# Frequency of EM waves in classical and quantum physics

donniedarko
in classical physics, when a charged particle oscillates, it emits an electromagnetic wave, and the frequency of the wave depends on the frequency with which the particle oscillates.
But in quantum physics, when an excited atom emits a photon, the energy of the photon depends on the magnitude of the quantum leaps that the emitting electron makes (if it jumps one level, the photon will have a certain energy; if it jumps two, a greater energy, and so on). So the frequency of the electromagnetic wave that corresponds to the photon will depend on the amplitude of the quantum leaps made by the electron.

I don't understand why these two cases are so different. In analogy with the classical case, shouldn't the frequency of the wave emitted by the atom depend on the frequency with which the electron makes quantum jumps? Or is there a quantum explanation of the classical case that I don't understand? I know that classical physics cannot be used to explain quantum phenomena, but it seems strange to me that there is this asymmetry in the two cases.

Sorry in advance if the question is dumb; I am approaching quantum physics because it is a topic that I am very passionate about, but I have a totally different background ... thanks if you can clarify me in a simple way!

## Answers and Replies

Science Advisor
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I know that classical physics cannot be used to explain quantum phenomena
Yep, this.
There is no "wave" when a single electron changes energy levels in an atom. The classical picture of an oscillating electron is not what electrons do in an atom.

vanhees71
WernerQH
shouldn't the frequency of the wave emitted by the atom depend on the frequency with which the electron makes quantum jumps?
No, the rate at which the electron makes quantum jumps determines the intensity of the emitted wave; the frequency is determined solely by the energy difference between the energy levels.
in classical physics, when a charged particle oscillates, it emits an electromagnetic wave, and the frequency of the wave depends on the frequency with which the particle oscillates.
Something similar is also true in quantum physics; it is referred to as the correspondence principle. But you have to be careful when you talk about the frequency or speed of electrons. They can also be described as waves, and one has to distinguish between phase and group velocity. A hydrogen atom can have highly excited states (Rydberg states), in which the electron moves "slowly", and the electron's energy levels are closely spaced, leading to "low" frequencies when the electron hops from one level to the next.

In a magnetic field an electron's motion is periodic, leading to (cyclotron) radiation at the fundamental gyro-frequency (## eB/2\pi m ##) as long as the motion is non-relativistic. The electron can jump only from one Landau level to the next. But at relativistic energies the radiation field is no longer sinusoidal, but extends to very high harmonics of the fundamental frequency ## eB/2\pi\gamma m ## (if you Fourier analyze the classical field). In the quantum picture this corresponds to electrons making big jumps involving a large change in the quantum number. At high energies (high ## \gamma ##) the Landau levels are more closely spaced.

gentzen
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The electron in an energy eigenstate doesn't move at all, because the energy eigenstates are the stationary states of the electron. That's what resolves, in contrastistinction to the ad-hoc solution by Bohr, the contradiction between the observed stability of atoms and the fact that electrons moving on some orbit around the nucleus must radiate "cyclotron radiation".

Further it's clear that there are no quantum jumps but transitions between from energy level to another through perturbations not included in the evaluation of these energy levels. E.g., for the hydrogen atom you just solve the time-independent Schrödinger equation for an electron and a proton including the Coulomb interaction between them (in the usual Coulomb gauge describing the em. field).

This neglects the coupling of the photons and the electrons to the electromagnetic radiation field, which can be taking into account with (time-dependent) perturbation theory. This "disturbance" leads to the conclusion that the before calculated approximate energy eigenstates are in fact not really stationary but due to the interaction with the radiation field there's some probability that the atom goes from an excited atomic state to a lower atomic state by spontaneous emission of a photon with an energy given by the energy difference between the two atomic levels.

Since all this state evolution is described by a differential equation there are no jumps there at all, just the time evolution of the quantum state through the full Hamiltonian.

Mentor
The electron in an energy eigenstate doesn't move at all, because the energy eigenstates are the stationary states of the electron.
This is not correct as you state it, because there are many energy eigenstates that are also momentum eigenstates with nonzero momentum. The electron does move in these states.

gentzen and vanhees71
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Well, yes, the qualification is "a bound energy eigenstate". Note that the momentum eigenstate of a free particle in fact is a generalized eigenstate in the continuous spectrum of ##\hat{H}##, it's not a proper Hilbert-space vector and thus does not describe a physically realizable state of a particle! A "moving" electron is described by a wave packet, which you can of course make as narrow in momentum space as you wish,
$$|\Psi \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 p \psi(\vec{p}) |\vec{p} \rangle,$$
where ##\psi \in L^2(\mathbb{R}^3)## being sharply peaked around some momentum ##\vec{p}_0## (e.g., a Gaussian), but that's not a true energy eigenstate, and that's why it describes a particle moving with a momentum of about ##\vec{p}_0##.

gentzen
WernerQH
The electron in an energy eigenstate doesn't move at all, because the energy eigenstates are the stationary states of the electron.
@vanhees71, you should know the difference between a statistical distribution and an expectation value. How can an electron that "doesn't move at all" contribute kinetic energy to the total energy of the atom? @donniedarko must be utterly confused.

weirdoguy
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Very simple: The kinetic energy of the electron in a bound state of the hydrogen atom doesn't take a determined value. Again: In a bound state the electron is described by a wave function
$$\psi(t,\vec{x}) = \exp(-\mathrm{i} E t/\hbar) u_E(\vec{x}),$$
where ##u_E## is a solution of the time-independent Schrödinger equation, i.e., a solution to the eigenvalue problem for the Hamiltonian with eigenvalue ##E##. The probability distribution is thus stationary
$$|\psi(t,\vec{x})|^2=u_E(\vec{x}).$$
In this sense the electron is "not moving".

weirdoguy
WernerQH
The kinetic energy of the electron in a bound state of the hydrogen atom doesn't take a determined value.
The Hamilton operator is a some of two terms: potential and kinetic energy. There's no problem at all taking expectation values for each term separately and arriving at a definite number for the (average) kinetic energy.

There is no doubt that the probability distribution is stationary. But this does not mean that the electron must be at rest. It's like saying that the Maxwellian velocity distribution implies that all molecules have the same velocity, namely the average, zero.

The textbook explanation for the particle in a one-dimensional box says that the wave function is a standing wave, a superposition of left and right running waves of equal amplitude, corresponding to a particle flipping back and forth. It's only on average that the particle is at rest. In the one-dimensional case the higher energy levels are increasingly widely spaced, proportional to the speed of particle (and to the frequency of the emitted radiation, if it is charged). The particle would be radiating at decreasing frequencies until it has reached the ground state. I find this quite intuitive, and surprising that you would withhold this from your students.

gentzen and docnet
Mentor
the qualification is "a bound energy eigenstate".
Those states aren't eigenstates of either momentum or velocity, so the question of whether the electron is "moving" in those states does not even have a well-defined answer.

Mentor
How can an electron that "doesn't move at all" contribute kinetic energy to the total energy of the atom?
Who said it did? I don't see such a claim being made anywhere in this thread.

There is no doubt that the probability distribution is stationary. But this does not mean that the electron must be at rest.
Nor does it mean the electron must be moving. As I said in response to @vanhees71 just now, the electron bound states in an atom aren't eigenstates of either momentum or velocity, so the question of whether or not the electrons are "moving" does not have a well-defined answer. In terms of energy, that means you cannot split the electron's contribution to the atom's total energy into "kinetic" and "potential" pieces in a well-defined way.

vanhees71
Mentor
I don't understand why these two cases are so different.
They're different because the classical case you are talking about involves free charged particles, whereas the quantum case you are talking about involves electrons bound in atoms. They're very different physical situations.

is there a quantum explanation of the classical case that I don't understand?
In the classical case you described, quantum effects are negligible; the charged particles and the EM field can be treated classically to a very good approximation, so there's nothing that QM adds to the explanation.

In the quantum case you described, classical physics can't even explain its existence--in classical physics atoms are impossible, the electrons would emit radiation continuously and spiral into the nucleus. So obviously QM is the only explanation for this case that we have.

vanhees71
WernerQH
Who said it did? I don't see such a claim being made anywhere in this thread.
You quoted vanhees71 yourself in #5.
Nor does it mean the electron must be moving. As I said in response to @vanhees71 just now, the electron bound states in an atom aren't eigenstates of either momentum or velocity, so the question of whether or not the electrons are "moving" does not have a well-defined answer. In terms of energy, that means you cannot split the electron's contribution to the atom's total energy into "kinetic" and "potential" pieces in a well-defined way.
I'm probably risking another thread ban, but I can't make sense of your objection. Are you saying that talk about electron velocities is meaningless unless the electrons are in velocity eigenstates? A statistical description of velocities should be meaningful even if the particles are not in plane wave states. I thought what I had written was nothing but conventional wisdom and intelligible for almost all physicists. Please point to a pertinent passage in a textbook if you think I'm mistaken.

BTW, classical and quantum physics are not so strictly separated as they appear to be in your mind. There's a delightful article that I read many years ago and that might be of interest also to the OP: "Classical Plasma Phenomena from a Quantum Mechanical Viewpoint" by E.G. Harris (Advan. Plasma Phys. 3: 157-241 (1969)). I can't see how wave-particle interactions can be discussed without referring to velocities, regardless of whether the particles are bound or free.

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It's of course not useless to talk about the momentum of an electron in a bound state in the Coulomb field of a proton (approximation where you fix the proton at a place due to its large mass and treat it as external Coulomb field, within which the electron is "moving"). It's clear that ##\langle E_{\text{kin}} \rangle= \langle \vec{p^2} \rangle/(2m)>0##. The important point is that ##\vec{p}## doesn't take determined values. In the ground state (s-wave) the average momentum is vanishing, and indeed in this sense the "electron doesn't move". What's however important for the question, whether the electron radiates is whether the charge-current distribution is time-dependent, which it is not, and that's why an electron doesn't radiate when bound by a (static) potential.

Mentor
You quoted vanhees71 yourself in #5.
He said the electron "doesn't move" (which I pushed back on), but he didn't say the electron that "doesn't move" contributes a definite kinetic energy to the atom's total energy. Only you are trying to claim that the electron in the atom has a well-defined "kinetic energy" contribution to the atom's total energy.

Are you saying that talk about electron velocities is meaningless unless the electrons are in velocity eigenstates?
Talk about electrons having definite velocities, and therefore definite kinetic energies, and therefore making well-defined "kinetic energy" contributions to the atom's total energy is, yes.

You can of course talk about expectation values of velocity (or kinetic energy) regardless of the state the electron is in, but those aren't sufficient to ground further claims about the electron contributing a definite "kinetic energy" to the atom's total energy.

vanhees71
Mentor
It's of course not useless to talk about the momentum of an electron in a bound state in the Coulomb field of a proton (approximation where you fix the proton at a place due to its large mass and treat it as external Coulomb field, within which the electron is "moving").
You can "talk about the momentum", sure, as long as you don't make any claims that require the momentum to have a definite value if the electron is not in a momentum eigenstate.

It's clear that ##\langle E_{\text{kin}} \rangle= \langle \vec{p^2} \rangle/(2m)>0##.
Yes, but this is an expectation value and does not justify a claim that the electron has a well-defined kinetic energy that it contributes to the atom's total energy. Nor, as your further comments make clear, do you need to make any such claim to answer questions about things like whether and under what conditions the electron radiates.

vanhees71
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That's what I'm stressing to whole time.

There is no state describing an electron with definite momentum. That would be momentum eigenstates, but these are the plane-wave modes,
$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p}),$$
using natural units with ##\hbar=1##. These are "generalized eigenfunctions" but not square integrable. It's because the momentum components all have an entirely continuous spectrum. All eigenvalues are ##\vec{p} \in \mathbb{R}^3##.

That are of course also generalized eigenvectors of the kinetic energy (i.e., the free-particle Hamiltonian), and indeed a particle prepared in such a state would "not move" in the sense that for all times ##|\psi(t,\vec{x})|^2=|u_{\vec{p}}(\vec{x})|^2=1/(2 \pi)^3## is time-independent, but it's not a physical state, because the plane-wave mode is not square integrable.

A true state with well-defined momentum has always a finite momentum width (which you can make arbitrarily small), and such a physical state thus is not an energy eigenstate and thus the free particle in fact "moves", but it has not a definite momentum. In position space you have a wave packet too with a width such that ##\Delta x_j \Delta p_j \geq \hbar/2## (Heisenberg uncertainty relation).

If you have a bound energy eigenstate of a particle in a potential, it's "not moving" as all energy eigenstates describe a situation where the particle doesn't move, but that also implies that the particle has no definite momentum nor a definite kinetic energy, and that's why the expectation value of the kinetic energy is not 0 although the "particle does not move".

WernerQH
What's however important for the question, whether the electron radiates is whether the charge-current distribution is time-dependent, which it is not, and that's why an electron doesn't radiate when bound by a (static) potential.
That's right, while an atom is in one of its stationary states, the velocities (currents) average out and the atom does not radiate. But during transitions between two levels there are non-vanishing velocity / current fluctuations which determine the angular dependence and polarization of the emitted radiation (and of course its frequency), just as in the classical theory.

Since all this state evolution is described by a differential equation there are no jumps there at all, just the time evolution of the quantum state through the full Hamiltonian.
Just like you seem to do, Schrödinger very much preferred to think of the emission of radiation as a continuous process, and of ## \nu = (E_2 - E_1)/h ## as the beat frequency of two electron "waves". But Bohr correctly emphasized that an excited atom loses its energy in discontinuous steps (quantum jumps!) and not continuously as in classical theory. This is, after all, what leads to spectra with discrete frequencies, rather than some kind of whining. Nowadays experimentalists can even count emission / absorption events. You may dislike quantum jumps (like Schrödinger!), but Bohr's model remains essentially intact. Only the language has been modernized: instead of orbits we now speak of orbitals.

If you have a bound energy eigenstate of a particle in a potential, it's "not moving" as all energy eigenstates describe a situation where the particle doesn't move, but that also implies that the particle has no definite momentum nor a definite kinetic energy, and that's why the expectation value of the kinetic energy is not 0 although the "particle does not move".
This is clearly a case of semantics, of what it means for a "quantum object" to move. If only we had a commonly accepted definition of the term "quantum object" that ties together the formalism with what happens in the real world. We use the same equations, but we look at them in quite different ways.

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That's right, while an atom is in one of its stationary states, the velocities (currents) average out and the atom does not radiate. But during transitions between two levels there are non-vanishing velocity / current fluctuations which determine the angular dependence and polarization of the emitted radiation (and of course its frequency), just as in the classical theory.
If there are transitions, you are not in an energy eigenstate. That's of course the usual way the problem is treated in the sense of perturbation theory: The unperturbed Hamiltonian is the electron in the Coulomb field of the proton (or in non-relativistic QM you can take the electron and proton as a two-body system with their mutual static Coulomb interaction). This defines your atomic bound states and the energy levels. Then you have some perturbation like an external electromagnetic wave field or just the quantized electromagnetic radiation field coupled to the electron (and proton). The latter more complete form includes then also spontaneous emission from an excited to a lower atomic energy level.
Just like you seem to do, Schrödinger very much preferred to think of the emission of radiation as a continuous process, and of ## \nu = (E_2 - E_1)/h ## as the beat frequency of two electron "waves". But Bohr correctly emphasized that an excited atom loses its energy in discontinuous steps (quantum jumps!) and not continuously as in classical theory. This is, after all, what leads to spectra with discrete frequencies, rather than some kind of whining. Nowadays experimentalists can even count emission / absorption events. You may dislike quantum jumps (like Schrödinger!), but Bohr's model remains essentially intact. Only the language has been modernized: instead of orbits we now speak of orbitals.
In quantum mechanics there are no jumps. It's all described by differential time-evolution equations. There are, however, very rapid changes, looking like jumps within the time resolution of your measurements. Of course, nowadays it's an exciting field to really observe atomic transitions, like experiments, where attosecond laser pulses are used.
This is clearly a case of semantics, of what it means for a "quantum object" to move. If only we had a commonly accepted definition of the term "quantum object" that ties together the formalism with what happens in the real world. We use the same equations, but we look at them in quite different ways.
Well, from a modern point of view everything is a "quantum object", and Q(F)T describes what happens in the real world or at least it predicts the probabilities for what we observe when measuring something.

Science Advisor
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I'm probably risking another thread ban, but I can't make sense of your objection. Are you saying that talk about electron velocities is meaningless unless the electrons are in velocity eigenstates?
It would seem to me that you have fallen into the same hole folks usually fall into via the Ehrenfest Theorem. $$m \frac d {dt} \langle x\rangle=\langle p\rangle$$
$$\frac d {dt} \langle p\rangle=-\langle V'(x)\rangle$$This is true, but $$\langle V'(x)\rangle \neq V'(\langle x\rangle)$$
And so the division into kinetic and potential energy pieces is no longer well defined. The question that is incorrect, not the answer.

.

PeterDonis
Motore
But Bohr correctly emphasized that an excited atom loses its energy in discontinuous steps (quantum jumps!)
Does it?
Here is an interesting article:
https://arxiv.org/abs/1803.00545

WernerQH
Here is an interesting article:
https://arxiv.org/abs/1803.00545
Yes, interesting. But also the details of such experiments are subject to interpretation!
Thanks for the reference.

WernerQH
If you have a bound energy eigenstate of a particle in a potential, it's "not moving" as all energy eigenstates describe a situation where the particle doesn't move, but that also implies that the particle has no definite momentum nor a definite kinetic energy, and that's why the expectation value of the kinetic energy is not 0 although the "particle does not move".
I'm wondering if you would consider an ion in a Penning trap to be in a "bound energy eigenstate". Can it "move", in your terminology? I'd think it always moves, and energy eigenstates are just idealizations that we use to ease our descriptions.

Science Advisor
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If it's in an energy eigenstate it "does not move", but since again the momentum is not determined, the momentum fluctuates.

Mentor
If it's in an energy eigenstate it "does not move"
If this is just a definition of what you mean by "does not move", then that's fine, but I don't think it's a good choice of words. "Does not move" implies that the electron has a definite velocity and momentum and they are both zero. That is not the case for an electron bound in an atom.

vanhees71 and gentzen
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That's of course just a matter of semantics. For me the quantum state describes, well what else, the state of the particle. An energy eigenstate ##\hat{\rho}_E=|u_E \rangle \langle u_E|## is time independent and thus for me a system prepared in such a state "does not move". This of course does not imply that the velocity or momentum must be both zero. As you say yourself, usually neither has determined values.

Mentor
thus for me a system prepared in such a state "does not move"
Yes, I understand what you mean by "does not move". I just don't think that's a good choice of words to describe what you're talking about (energy eigenstates).

vanhees71