# How does the Schrödinger model of the hydrogen atom take into account radiation friction?

When one first encounters quantum mechanics, he learns about Bohr's model of the hydrogen atom and one of his biggest problems - electrons were accelerating and not emitting EM radiation (which is sometimes referred to as "radiation friction"). Then when you solve Schrödinger's equation with the following Hamiltonian: $$H = \frac{p^2}{2m} - \frac{e^2}{4\pi \varepsilon_0 r}$$ You get stationary states of the electron, eigenfunctions of the Hamiltonian, and no radiation friction is emitted.

However, solving this Hamiltonian classically, also doesn't give any evidence for radiation friction, because it doesn't describe the interaction between the electron and the EM field.

So my question is, why is this Hamiltonian taken to explain the lack of radiation friction in QM, whereas in classical electrodynamics it is an invalid Hamiltonian for that exact same reason? Hope my question was clear, thank you!

• “Radiation friction” was needed because Bohr’s model was wrong. – Jon Custer Oct 26 '18 at 13:31
• Of course it was wrong, but the question is about Schrodinger's model not addressing the problem. – Ofek Gillon Oct 26 '18 at 14:21
• But the 'problem' was only a 'problem' because the model was wrong - to avoid classical orbiting electrons from emitting radiation from the acceleration, one had to magically suppress the radiation. In QM the electrons aren't classical and don't orbit so it isn't a problem that needs to be addressed. – Jon Custer Oct 26 '18 at 14:24
• I'm sorry sir, you are wrong. radiation friction is a huge deal in classical EM theory, that's how antennas work. Please read more here: en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force . This is exactly why the Bohr model was considered false – Ofek Gillon Oct 26 '18 at 14:39
• Is there a dfiference between "radiation friction" and "radiation reaction"? – R. Romero Oct 26 '18 at 16:34

## 3 Answers

why is this Hamiltonian taken to explain the lack of radiation friction in QM, whereas in classical electrodynamics it is an invalid Hamiltonian for that exact same reason?

That Hamiltonian is not an explanation of "lack of radiation friction in QM". It is an Hamiltonian that does not manifest that friction, neither in classical nor in the quantum theory, because it is based on a fiction - an instantaneous Coulombic interaction, where no EM radiation exists.

The reasons textbooks suggest Schroedinger's model resolved the problem with stability of the atom are not entirely clear to me, but it is probably partially because:

• Schroedinger's model proved to be very general and successful, not only for atoms, but molecules;

• Schroedinger's model has a ground state, which the older classical model based on EM theory has not.

With these observations, it is natural to expect that whatever happens with EM interaction in the atom, the atom cannot collapse in quantum theory, because the immensely successful Schroedinger's model says its energy cannot go below certain value.

Of course, careful student will notice that this argument is unsatisfactory, because the Hamiltonian used is simplistic. It does not even obey special relativity, far from taking into account fine details of EM interaction such as EM radiation.

To answer questions related to stability of the atom, both in classical and quantum theory, in a satisfactory way, one has to include the revelations of special relativity, such as the fact that the interaction cannot be instantaneous. The most plausible direction then is to assume Maxwell's equations hold down to the atomic level and go from there, but the calculations aren't easy.

The question of stability also needs to be stated in a more specific way, including some specification of the environment the system is in, for example by stating the state of the external EM field, and how exactly are the system's particles' fields connected to their motion - whether they are retarded, advanced, or some mix of the two plus some free field component. What the old and common accounts of this problem forget (including Bohr's) is that real atoms are not it empty featureless vacuum, but they are under constant action of background EM fields, from other atoms nearby and EM radiation coming in from far distances. A molecule of hydrogen in an empty universe filled with thermal radiation may not be stable (because the radiation will break it down and the parts will go their separate ways), but in a chamber full of high pressure hydrogen it may be (because the parts cannot easily go their separate ways, due to presence of other molecules).

• Thank you for your answer! Can you please elaborate on what you wrote at the last paragraph? – Ofek Gillon Oct 26 '18 at 22:14
• @OfekGillon I've edited the answer. – Ján Lalinský Oct 27 '18 at 22:07

The Schrödinger hydrogenic hamiltonian, which does not contemplate any interaction with the electromagnetic field, is important because its spectrum is bounded from below, which means that the electron cannot decay by emitting energy if it is already at the ground state. That means that, whatever form the quantized EM field takes, the ground state cannot decay from the quantized equivalent of 'radiation friction'.

Moreover, the Schrödinger hamiltonian is important because it predicts the existence of stationary states at precisely the energies that give rise to the observed spectra using level-to-level transitions. It doesn't explain how those transitions happen, but it does produce the correct frequencies for those transitions.

If you want a full quantum-mechanical account for radiation friction, then you need to fully quantize both the atom and the radiation field, and the result (known as Quantum Electrodynamics, or QED) is an extremely functional theory, though it is not normally taught at undergraduate level because its complexity puts it a bit beyond those tools.

Once you do that, though, the resulting structure directly solves one of the biggest problems of the Schrödinger hamiltonian: to wit, the existence of excited stationary states. Those excited states, within the plain Schrödinger theory, are stationary, which means that they don't evolve and therefore that they don't decay down to the ground state, and this is in direct contradiction to experiments. In QED, on the other hand, those states morph into reasonably-well-defined resonances, but they are not eigenstates and they do not stay constant in time: instead, if the atom is in free space, those states start off with all the energy on the atom but they transfer that to field excitations (i.e. photons) that carry that energy away.

Or, in other words, the quantum mechanical equivalent to 'radiative friction'.

• Note that Schrödinger and more so, Dirac, theory also predicts the correct probabilities for the transitions. QED just produces significant, accurate but minor corrections to these results. – my2cts Oct 26 '18 at 17:49
• @my2cts Not particularly, no. There does exist a semiclassical theory (quantized matter, classical field) which correctly predicts transition probabilities, but it cannot be understood as a coherent account of the light-matter interactions in that configuration, as the energy loss isn't "fed back" to the EM field in any meaningful way. The shift from that semiclassical theory to QED isn't to provide "minor corrections" - it's required to provide anything that looks like a solid conceptual foundation. – Emilio Pisanty Oct 26 '18 at 18:14

There is no need to consider radiation friction because both classical mechanics and classical electrodynamics are wrong. It turns out that electrons can't be described as accelerated particles, but have to be described by wave function in quantum mechanics.

Classical electrodynamics is also modified in Quantum mechanics. The correct quantum mechanical description of electrodynamics is given by QED. It turns out that electromagnetic field is also quantised. You are right in saying that the Hamiltonian that you wrote is incomplete. Even in Quantum mechanics, you have to include a term describing interaction of electron with EM field. However since electromagnetic radiation is quantised, an atom cannot continuously emit radiation, so there is no problem of radiation friction.

Edit : We can describe an atom as isolated electron nucleus system with an external potential describing interaction with EM field. An atom has quantised states and it is impossible to go below n=1 state, called ground state, because no such state exists. However even if an atom is in excited state, it cannot continuously emit radiation, because electromagnetic field is quantised.

• I think this answer may be misleading. Including the interaction term between electron and electromagnetic field does describe the process wherein an electron falls from one level to another with the emission of a photon. The reason we don't get the electron radiating until it falls into the nucleus is simply that there are no states of the system below the $1s$ orbital, i.e. there's nowhere for the electron to fall! – DanielSank Oct 26 '18 at 16:07
• That's true, and maybe I should have mentioned that, but I wanted to avoid the simplistic explanation I have seen that since electrons are in stationary states, they don't emit radiation. – Rohit Gupta Oct 26 '18 at 16:10