# Prove that an electron in a hydrogen atom doesn't emit radiation [duplicate]

According to electrodynamics, accelerating charged particles emit electromagnetic radiation.

I'm asking myself if the electron in an hydrogen atom emits such radiation. In How can one describe electron motion around hydrogen atom?, Murod Abdukhakimov says the total momentum of the electron is zero, hence it does not emit radiation. Could someone prove this statement ?

It may be an obvious question, but I can't figure out why the total momentum of the electron should always be zero, in any energy state.

• Possibly a duplicate of physics.stackexchange.com/q/20003. Jun 27, 2016 at 13:49
• I'm looking for a proof of the fact that the electron of an hydrogen atom doesn't emit radiation ! Can't we apply electrodynamics to the electron by taking the exepectation value of its momentum and/or acceleration ? If we can do this, we shall prove that these expectation values are zero for any eigenstate. Jun 27, 2016 at 14:00
• Possible duplicate of Where did Schrödinger solve the radiating problem of Bohr's model? Jun 27, 2016 at 14:23

## 6 Answers

You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following:

1. Construct a sample of ground-state neutral hydrogen atoms.
2. Place this sample near a detector which is sensitive to the sort of EM radiation you expect.
3. Die of old age waiting for a signal, because ground-state hydrogen doesn't emit radiation.

This experimental evidence demonstrates that classical electromagnetism, in which accelerating charges emit radiation, does not describe the hydrogen atom.

• What if your detector is so sensitive to the EM of hydrogen it reflects back the EM to the atoms? What if the EM from the hydrogen atoms is outside the spectrum that your device can detect?
– user121963
Jun 27, 2016 at 15:28
• Well if you haven't died of old age yet, you can design a different detector with a different set of strengths and weaknesses. In the meantime develop quantum electrodynamics which describes various features of the hydrogen atom spectrum to eight or ten significant figures, and discover that in QED there is no reason to predict that the ground state of hydrogen should radiate.
– rob
Jun 27, 2016 at 22:26
• I don't think this answer is useful. The question clearly asks for a theoretical explanation which QM is perfectly able to provide, even if the OP doesn't know that. Jun 27, 2016 at 23:42
• I think this answer is useful. I don't know if the OP has misconceptions on how physics as a science works, in which case this is a good explanation, but at least it looks like they haven't heard of quantum mechanics, in which case giving a QM-based explanation would be way too detailed and a simple "classical electromagnetism doesn't work here, you need a thing called QM to explain it" is better.
– JiK
Jun 28, 2016 at 12:01
• @Javier I think it is useful. The question asks for a "proof" that hydrogen atoms don't radiate. There is no such proof because mathematics can't prove things about hydrogen atoms; it can only prove things about our models of hydrogen atoms. If you want to know something about hydrogen atoms, you must look at actual hydrogen. Jun 28, 2016 at 12:09

The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation.

If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus.

If you accept that electrodynamics applies, then you have to accept that atoms cannot exist - since they do, electrodynamics must not be the whole story.

Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects), and spherically symmetric charge distributions do not radiate (there's no special direction). Accelerated charges do not always radiate em radiation. See also How to find the magnetic field due to a revolving electron of hydrogen atom in first orbit

• But what about the excited $s$ states? They're spherical too, and they can decay. Jun 27, 2016 at 18:44
• Excited $s$ states must decay to $p$ states (or higher).
– rob
Jun 27, 2016 at 22:30
• @Javier Good point, I'm trying to answer this.
– jim
Jun 27, 2016 at 23:36

I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are now incorporated into a different algorithm, the Schroedinger equation (if we limit the expansion to semi-classical quantum mechanics), and the measurement formalism of quantum mechanics. As with many theories that have been generalized, they still work well in many realms, in this case you do not need to use quantum mechanics when dealing with the electromagnetic properties of "most" macroscopic objects, but the reverse is not true, classical electrodynamic predictions fail often when you deal with quantum objects.

In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather than something physically interesting. Here's a second answer to hopefully address that concern.

In Schrödinger's quantum mechanics the probability density $\psi$ for finding the electron in some small volume near the nucleus (charge $Z$, mass $m_\text{nuc}^{-1} = \mu^{-1} - m_e^{-1}$), obeys the differential equation $$\left( \frac{\hbar^2}{2\mu} \vec\nabla^2 - \frac{Z\alpha\hbar c}r \right) \psi = E\psi. \tag 1$$ It turns out that this equation has bound solutions with $E<0$ if, and only if, you introduce some integer parameters $n,\ell,m$ subject to some constraints: $1\leq n$, $\ell<n$, and $|m|\leq \ell$. The energies associated with these quantum numbers are $$E_{n\ell m} = -\frac{\mu c^2\alpha^2Z^2}{2n^2} = Z^2 \cdot \frac{-13.6\rm\,eV}{n^2}. \tag 2$$ Critically for our discussion, this means that there is a state with $n=1$ that has the minimum possible energy for an electron interacting with a proton. This is totally different from the unbound case, or the interaction between two like-charged particles, in which you can give your mobile particle any (positive) total energy that you like and inquire about its motion. If the total energy doesn't satisfy (2), it's simply impossible for the system to obey the equation of motion (1).

You compute transition rates in quantum mechanics using Fermi's Golden Rule: a transition between an initial state $i$ and a final state $f$ occurs in some time interval $\tau_{if} = 1/\lambda_{if}$ with probability $1/e$, where the decay constant $\lambda_{if}$ is $$\lambda_{if} = \frac{2\pi}\hbar \left| M_{if} \right|^2 \rho_f.$$ The density of final states $\rho_f$ is interesting if there are multiple final states with the same energy. (For instance, in hydrogen there are generally several degenerate final states with given $n,\ell$ but varying $m$.) The matrix element measures the overlap of the initial and final state given some interaction operator $U$: $$M_{if} = \int d^3x\ \psi_f^* U \psi_i$$ For electric dipole radiation the operator is $U_{E1} = e\vec r$; for magnetic dipole radiation, $U_{M1} = {e}\vec L/{2\mu}$; for quadrupole etc. radiation there are other operators. You could also couple to multiple photons: for instanced the $n=2,\ell=0$ state cannot decay to the ground state by emitting a single photon, since the photon carries angular momentum, but can decay by emitting two dipole photons at the same time. This forbidden transition has lifetime $\sim 0.1\rm\,s$, compared with nanoseconds for the $n=2,\ell=1$ states at the same energy. Computing matrix elements gives you some hairy integrals, so generally you let someone else do them.

You can in principle use these arguments and the Golden Rule to calculate the radiation emitted in three cases:

1. From a free electron with $E_i>0$ to a free electron traveling in a different direction with a different energy $E_f>0$. This should give a result most similar to the classical case, where you can get continuous radiation from an accelerating charge.

2. From a free electron with $E_i>0$ transitioning to a bound electron with $E_f<0$.

3. From one bound electron state to another.

It's this final option, transitions between bound states, that interests you. The salient feature, unique to quantum mechanics, is that the energies of the bound states are quantized. Unlike in classical mechanics, in quantum theory the equation of motion has no solutions with $E<E_1$. Even if you made up some trial sub-ground-state wavefunction to compute the matrix element for the transition (which can't be done, since the existing wavefunctions form a complete set), you'd find that the density of states at your hypothetical lower energy is $\rho_f=0$, so the time before the transition occurs is, on average, infinitely long.

The classical theory predicts radiation when a charge accelerates from one continuum momentum to another. So does the quantum theory. But the quantum theory also predicts bound states with quantized energies. Non-transitions from a state to itself have zero matrix element, therefore never occur; transitions from one state to another can only occur if there's a final state available.

The answers posted so far repeat the common fallacy that Maxwell's Equations do not apply to the hydrogen atom. They may not work for the Bohr atom, but they certainly explain everything the hydrogen atom does in terms of its emission and absorption of radiation. In the Schroedinger equation there is a charge density, and for the eigenfunctions of the hydrogen atom that charge density is stationary. It is therefore consistent with Maxwell's equations that the atom does not radiate. However, when the atom is in a superposition of two or more eignenstates, there is in general time-varying charge density. And so the atom emits or absorbs radiation. Furthermore, the rate at which it emits or absorbs radiation is exactly given by applying Maxwell's equation to the time-varying charge distribution. The atom behaves exactly as a tiny radio antenna would. See my blogpost "There Are No Pea-Shooters for Photons".