How exactly does an electron falling back to its original state from an excited state produce electromagnetic waves?

Question

TL;DR Alternating currents create EM waves, that is quite clear. But why do electrons falling back to their original state create EM waves? Is there are clear explanation like alternating currents? Or is it just what we have observed?

Thanks in advance.

Both in Chemistry and Physics class, I have been told that when the right amount of energy is given to an atom, its valence electrons absorb the energy and jump into a higher energy level. When it comes back down, it has to lose its energy, and emits it in the form of electromagnetic waves, which may or may not be in the visible range.

While my teachers move on to the next thing, my question of course is why? I know that alternating currents in a wire is one thing that creates electron magnetic waves, and that is quite clear: currents create magnetic field, alternating currents create alternating magnetic waves, and alternating magnetic waves propagate EM waves.

But why for the case of falling electrons? My only possible theory is that electromagnetic waves are produced when charged particles move, but then EMs should be emitted when it jumps to a higher level.

Answer 1

This is a nice question and has a rather illuminating answer.

Let's start simply, if I may, by considering 1 dimension rather than 3 and an infinite square well potential rather than the $1/r$ Coulomb field. If the well goes from $-a/2$ to $+a/2$ then it has solutions $\psi_n(x)=\sqrt{2 \over a} \cos{n \pi x\over a}$ when $n$ is an odd positive integer and $\psi_n(x)=\sqrt{2 \over a} \sin{n \pi x\over a}$ when $n$ is an even positive integer, and these have energies $E_n={\hbar^2 n^2 \pi^2 \over 2 m a^2}$ and the full wavefunctions including the time dependence are $\Psi_n(x,t)=\psi_n(x)e^{-iE_n t /\hbar}$. Standard stuff.

Notice - it's trivial but important - that the mean position of the electron $\langle x \rangle =\int {\Psi_n}^*(x,t) x \Psi_n(x,t) dx$ is zero for all $n$, as $x$ is odd and $\psi(x)^2$ is even .

Now consider an electron which starts in an excited state, say state 2, and decays to a lower state, say state 1, the ground state. Initially it is in $\Psi_2(x,t)$ and finally in $\Psi_1(x,t)$. In the middle it is in some superposition of the two. $\Psi(x,t)=A\Psi_1(x,t)+B\Psi_2(x,t)$. The energy during this (short) intermediate period is not defined but that's OK because of the uncertainty principle. $A$ and $B$ are functions of time and normalised to one, but these details don't concern us right now.

Now $\langle x\rangle$ is $\int \Psi^*(x,t) x \Psi(x,t) dx$ which is $\int (A{\Psi_1}^*(x,t)+B{\Psi_2}^*(x,t)) x (A{\Psi_1}(x,t)+B{\Psi_2}(x,t)) dx$

This contains terms $|A^2||\psi_1^2|x$ and $|B^2||\psi_2^2|x$ which vanish as before, but it also contains a couple of cross terms

$[AB {\Psi_1}^*(x,t){\Psi_2}(x,t)+ AB {\Psi_1}(x,t){\Psi_2}^*(x,t)]x$

Putting in the expressions for $\Psi_1$ and $\Psi_2$ turns this into

${2 AB\over a}[e^{i(E_2-E_1)t/\hbar} + e^{-i(E_2-E_1)t/\hbar}]x \cos{\pi x \over a} \sin{2 \pi x \over a}$

The space integral does not vanish, as it is the product of an even function and two odd functions. The time dependence looks like $\cos(E_2-E_1)t/\hbar$.

Now remember that the electron has a charge. What the maths tells us is that during the transition there is a dipole moment which oscillates with frequency $\omega = (E_2-E_1)/\hbar$, i.e.$f = (E_2-E_1)/h$. It is behaving as a little dipole radiator oscillating at just the right frequency to emit the EM radiation corresponding to the energy transition (i.e. the photon)

Notice that we pick up for free the selection rule that such transitions can only occur between odd and even states. In 3D such rules are more complicated but are basically just saying that the integral of the product of the space parts of the two states involved must not vanish.

Answer 2

This is really a classical question about a non-classical process.

In regular quantum mechanics, an electron doesn't need to move physically to change states, since for some $x$, $\psi_i(x)$ and $\psi_f(x)$ can both be non-zero, so if you say the electron is at $x$, it can be in any either $\psi$. In fact, that's generally how QM is calculated under a change of potential: you just project the old state onto the new states without any change of $\psi(x)$.

In our best description of these things, quantum electrodynamics, there is an initial state (excited atom) and final state (atom in the groundstate + a photon). What happens in-between is: everything, but that is not tractable, so we take approximations.

If you look at the 1st order term in the interaction that causes the atom to relax, you see 3 things: a photon creation operator, an electron destruction operator and an electron creation operator.

So, according to that, the electron doesn't fall, nor does it jump. It is destroyed, and a "new" electron is created in the ground state, along with a proper photon.

I put "new" in quotes because no electron has an identity versus any other electron.

In the absorbing case, the incoming photon is annihilated by the appropriate operator.

Answer 3

Both in Chemistry and Physics class, I have been told that when the right amount of energy is given to an atom,

A photon is an elementary point particle in the standard model of particle physics. It has to be the right energy photon, where $E=hν$ where $h$ is the Planck constant an $ν$ is the frequency of the classical electromagnetic wave that would emerge from millions of such photons. To see experimental evidence of the difference between photons, which are quantum mechanical point particles, and light see this answer of mine.

its valence electrons absorb the energy and jump into a higher energy level.

This is not what happens. The whole atom absorbs the energy with the result to find the electron at the higher energy state.

When it comes back down, it has to lose its energy,

Means that the atom deexcites, and the lower energy level is occupied by the electron.

and emits it in the form of electromagnetic waves,

This is wrong. A single atom does not emit electromagnetic waves. It emits one photon. You should know that in matter there are of order $10^{23}$ atoms in a mole. Light, classical electromagnetic waves, emerge from a confluence of the zillions of photons from bulk matter. This can be shown mathematically using field theory, but in the link I gave you there are experiments showing how the light behavior comes out of superposition of individual photons.

which may or may not be in the visible range.

This is correct.

But why for the case of falling electrons?

The electrons do not rise or fall. Atoms are described by quantum mechanical equations which give solutions in terms of probability . When the photon energy is in the difference between two atomic energy levels, the atom absorbs the energy and the electron is at a higher energy level, not in an orbit, but an orbital, a probability locus.

When the atom goes back to the lower energy level, there is a calculable time in QM for this to happen, a photon is emitted, as described above. See how the energy levels of the hydrogen atom are.

My only possible theory is that electromagnetic waves are produced when charged particles move, but then EMs should be emitted when it jumps to a higher level.

The first quantum mechanical model , the Bohr model of the atom, considered such a semi classical scheme . To make a planet like model for the hydrogen atom, for example , would be very unstable, the electron as it is attracted to the proton would fall down emitting continuous radiation (as you say) and no hydrogen atom would exist.

The data of hydrogen light though showed very specific spectral lines, that could mathematically be fitted with known series. So Bohr introduced the hypothesis that angular momentum was quantized so only certain stable energy levels existed. This was expanded with Schrodinger's equation and Quantum Mechanics theory was born.

To answer the title , one excited atom does not produce an electromagnetic wave, but a single photon. Electromagneitc waves are emergent from zillions of photons.

Answer 4

consider a cup in height , if I drooped it, its potential energy change to kinetic energy in the path to ground when it hits ground its kinetic energy goes to ground and ground give back equal energy to it backward and cup will broke .

an electron in higher level of energy when goes back to lower level of energy what would happens to its extra energy ? it could not like a cup break apart because it is a elementary particle and can not break .

so his extra energy will come out of it as a photon (electromagnetic wave)

Answer 5

It doesn't. The wave is already there.

We know it's already there because it induces downward transitions from excited states. The wave field might be in its ground state, the state with no excitation, but there are still EM fluctuations ... the "zero point" field. This field induces a downward transition even though there is no applied wave. The zero point wave is always there. This is spontaneous emission, but in one way of looking at it it is stimulated emission driven by the zero point field. The transition exchanges energy from the atom to the already-existing wave.

The next question is "How is the energy transferred from the atom to the field?" No one has an answer to that, although I think I see people approaching the question. Undoubtedly when we get the answer to that question it will beg another "How is ..." question. I have a feeling we'll never get to the end of that chain of questions ... we'll never be able to answer "How is ..." fully satisfactorily. Physics is currently set up to predict the results of experiment, not to answer "Why" or "How".

Answer 6

This is a good question. It's true that in those sort of courses, we generally don't explain the detailed mechanism of what is going on in these atomic transitions. There are two reasons for that:

1. The actual process is complicated.
2. You don't need to know the answer to do a lot of interesting stuff with the wavelength of like that excite or are emitted from an atom.

I think what you are looking for is some physical intuition as to what is happening with the electron to produce a electromagnetic wave. I'll give some heuristic descriptions here with the disclaimer that the actual details are more complicated.

Heuristic 1: Resonance

Let's take a hydrogen atom. The electron can only have certain energy levels $E=-13.6 {\rm eV} /n^2$. Let's start in the ground state $n=1$. The final state has to be one of the other eigenstates $n=2,3...$. Therefore the only way the atom can adsorb a photon is if it can accept exactly all the energy the photon has. Therefore the photon must have an energy that corresponds to one of those differences between $E(n=2)-E(n=1)$ (for example).

Heuristic 2: Accelerating charges

You are correct that changing currents produce changing magnetic fields. A more sophisticated version of that statement is that accelerating charges give off radiation. One example is a radio antenna. To broadcast radio waves, you are accelerating charges up and down the radio antenna. You could also send out EM waves by grabbing a charged ball and shaking it up and down. Consider an electron jumping from the $n=2$ state to the $n=1$ state and emitting a photon. During the emission of a photon, the electron is temporarily in neither $\psi_1$ nor $\psi_2$, but is sort of jumping back and forth between them, accelerating back and forth to radiate light.

Note of caution: this is a fundamentally quantum phenomenon, so this classical explanation can provide some intuition, but it's not the whole truth.