohr model was to give a correct physical justification of the use of electron wavelength - leading to the wave equation. In this regard, the Bohr model was a confused mess, but it could be tweaked to be non-radiating per my above argument, which entails non-localization of the electron along the circumference of orbit. Obviously this is nonsense, because if it's non-local along that line, then why isn't it non-local along other dimensions as well? QM answered this correctly.

share|improve this answer
I'm losing you from the classical analog. But my question rather about the solution of Schrödinger, than the solution of Bohr. As I mentioned in a comment on an answer below, one is able to calculate (quantummechanically) a current density in the hydrogen atom, so charge is moving (and thus accelerating, since otherwise it would escape the atom). Do you mean with your classical analog, that a current flowing in a ring won't radiate either? And if yes, why is this so? Thanks already for your elaborate answer! –  BNJMNDDNN Jun 17 '13 at 21:11
@BNJMNDDNN The latter, meaning "current flowing in a ring will not radiate". One problem is that a single electron flowing in a ring will radiate. Do you see the issue? So then the problem becomes determining what the nature of the electron-wave in the Bohr model is. If you follow the Wikipedia article on non-radiating condition they argue that the Bohr model did assume a non-local electron, and thus, never had the problem to begin with. I have heard others make the same argument as the OP, so the interpretation of the Bohr model might not be universal. –  AlanSE Jun 17 '13 at 21:16
"OP"? So when stating that the Bohr model isn't correct because of Larmor's law (as is done in almost every introduction in quantum mechanics or chemistry), is actually not really correct? And the actual problem with Bohr's model is that it cannot describe other atoms/molecular systems? (I don't think the angular momentum of the electron is an issue, since I don't know how they'd measure experimentally that it hasn't got one.) And the intrinsic problem is actually that the classical particle concept just leads to wrong conclusions, and therefore one needs a wave description? –  BNJMNDDNN Jun 17 '13 at 21:46
@BNJMNDDNN, for hydrogen atom states of definite energy (stationary states), the probability density is constant with time. Indeed, in the Bohmian picture, the electron doesn't move for a stationary state. –  Alfred Centauri Jun 17 '13 at 22:09
@BNJMNDDNN That means that the quantum mechanics textbooks use a different definition for Bohr's atomic model than the Wikipedia article does. The Larmor's law argument must assume a local electron because the equation for the radiated energy requires a time-dependent charge distribution. I was correct to be careful about describing Bohr's model. I can't pick a winner among the different sources because this is definitional. It's a question for historians. "OP" means opening post, which refers to the text of your question. –  AlanSE Jun 17 '13 at 23:38
add comment

In any proper quantum mechanical understanding of the atom, a bound electron does not have a position and follow a path (i.e. have a time-varying position) in the sense that it would have in a classical or semi-classical theory.

Instead the electron "has a state" or "occupies an orbital" (an orbital not a orbit!), and because there is not a path there is not an acceleration associated with the path.

This poses a problem when you ask "Well, does it radiate or not?" because at first there is no theory for interaction of electromagnetic fields with "orbitals". You need to develop a new theory (eventually QED).

So the answer is that Schrödinger didn't fully solve the problem. He just said, "it doesn't have to an acceleration in a classical sense" and left it at that.

share|improve this answer
But what actually bothers me too, is that one is able to calculate a current density in the hydrogen-atom (with j=h-bar/(2.m.i).(phi*.\nabla phi - (\nabla phi*).phi), with phi ofcourse the wave function). Solving this with spherical coordinate system will return an azimutal current, which actually corresponds with the Bohr-model in which the electron orbits around the nucleus. So there is some charge moving around the nucleus. To keep the charge around the nucleus, it should have some acceleration towards the nucleus sometimes (otherwise it would escape). –  BNJMNDDNN Jun 17 '13 at 20:44
@BNJMNDDNN: The current density isn't time-varying. See this answer physics.stackexchange.com/a/70201/4552 , which unfortunately was posted on a duplicate question. –  Ben Crowell Jul 6 '13 at 23:01
I know it's not time-varying, but a loopcurrent doesn't have to be time-varying to have accelerating electrons, since they'll have a centripetal acceleration while travelling in a circle. –  BNJMNDDNN Jul 9 '13 at 9:19
@BNJMNDDNN Classically a loop has an acceleration. That's fine, but this is quantum mechanics and you have to consider the 2nd partial time derivative of state, which is zero because the state is independent of time even in those states that have non-zero angular momentum (s-states don't even have that). –  dmckee Jul 9 '13 at 14:32
add comment

Schrodinger describes the electron in an atom as a standing wave pattern. Essentially it's a particle in a box. A particle in a box has a maximum wavelength, and therefore a minimum kinetic energy. This is the ground state. In an actual atom, the details differ, but you still have a ground state, which is a state of minimum energy.

Due to conservation of energy, the ground state can't radiate. If an electron in the ground state were to radiate, it would be radiating energy, but it doesn't have a state of lower energy that it can go to.

share|improve this answer
add comment

To add yet another voice, I believe that Bohr got around the radiation problem by simply stipulating that, for reasons not yet understood, electrons in certain special orbits simply did not radiate.

His theory certainly allowed radiation as when an electron moved from one orbit to another. But the predicted orbits in Bohr theory were to be "special".

share|improve this answer
Yes I know that, but I think that's kind of a cheap way to solve such a fundamental problem, don't you? Anyway, thanks for your answer! –  BNJMNDDNN Jun 18 '13 at 6:39
@BNJMNDDNN: I understand you feeling of "cheapness", but really, this is the difference between physics and mathematics. Even in the latter you take axiom systems as a starting point. Sometimes this "cheapness" is just the right thing to do and becomes genius! –  WetSavannaAnimal aka Rod Vance Jul 7 '13 at 2:53
Ofcourse but it's always best not to postulate your things, but solve them. Instead of the postulations of Bohr with his orbits, Schrödinger came (postulated) with a new equation which solved a lot of those postulates. And since there are no other allowed energy states, it's kind of logic there can't be a transition in state or radiation (just like the energy gap hasn't got allowed states in a semiconductor). The difference is that Bohr postulated this, and Schrödinger gave a sort of "proof" (although he had to postulate something too). Thank you all –  BNJMNDDNN Jul 9 '13 at 9:31
add comment

As discussed in other answers, the short answer is "he didn't", in the sense you seem to be asking. The electron is simply not a classical object - it's behaviour is something wholly new (as of 1926 that is, when Schrödinger published his equation). The "solution" was that his approach foretold the hydrogen spectrum to an accuracy that was much greater than Bohr's model: in short the "solution" ultimately came experimental verification. You may benefit simply by reading more on the history of QM if you have not already done so so that you understand that great minds have had almost exactly the same kind of misgivings as you seem to be having. To invoke Feynman here (the exact words are on the audio version of his QM lectures) how does one come up with a new theory? - step 1 - you guess it, step 2 you do experiments to test it, step 3, if the experiments tell against the theory, no matter how clever or appealing it is, then it's wrong and you go back to step 1.

As an aside, it might help (I know this too might seem highly artificial) to know that there is a formulation of QM (the Heisenberg picture) wherein the electron is perfectly "still": its state does not change and instead the "observables" - the operators + the special "recipe: which tells you how to interpret them and whose eigenvalues are the possible measurements - are the things that evolve with time. This Heisenberg "matrix mechanics" is made equivalent to the Schrödinger picture through a unitary (i.e. roughly something that preserves probability distributions) transformation that evolves with time. The Heisenberg picture is kind of analogous to doing mechanics in a rotating frame, but it may help to know of this approach and anyway who's to say, without further experimental justification, which is the rotating frame!

The "solution" discussed took a long time to be truly embraced. Schrödinger came up with his famous cat thought experiment because he thought the Copenhagen interpretation of his and Heisenberg's theory (Heisenberg's matrix mechanics and Schrödinger's wave mechanics were shown to be the same) was crazy and needed to be repudiated - he ultimately left the field; Rutherford was aghast at what he thought was the possibility of "half an electron" being somewhere and the other half somewhere else (these thoughts provoked discussion as to whether even our conception of space as being modelled by a smooth manifold were even valid at the atomic level) and Einstein famously fought and thought so hard against what QM seemed to be saying that he (and Podolsky and Rosen) came up with the famous EPR paradox which they thought repudiated the prevailing QM interpretations but instead led to the discovery of quantum entanglement. I find it ironic that even if Einstein had done nothing else than try to tear the probabilistic interpretation of QM down, he would still have been one of the greatest physicists of the 20th century.

Lastly, as animals we have evolved to recognise and understand patterns that arose in our birth home, to wit the wet savannas of Eastern Africa. From a evolutionary biology standpoint, there is absolutely no reason why we should understand things like electrons, let alone why they should fit into our "Wet Savannah World" view of the world studied in classical physics. You just have to accept that the electron is standing still, de-localized and so somehow spread over all the points in its orbital at once and not shifting from one to the next. The weirdness and psychological jarring that you feel in letting go of the idea of an electron as a point buzzing around reflects that your evolutionary forebears came across nothing truly analogous to the delocalised electron in their wet Savannah home. This is not to say we give up because it's too hard - we simply have to heed that we have certain biologically programmed prejudices about the World that sometimes help us in phyics as intuitional insight, but other times they hinder us and point the wrong way.

You may find it helpful not to think of separate electrons as the fundamental things, but rather the oneness of the quantum electron field: the electrons themselves are like the discrete "data packets" whereby this field communicates with and interacts with the other quantum fields in the World and they don't needfully have to be anywhere in particular - but again, you have to be wary of leaning too hard on any analogy. Please see this excellent elementary level video http://www.youtube.com/watch?v=Fxeb3Pc4PA4&list=UUUHW94eEFW7hkUMVaZz4eDg.

Further Reading: Heisenberg very much chose to concentrate on what could be measured rather than what was happenning to give rise to the measurement and this "quantum measurement theory" approach is something some people find appealing. This is where one goes back to a set of experimental observations about "observables" (the so-called "operators") and uses them as axioms without trying to get too caught up in other ideas. I have a set of undergrad notes along these lines by Hideo Mabuchi which I thought were excellent - I couldn't find them on the web but you might like to contact him through the link.

share|improve this answer
add comment

I will put in my two cents:

The Bohr model as such can be saved by postulating standing waves for the electrons. The contrast with the Schrodinger formalism lies not only, as observed by others , that the solutions of the Schrodinger equations are more accurate and can be generalized to complicated potentials, but that the Bohr model is only one step higher than the numerology/data-fitting of the Lyman and Balmer series and the Rydberg formula. It is not a theory.

What made Schrodinger's equation lead to a theory, was the postulate that the solutions of his wave equation were to be squared and interpreted as a probability distribution of the possible position of the electron. This is that led to a theory of quantum mechanics, rather than a model of quantized states. The position of the electron was interpreted as a probable one, leading to orbitals, and not orbits. In addition within this theory the eigenstates of these orbitals were stable, and thus not radiating.

share|improve this answer
add comment

Here is my answer to this very difficult issue. In my opinion the elementary Schroedinger approach does not solve the problem of radiation. The electron still radiates when it changes its energy level and this process is not described in the elementary Schroedinger model based on the Coulomb potential only. Experiments prove that every level is not stable, only the fundamental level is stable. This is the only true difference with respect the classical description, where there is no lower bound to the energy of the electron.

Let me be more clear. If you consider the classical model neglecting electrodynamics, but only considering Coulomb interaction, even the classical model is stable. The same happens to the Coulomb-Schroedinger model neglecting the interaction with the quantized electromagnetic field (in the vacuum state in general). The energy levels seem to be stable, but they are not as experiments show. In fact, also theoretically, once you switch on the interaction with the E.M. field, exactly as in classical electrodynamics, all levels but the fundamental one become unstable in accordance with experiments. The only true difference form the classical description including all electrodynamics (barring the fact that the allowed energies are now discrete) is the existence of the ground state in the quantum picture.

So a more right question would be instead "why, differently form classical physics, there is a ground state for the Coulomb attractive potential in the Schroedinger model?"

share|improve this answer
add comment

Actually Bohr's model solved the radiation problem by setting a minimal value for the angular momentum of the electron. Hence, in his model, radiation or not, the system cannot go lower in energy.

Now Bohr was wrong in his model mainly because for instance it was completely incompatible with a zero angular momentum which incindently happens to be the ground state of the hydrogen atom. In this respect it was an epic fail but otherwise it was a very nice model that solved many problems.

share|improve this answer
That formally answers the question, but like many physics questions, people thirst for a specific retort to "why doesn't the moving charge emit radiation". I'm not savvy enough on E&M specifically, but my thinking derived from gravitational radiation problems is that an unchanging charge density function allows there to be no radiation, even if the insides are moving. If you had 2 wires tapped together with opposing currents they could cancel the magnetic field out. I think this is more satisfactory for the specific objection. But I could be wrong. –  AlanSE Jun 17 '13 at 18:39
It is also a minimal value for the change of angular momentum. –  arivero Jun 17 '13 at 18:43
In my opinion, Bohr didn't really "solve" the problem by setting the minimal value. He just "postulates" what's wrong and makes it right by convention (kind of a silly trick which is morely done in physics). But as Larmor states, any charged particle which accelerates, will radiate away energy in the form of electromagnetic waves. So why won't the electron radiate in this case? I thought the electron still has angular momentum in the ground state, and thus centripetal acceleration... –  BNJMNDDNN Jun 17 '13 at 18:50
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.