# Radiation from free electron with small acceleration

1. Feb 19, 2010

### DaTario

Hi All,

We know that, classically (and quantum mechanically as well), accelerated charges produce radiation. We also know radiation expresses itself by means of quantized way, i.e., by means of photons.
Large value of acceleration of an electron may be treated classically and the corresponding spectrum may be obtained by Classical Electrodynamics methods.
But how about an electron which accelerates with a very small value of acceleration?
It is intuitive that only a small number of photons will come out in a unitary interval of time.
What about the frequency of these photons?
Will it resemble the classical patern for this value of acceleration with a QM true statistical behavior, i.e., photons of several different frequencies coming out with statistical weight given by the spectrum ?
Is this statistics presented in some reference ?

Best Regards,

DaTario

2. Feb 20, 2010

### ytuab

I saw some new threads of yours like this. You often use the phrase "classically accelerated charges produce radiation".
First, what atomic model do you imagine each time you use this phrase?

As far as I know, there were no atomic models using only the Maxwell's laws (not using the de Broglie's theory) since 1913.

For example, can the two-slit(or Davisson-Germer experiment) behavior of the electrons be explained only by the Maxwell's law? What force do you think arrange the electrons' positions unequally in this experiment ?
Of course, in the free electrons, this force(which causes the interference) exists.

And I want to know how you think about the below?

Last edited: Feb 20, 2010
3. Feb 20, 2010

### meopemuk

Photons are massless, so their energy can go all the way down to zero. When a charged particle accelerates it can emit an infinite number of "soft" photons, whose total energy is still negligible. This "infrared catastrophe" is discussed in QFT textbooks.

Eugene.

4. Feb 20, 2010

### DaTario

Hi, in the case you cited it is just the standard interpretation. Classically, i.e., chapter 14 of Jackson (second edition).
Your criticism to my use of words in other posts is appreciated. But I would ask you to be more specific, so that I can better defend myself.

I agree with you in many aspects. But classical physics still have importance as a name (even in quantum context), for it refers to a huge number of theoretical results that fit well with experimental data and phenomena. So, as I have received other critics like yours, I am inclined to think that this attitudeof mine is only shared by a fraction of the physicists that work in quantum theory, as for instance, Aspect, Scully, Zubairy, Orzag, Tannoudji, Haroche, Wineland, Davidovich, Fabre, Walls, Milburn, Grynberg, among just few others.

The text you redirected to me is interesting and present a fact that I was unaware, namely, that using the finite value of proton mass yields a more precise result in the quantum theory of the H atom. But I don't think that the conclusion that the electron is really moving around the nucleus is the only reasonable that can be drawn from this.

Finally I appreciate your comment, but my doubt is standing up in front of me, untouched. According to the typical quantum style of explanation, the answer to this post would be that the electron emits randomly photons with frequecies belonging to the classically calculated spectrum, with the addendum that the number of times a certain frequency interval appears is proportional to the specific value of intensity in the spectrum for the interval's central frequency. I am correct?

Best Regards,

DaTario

5. Feb 20, 2010

### DaTario

Ok, Thank you, but will the collection of all these "soft photons", as you named it, mount a statistic histogram that corresponds to the classically calculated spectrum for the accelerated electron?

Sincerely,

DaTario

6. Feb 20, 2010

### meopemuk

I am not an expert on this, but I guess there should be an agreement between classical and quantum emission spectra.

Eugene.

7. Feb 21, 2010

### ytuab

Then, what does the reduced mass of the electron actually mean in QM?.

(Sorry for my persistent question. But I think that the interpretaion of this reduced mass in QM may influence the basic conception of the "fields" including all the electronic, magnetic, (photons), and de Broglie's waves.)

May I ask you one more question?
Do you think the meanings of the word "accelerated" are the same in all cases?

For example, in the Bohr orbit, the electron is moving around the nucleus on the circular orbit. Of course, this electron is accelerated by the nucleus. But the orbital length is always a integer times the de Broglie's wavelength, so the field is stable.
(Also in the Schroedinger equation, this de Broglie's wave is used, and if we increase (or decrease) $$\phi$$ by $$2\pi$$, we always get the same wave function.)

When the electron is freely and randomly moving, and then accelerated, it radiates the electromagnetic waves like synchrotron. But in this case, the idea of the integer times de Broglie's wavelength is not used.

So the meanings of the word "accelerated" are the same in both the cases, excluding the stability of the de Broglie's waves. How do you think about it? As I said, the de Broglie's waves are influencing the electron's movement as shown in the interference experiments.

About the electron's movement, this site about the Schroedinger equation says as follows,
-------------------------------------
Here, $$\mu=m_{e}m_{p}/(m_{e}+m_{p})$$ is the reduced mass, which takes into account the fact that the electron and the proton both rotate about a common centre, which is equivalent to a particle of mass $$\mu$$ rotating about a fixed point.
-------------------------------------

This equation of the reduced mass can be obtained only by the "classical mechanical" methods.
Now it has been over 80 years since the quantum mechanics was born. But even now we can not know the real meaning of this reduced mass in QM? (Also about the relativistic effect in QM?)

8. Feb 21, 2010

### SpectraCat

It's a little hard to tell from your post, but you seem to be arguing that an electron in an atom "rotates" around the nucleus .. or that the electron and nucleus rotate around a center. I have a little trouble with that description in the ground state (and I guess this is at least part of DaTario's problem), which is that the ground state electron in the H-atom has ZERO angular momentum. It is hard to imagine that it can be correct to describe a system with exactly zero angular momentum as undergoing "rotation".

On the other hand, the electron has a non-zero kinetic energy, which for a classical particle would mean that it was moving (this is also part of what is bothering DaTario I think). However, it is easy to show from the HUP that any particle confined in a potential will have a non-zero kinetic energy. So, to me, all that we can say with certainty about a Q.M. system with a non-zero K.E. is that there must be a finite uncertainty in the position. We cannot say anything about motion in a classical sense, because that would indicate a well-defined value for the momentum of the system, and all we have is an uncertainty. As far as the relativistic argument you quoted is concerned, I think it is a non-issue. Once again, the classical bias about relativity having to do with "speed" and "motion" is the source of the confusion .. it is completely consistent to frame the result entirely in terms of the kinetic energy, and we have already seen that QM systems can have KE without having any well-defined motion in the classical sense.

I actually think the angular momentum issue is a bigger problem for people who want to conclude that the electron is moving in some classical sense, because there is no uncertainty or ambiguity in the interpretation of that result. The eigenvalue of the angular momentum is zero for the ground state, so there cannot be any net rotation about a common center for the electron and nucleus in the classical sense. Furthermore, since the Coulomb potential is a central potential, I think you will find it hard to construct a classical model that involves relative "motion" of the electron an nucleus without angular momentum, and that still matches the results of experiments with atoms.

Also, I don't think it is correct to say that the reduced mass is somehow inherently a classical concept. The reduced mass just represents a transformation to a particular inertial reference frame, and is valid for any problem where two particles are bound in a potential. It is only dependent on the masses of the objects in question, and not on their shapes or spatial distributions, so I don't think there is anything that is either profound or limiting about using the reduced mass to describe a QM system.

Last edited: Feb 21, 2010
9. Feb 21, 2010

### DaTario

I can, to a certain extent, deal with that. It is just to think of several states of rotation coherently superposed yielding quantum statistically a zero for angular momentum. Quamtum interference plays an important role here.

But [\B] I would like to emphasize that my question is related not to a atomic electron, but to a free one.

Best Regards,

DaTario

10. Feb 21, 2010

### DaTario

Let me try to clean up the area. It is becoming a little messy.

What you both sometimes call classical concept may be rephrased as:

a concept that was created (identified) before QM.

But note that Schroedinger equation calls for a potential function and, in Harmonic Oscilation, we use the harmonic potential which is classic. Coulomb potential is classic. So, what seems clear to me is that Schroedinger equation is made to receive classical potential functions and yield quantum wave functions.
One step further: Schoredinger equation will produce better and better results as long as you insert better and better potential functions, i.e., functions that represent with increasing quality and precision, the forces experienced by a point like particle in the specific situation under attention.

I can, to a certain extent, deal with that. It is just to think of several states of rotation coherently superposed yielding quantum statistically a zero for angular momentum. Quamtum interference plays an important role here.

But I would like to emphasize that my question is related not to a atomic electron, but to a free one. Perhaps, it will be fine to consider first a linear and constant acceleration.

Best Regards,

DaTario

11. Feb 21, 2010

### SpectraCat

There is nothing inherently classical about the harmonic or coulomb potentials ... what makes you think that there is? What potentials would you consider to be "classical"?

Ok .. so I guess I see what you mean by "classical" potential now .. you mean a local potential, that is, one that depends only on a single point in space. However, the Schrodinger equation is not in any way restricted to local potentials. We have just found that local potentials are often sufficient to reproduce experimental results. (Try googling non-local potential sometime to see how they are being used in Q.M.)

No, this is not correct. The situation you describe does not correspond to a system with an eigenvalue of zero for the angular momentum. If the ground state were somehow a superposition of states with non-zero angular momentum, then any single measurement would yield a non-zero angular momentum.

Ok, so what is your argument/concern? Please set up a framework for the discussion.

12. Feb 21, 2010

### DaTario

I agree they are potentials, serving both classical and quantum formalism. But they were born in classical era.

Now paying some attention to your question I would say that those potentials which takes into account spin are certainly not classical.

I didn't go that far, but let's work with this assumption. It seems to be a good fixed point in the discussion.

This argument of yours seems funny to me. Measuring position and momentum of an electron (under the action HUP) as simultaneously as possible will yield experimental detection of angular momentum, and it is typically not zero.

If you are not able to describe how orbits are superposing coherently in QM, it doesn't mean they are not.

I think my OP is suficiently clear.

Best Wishes

DaTario

13. Feb 21, 2010

### SpectraCat

But that is not correct, at least not in the general case. Many potential terms involving spin (such as Zeeman splitting in a magnetic field) are just as "classical" as the coulomb potential .. they have the same form as if the electron were a small, classical spinning magnet, and this is sufficient to reproduce the experimental results. In other words, the "quantum" results of the problem come from the wavefunction, not the potential.

First of all, the treatment you suggest above is not valid for a quantum system .. you cannot just measure the position and momentum separately and take the cross product. That fact that you would event suggest this makes me wonder how well you understand quantum mechanical operators.

Second, I also wonder how well you understand the concept of an eigenvalue. The solutions to the H-atom are eigenstates of the angular momentum operator ... that means that they have perfectly precise values of angular momentum, and can be measured simultaneously with other commuting observables (e.g. energy, projection of angular momentum on space fixed axis), with NO fundamental restriction on the uncertainty from the HUP. From the Schrodinger equation, the ground state of H-atom is a 1s orbital, where the electron has ZERO orbital angular momentum ... this is completely consistent with all experimental measurements.

It's not a question of me not being able to describe it, it's that it is an experimental FACT that it does not happen.

Your questions are clear, but lack specificity .. I don't understand what is confusing you about the situation. The frequency spectrum of the accelerated free-electron will be continuous. The mean frequency of the distribution will be proportional to the magnitude of the acceleration. Google "bremsstrahlung" for more information and the references you seek. One thing that would help sort out your issues is if you tried to give a mathematical description of your issue.

P.S. One thing to keep in mind is that a free-electron is NOT free when it is under acceleration. If there is a force on the particle, then the potential term in the Hamiltonian is not zero, which means that it is not a free-electron.

14. Feb 21, 2010

### DaTario

your ability to attain to irrelevant questions also makes me wonder things relative to your formal and intelectual background, but if we are here, it is desirable that we get enrolled in productive discussions. Of course the name "free electron" is not to be taken after the interaction begins, It is just to create the oposition to atomic electron.

Tell me one thing: Does the expression "free electron laser" bother you ?

Eigenvalues and eigenvectors for H atom are beautiful constructions of quantum theory, but if I were you I would start to exercise seeing them as you see plane waves (which, most of the time is not good representation of what really happens in nature).

In order to equal our references, are you familiar with quantum trajectories? They may explain the electron cloud as quantum and coherent superpositions of orbits.

It seems you are as capable of guessing some possible answers as I am. So we have to pay some respect to each other. We are waiting form someone who knows more than us.

Best wishes

DaTario

15. Feb 22, 2010

### PhilDSP

It may be that in the hydrogen atom ground state, the electron's motion is so sporadic and chaotic that it really can't define a state of angular momentum when integrated over a reasonable period of time. Einstein wrote a very interesting paper analyzing the mathematical models of the time and attempts to extent them. Sorry I no longer have a reference but if the interest is large enough I could probably dig it up.

16. Feb 22, 2010

### Bob S

There is electromagnetic (photon) radiation from two distinct types of classical acceleration of free electrons (excluding bremsstrahlung), as discussed in the OP. These two types of radiation have nothing to do with hydrogen atoms or protons or nuclear scattering. There is acceleration perpendicular to the velocity, such as electrons travelling in circular orbits in magnets, and acceleration parallel to the velocity.

Acceleration perpendicular to velocity is often called synchrotron radiation. See

http://geant4.web.cern.ch/geant4/UserDocumentation/UsersGuides/PhysicsReferenceManual/BackupVersions/V9.1/html/node49.html [Broken]

Synchrotron radiation is usually x-ray energies in circular electron accelerators, but at lower electron energies the synchrotron radiation critical (mean) energies are in UV or visible light. The above URL has plots of photon energy spectra (Fig. 8.4), which approach classical spectral densities.

The classical radiation from a particle being accelerated parallel to its velocity, and perpendicular to its velocity, is derived from the Leinard-Wiechart potentials on pages 301-306 and 306-307 in Panofsky and Phillips, Classical Electricity and Magnetism (first edition).

Bob S

Last edited by a moderator: May 4, 2017
17. Feb 22, 2010

### SpectraCat

Which parts of my posts "attain to irrelevant questions"? You are focusing in on a post-script comment I added to make sure that we were clear on an important (but as you say obvious) aspect of the discussion. It looks like we are .. so good.

Nope, because I understand the context in which it is used ... now I understand that we agree on the context of "free electron" in the current discussion as well, so as you say, we can move on.

Why should I do that? Where is the evidence that the eigenstates of the H-atom are not good representations of the actual system, especially with regard to the angular momentum? More specifically, where is the experimental evidence that your picture of interfering states with non-zero angular momenta has any validity?

I am somewhat familiar with quantum trajectories in the context of Bohmian mechanics, and they definitely don't seem consistent with the explanation you mentioned involving interference between states with non-zero angular momentum. In the Bohmian picture, the electron is STATIONARY in the hydrogen ground state .. it's classical velocity is zero. Therefore the proof that it's angular momentum is also zero is trivial in Bohmian mechanics, without any requirement for quantum interference. So again, I don't see any reason to accept your hypothetical situation in the case of the H-atom ground state as a possibility.

I cannot think of any experimental results dealing with radiation from electrons under very small accelerations. All I can do is suggest looking at the limits of experiments done for larger accelerations (e.g. Compton scattering, bremsstralung, etc). With regard to a theoretical answer to your question, the relativistic and non-relativistic matrix elements for a free electron interacting with a quantized field can be found in many textbooks that deal with the quantum theory of radiation (e.g. Heitler's "The quantum theory of radiation"). For the purposes of this discussion, the relevant features are that these matrix elements show that the probability of a transition coupling states A and B (defined below) of the electron and field is proportional to the wavelength $$\lambda$$, and the number of photons with wavelength $$\lambda$$. So your earlier statement,

appears correct in this context.

State A describes an electron with momentum $$p$$, and a radiation field containing $$n$$ photons with wavelength $$\lambda$$

State B describes an electron with momentum $$p + k_{\lambda}$$, and a radiation field containing $$n-1$$ photons with wavelength $$\lambda$$

18. Feb 27, 2010

### DaTario

Thank you Bob S.

Thak you All too

DaTario

19. Feb 27, 2010

### DaTario

Thank you SpectraCat for Heitler's reference. I'll try to find it.

Best wishes

DaTario