# Why doesn't the electron radiate?

1. Jan 11, 2007

### Quantum River

In Bohr model of Hydrogen, the electron will not slowly lose energy as they travel, and hence will remain in stable, non-decaying orbits.

Let me ask a "silly" question. Is the electron in Hydrogen ground state stationary or moving? If it is moving, surely it will accelerate and it will radiate according to Maxwell equations?

So why Bohr assumption above works? Please tell me why the Maxwell equations do not work any more in the hydrogen atom.

2. Jan 11, 2007

### Quantum River

Einstein's peculiar reasoning of spontaneous emission?

Einstein's peculiar reasoning of spontaneous emission
Spontaneous emission is the result of quantum mechanics. So we need not use some theorems of thermodynamics to duduce spontaneous emission.
But in Einstein's reasoning of spontaneous emission, Einstein used Boltzmann distribution and Planck black-body radiation law. But the temperature T part cancels each other of both sides of the equation. So at last in the formula of spontaneous emission the temperature T will not appear any more.
But is it proper to deduce a quantum formula using thermodynamics? Remember Boltzmann distribution is definitely a theorem of thermodynamics.

Last edited: Jan 11, 2007
3. Jan 11, 2007

### vanesch

Staff Emeritus
I would say that the quantum state is not changing. That is: the point in projective state space is not changing for a stationary state: there is only a non-physical phase factor to be added to the hilbert vector but this doesn't change the point in state space, which is the ray in hilbert space.

4. Jan 12, 2007

### Staff: Mentor

Or, to put it in other words, when the electron is in one of the Bohr energy levels, its probability distribution $\Psi^*\Psi$ does not change shape with time.

When the electron "jumps" from energy level m to energy level n, we can think of it as being briefly in a superposition of the two wave functions: $a\Psi_m + b\Psi_n$. This superposition has a probability distribution that does change shape with time. In fact, it pulsates or "sloshes" periodically with a frequency equal to that of the light emitted or absorbed in the transition!

5. Jan 12, 2007

### dextercioby

The trouble with this view is that Maxwell's equations describe the classical em field produced by a classical particle. For the H atom, the electron is NOT a classical particle anymore.

The Maxwell's eqns have nothing to do with the H atom when the latter is described using the QM formalism.

Daniel.

6. Jan 17, 2007

### reilly

For my money, Bohr's original theory of the H atom is the single most creative bit of physics of the last century -- Einstein's explanation of the photoelectric effect and Special Relativity are close seconds.

Bohr recognized that the physics known at the time -- Maxwell, Newton, Lorentz,.... -- was at total odds with observed phenomena like atomic spectra. So, basically, he worked backwards -- what would have to be "true" in order for hydrogen to behave as it did and still does. Out goes Maxwell because Maxwell ruled against atomic stability for the Rutherford nuclear model of the atom. So Bohr assumed the totally crazy idea that electrons moved in orbits without any radiation -- again, the operative word is assumed. Within this approach, the only way an atom could radiate is to change energy by means of an electron "jumping" from orbit to orbit -- a jump down in energy allows a photon to take said energy and thus create the atomic spectra.

While he used classical physics to relate the orbits to the Coloumb forces involved, he threw out the classical theory of radiation. That is, he said that Maxwell does not apply to atomic radiation. Why? Because ultimately his assumptions lead to brilliant agreement with the spectra of atomic hydrogen. He made it all up.

Although today we have a far more sophisticated quantum theory of atomic radiation, which has resurrected Maxwell in the quantum world, the basics of Bohr's ideas still hold today -- stationary states, radiation from transitions between states. and so on.

Regards,
Reilly Atkinson

7. Jan 17, 2007

### Sojourner01

Does it then follow that the 'wavetrain' of a single photon - that is, the spatial distance between where the photon 'begins' and 'ends' is precisely equal to c * the duration of the intermittent state?

*lightbulb winks on* ;)

8. Jan 17, 2007

### ZapperZ

Staff Emeritus
But I think the value of the Bohr model is significant only in the historical sense. It is similar to one coming up with the caloric theory of heat and happens to get some of the thermodynamics properties right. I still don't think that it should be introduced in school as anything other than history. We can already see the level of confusion and wrong impression it creates not only on here, but everywhere else. As an ad hoc model, it served its purpose at that time, the same with the caloric model. We should let both of them rest in peace knowing that their roles in advancing physics are secure.

Zz.

9. Jan 18, 2007

### vanesch

Staff Emeritus
In a way, yes.

Another way of seeing that, is that the EM bandwidth of the emitted radiation is (by a property of Fourier transforms) related to the average length of the wavetrain in time (inversely proportional). And the bandwidth of the EM radiation also gives us the "lifetime" of the exited state (in other words, the time that this excited state is still present in the overall state) through an inversely proportional relationship.

This is a bit handwaving when stated this way, but the essence is there.
You can picture somehow that the excited state in superposition with the ground state "wobbles the charge density", which emits some EM radiation, the time it takes for this excited state to die out, at which point the charge "doesn't wobble anymore" and the EM radiation train is finished.
This is a semi-classical picture of course (we treat the EM field here classically, with a source which is the probability density of the charge, which is not entirely correct, but a good semiclassical approximation). But it comes out correctly concerning lifetimes, wavetrains, and bandwidths.

10. Jan 31, 2010

### namboori007

hoo :-?...is it accelerated charge or oscillating dipole that radiates??...

11. Jan 31, 2010

### bobydbcn

Your explanation is very intrinsical! I can understand that the state of classical particle and quantum object is different. the reason why classical particle and quantum object is different. Maybe quantum object is more appropriate to the real world.