Change in the EM field energy due to dispersion of a wave packet

[SergioPL]

Let’s suppose we have an electron with a Gaussian eigenstate, as the time runs, the wave spreads in space without changing its energy, however, the induced EM field caused by the particle decreases its energy. I assert this from the classical electromagnetism result in which the more concentrated the charge is, the more energy the field has.

I suppose this energy lose in the EM field must be compensated somehow, maybe photons are generated but how? Does anybody can explain me how this process could work?

In this link to Wikipedia there is an explanation about wave packets.

https://en.wikipedia.org/wiki/Wave_packet

And in this other link there is a definition of the Hamiltonian of the EM field:

https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field


Thanks,

Sergio

 

[dextercioby]

Is the electron initially in a state of known/well determined energy? If not, how would you assess it losing energy?

 

[PeterDonis]

Let’s suppose we have an electron with a Gaussian eigenstate, as the time runs, the wave spreads in space without changing its energy

This would mean the electron's state would be an energy eigenstate. Is a Gaussian an eigenstate of energy? (Hint: I think the answer is "no".)

the induced EM field caused by the particle decreases its energy. I assert this from the classical electromagnetism result

You can't use classical EM to model the field if you're trying to analyze the quantum behavior of the electron. Unless you want to ignore the electron's quantum details and use the expectation value of the electron's charge-current density as the source of the classical EM field. But that's not what you're doing.

 

[SergioPL]

This would mean the electron's state would be an energy eigenstate. Is a Gaussian an eigenstate of energy? (Hint: I think the answer is "no".)

Sorry, I expressed wrong, I did not mean that the electron is in a Gaussian eigenstate but simply that the electron is taking a Gaussian shape. I don't know of any situation where a Gaussian can be an eigenstate...

You can't use classical EM to model the field if you're trying to analyze the quantum behavior of the electron. Unless you want to ignore the electron's quantum details and use the expectation value of the electron's charge-current density as the source of the classical EM field. But that's not what you're doing.

No, I don't want to ignore the quantum effects. I suppose that the problem is what you say, that with the classical approach I would only get in the best of the cases the expectated value of the electromagnetic field. Does it means that the field generated by the charge can only be determined statistically?

 

[PeterDonis]

I don't know of any situation where a Gaussian can be an eigenstate...

I don't either, and that's why I said that your statement that "the wave spreads in space without changing its energy" cannot be correct. "Without changing its energy" can only be correct if the electron's state is an eigenstate of energy.

I suppose that the problem is what you say, that with the classical approach I would only get in the best of the cases the expectated value of the electromagnetic field. Does it means that the field generated by the charge can only be determined statistically?

No. The expectation value of the electron's charge-current density is not "determined statistically"; it's a single number that you calculate based on the electron's state. It becomes the source on the RHS of Maxwell's Equations, if you want to use the classical approach.

If you want to use the quantum approach, there is no such thing as "the field generated by the charge". There is the electron, and there is the electromagnetic field, and you have to model the two together as a quantum mechanical system and then model the interaction between them as a part of the Hamiltonian of the overall system.

 

[SergioPL]

I don't either, and that's why I said that your statement that "the wave spreads in space without changing its energy" cannot be correct. "Without changing its energy" can only be correct if the electron's state is an eigenstate of energy.

Can be the electron in a superposition of plane waves (assuming there are not external potentials) that form a Gaussian? If that's not possible then my question is wrongly focused. If it can be, then I expect that the Gaussian wave will spread as time advances.

No. The expectation value of the electron's charge-current density is not "determined statistically"; it's a single number that you calculate based on the electron's state. It becomes the source on the RHS of Maxwell's Equations, if you want to use the classical approach.

If you want to use the quantum approach, there is no such thing as "the field generated by the charge". There is the electron, and there is the electromagnetic field, and you have to model the two together as a quantum mechanical system and then model the interaction between them as a part of the Hamiltonian of the overall system.

So the classical charge-current can be obtained from the expectation value of the quantum charge and current. In order to measure the quantum effects over the EM field it will be more accurate to use the Hamiltonian instead of the classical energy, however the Hamiltonian depends on the existing EM field so I need the classical charge-current to calculate it.

For a plain wave I think its induced electric field has zero energy since its charge is dispersed through an infinite volume but for "compact" waves I guess the field should have some energy and it may change, but my skills to calculate it are poor.

 

[PeterDonis]

Can be the electron in a superposition of plane waves (assuming there are not external potentials) that form a Gaussian?

Yes. But again, such a state will not be an eigenstate of energy.

I expect that the Gaussian wave will spread as time advances.

Yes.

 

[blue_leaf77]

I don't know of any situation where a Gaussian can be an eigenstate.

The ground state of a harmonic oscillator as well as the so-called coherent state of photons which is the eigenstate of photon lowering operator, but the last is not a function in position space.

I suppose this energy lose in the EM field must be compensated somehow, maybe photons are generated but how?

Take a look at Bremsstrahlung effect.

 

[dextercioby]

[...]Take a look at Bremmstrahlung effect.

It is Bremsstrahlung.

 

[blue_leaf77]

Ah thank you for the correction.

 

[SergioPL]

The ground state of a harmonic oscillator as well as the so-called coherent state of photons which is the eigenstate of photon lowering operator, but the last is not a function in position space.

Take a look at Bremsstrahlung effect.

I have read about Bremsstrahlung in a QED textbook but I see an important difference respect the situation shown here, as far as I know, Bremsstrahlung is caused by an external potential, for example a proton, while here we only have an electron Gaussian wave moving freely so there is no potential acting on the electron.

Maybe the electron can interact with the vacuum generating photons so that the energy of the electron induced field turns into photons... Is that what you mean?

 

[blue_leaf77]

here we only have an electron Gaussian wave moving freely so there is no potential acting on the electron.

In your first post you did not specify that the wavepacket moves in free space.
If it's in free space then according to QM the momentum, and hence the energy distribution of the particle will not change as the wave packet moves.

 

[SergioPL]

In your first post you did not specify that the wavepacket moves in free space.

I should have specified that, my doubt is not restricted to a wave packet moving in free space but that's the most significative and simple case.

If it's in free space then according to QM the momentum, and hence energy distribution of the particle will not change as the wave packet moves.

Yes, that's the answer that any QM textbook would give, but my concern comes because as the wave packet moves it will spreads more and more and therefore it's charge density will decrease and the energy of its induce field will decrease. Therefore, it looks like the energy of the wave packet keeps constant but the energy of the field decreases and therefore there should be elsewhere where this energy goes.

 

[blue_leaf77]

What do you mean by the induced field?
The total energy of the system is constant, which is guaranteed by the time evolution operator.

 

[DrClaude]

I don't either, and that's why I said that your statement that "the wave spreads in space without changing its energy" cannot be correct. "Without changing its energy" can only be correct if the electron's state is an eigenstate of energy.

Well, it depends on what you mean by "changing its energy." I would interpret that as the expectation value of energy being constant, which is the case for any state if the Hamiltonian is independent of time. Otherwise, it would be hard to reconcile QM with conservation of energy

 

[SergioPL]

What do you mean by the induced field?
The total energy of the system is constant, which is guaranteed by the time evolution operator.

For induced field I mean the electromagnetic field created by the wave packet. I assume that the quantum wave density and quantum current density behave like classical distributions of charge and current. This approach omits some quantum effects, I would use it as starting point.

All I have read about QM Gaussian wave packets show how the wave packet evolves in space, in the case with no external potential, the momentum representation is not affected ( |Ψ(p)|² will remain constant), which is an absolutely expected result.

However, the textbooks does not include the EM Hamiltonian in the formulae. with the classical approach the overall electric field's energy will decrease as wave spreads because the induced field losses energy and there is no radiation.

What I’m asking is if this approach is correct or not and, in case it is, where can this field energy be going.

 

[blue_leaf77]

Why do you think that the electron described by a wavepacket will radiate an EM field? It does have electric field since an electron is charged but even if you mean this DC field it's not like the electron interacts with its own electric field, at least in the non-relativistic limit.

the textbooks does not include the EM Hamiltonian in the formulae.

Because there is no electromagnetic radiation involved in the motion of a wavepacket in free space. If you arbitrarily include EM potential into the Hamiltonian, it would mean an entirely different system. Such system means that the wavepacket does not move in a field-free space and the momentum space wavefunction should not be constant of time anymore.

 

[SergioPL]

Because there is no electromagnetic radiation involved in the motion of a wavepacket in free space. If you arbitrarily include EM potential into the Hamiltonian, it would mean an entirely different system. Such system means that the wavepacket does not move in a field-free space and the momentum space wavefunction should not be constant of time anymore.

I don't pretend to include any potential in the electron's Hamiltonian.

For a classical electrostatic system it's proven that the energy needed to put together a distribution of charge (its potential energy) is the same that the energy stored in the electric field. Electrons have a discrete (not infinitesimal) charge and therefore its electric field has some energy, but we cannot talk about “putting together the charge” of an electron: the electron does not repel itself, so there is no self-potential term on the electron Hamiltonian.

If the charge distribution changes, the field energy changes, where does this energy go?

 

[blue_leaf77]

If the charge distribution changes, the field energy changes, where does this energy go?

Now I get what you mean by the energy of the charge distribution. The answer lies in the very paragraph you wrote "the electron does not repel itself, so there is no self-potential term on the electron Hamiltonian." The probability density of a single electron shouldn't be interpreted the same way as any continuous classical charge distribution. The former is associated to the probability of finding a single electron, it's not supposed to be viewed as an actual charge distribution as you would in the classical counterpart. Consequently, no energy is stored in the quantum mechanical single-particle charge distribution because as you know a particle does not interact with itself.

 

[SergioPL]

Now I get what you mean by the energy of the charge distribution. The answer lies in the very paragraph you wrote "the electron does not repel itself, so there is no self-potential term on the electron Hamiltonian." The probability density of a single electron shouldn't be interpreted the same way as any continuous classical charge distribution. The former is associated to the probability of finding a single electron, it's not supposed to be viewed as an actual charge distribution as you would in the classical counterpart. Consequently, no energy is stored in the quantum mechanical single-particle charge distribution because as you know a particle does not interact with itself.

Ok, so the probability density of a single electron should not be interpreted as a continuous charge distribution, and therefore I cannot calculate the electric field as if it were generated from such distribution of charge.

Thanks!

 

[blue_leaf77]

Ok, so the probability density of a single electron should not be interpreted as a continuous charge distribution, and therefore I cannot calculate the electric field as if it were generated from such distribution of charge.

Yes.