Questions about photon/electron interaction

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In summary: It's not a true statement, although it's a bit more complicated than that. In general, if the mass of the system decreased by a certain amount then some nucleons would disappear - but this would be in violation of the conservation of mass.
  • #1
davidbenari
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I won't go into any details because it is really not important.

In class we handled a problem dealing with an incoming photon that crashes into an electron, the electron gains momentum and a new photon is emitted too.

My first question is about Maxwellian E&M. I understand that according to classical EM an accelerating particle produces an EM wave. And according to the photon theory of light, the wavefront of any such wave is replete with bundles of energy called photons. So in this particle interaction I mentioned above, why has the electron emitted a single photon and not the massive amount of photons that an EM wave would predict?

My second question deals with the mathematical construction of such interactions. It was assumed in this collision that the mass of the system was invariant in the before-after scenarios. This is not the typical assumption for nuclear disintegration, for example. The reason why I think the mass is assumed invariant is because the identity of the particle traveling is known in both cases. Namely, it is an "electron" and electrons have mass ##m##. This makes perfect sense, but is it known for a fact that all electrons have the exact same mass?

In the case of protons and neutrons, I guess it is also assumed they all have the same mass (not implying here that m-proton=m-neutron). But I don't like this idea because it suggests that whenever I did problems on nuclear disintegration the mass difference (before vs after) was always a linear combination (with integer coefficients) of the masses of the proton and neutron. I think this is assuming too much, and perhaps not even true when I did in fact do such problems.

Thanks for reading so far! Hehe.
 
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  • #2
davidbenari said:
And according to the photon theory of light, the wavefront of any such wave is replete with bundles of energy called photons. So in this particle interaction I mentioned above, why has the electron emitted a single photon and not the massive amount of photons that an EM wave would predict?
Your understanding of photons and light is incorrect. If you search the quantum mechanics forum here, you'll find a number of threads discussing how photons interact with matter. (Although if the incoming photon is sufficiently energetic, you may get multiple photons out of the interaction - but that's not what you're describing here).

My second question deals with the mathematical construction of such interactions. It was assumed in this collision that the mass of the system was invariant in the before-after scenarios. This is not the typical assumption for nuclear disintegration, for example. The reason why I think the mass is assumed invariant is because the identity of the particle traveling is known in both cases. Namely, it is an "electron" and electrons have mass ##m##. This makes perfect sense, but is it known for a fact that all electrons have the exact same mass?
We know that if all electrons do not have the same mass the difference doesn't show up in the first ten decimal places - we've measured the mass of electrons to that degree of accuracy without seeing any difference. It's also a prediction of quantum field theory that all electrons will have the same mass, and QFT works so well in so many places that it would be bizarre not to trust it here too.

However, this is all a bit of a digression. The energy of the system is conserved across the interaction, not the mass. It just so happens that if the energy of the incoming photon is small compared to ##m_ec^2## then the rest mass of the system before and after will be so close to ##m_e##, the mass of the electron, that we can use conservation of mass as a very good approximation.

In the case of protons and neutrons, I guess it is also assumed they all have the same mass (not implying here that m-proton=m-neutron). But I don't like this idea because it suggests that whenever I did problems on nuclear disintegration the mass difference (before vs after) was always a linear combination (with integer coefficients) of the masses of the proton and neutron. I think this is assuming too much, and perhaps not even true when I did in fact do such problems.
It's the same general principle - we've measured the masses to nine decimal places without seeing any variation from one proton to another. The QFT argument also applies, although it is a bit more involved because the nucleons are not elementary particles - they're composed of quarks which are.
 
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Is the thing I said about a linear combinations of the masses of the nucleons (with integer coefficients) a true statement? So that if the mass of the system decreased then necessarily some nucleons disappeared, and its not that some nucleons now "weigh less" ?

I am having trouble seeing how my interpretation of the photon theory of light is incorrect. My textbook (Serway, Modern Physics) says this

""He maintained that the energy of light is not distributed evenly over the classical wavefront, but is concen- trated in discrete regions (or in “bundles”), called quanta, each con- taining energy, hf. ""

I think of this as the EM wavefront but whose energy is distributed over the wavefront in a granular way.

The electron gained momentum, and it accelerated. Why didn't it produce an ordinary Maxwellian wave?
 
  • #4
davidbenari said:
Is the thing I said about a linear combinations of the masses of the nucleons (with integer coefficients) a true statement? So that if the mass of the system decreased then necessarily some nucleons disappeared, and its not that some nucleons now "weigh less" ?

The nucleons do not weight less, but there is also a binding energy involved. The binding energies are large enough to make a difference.

davidbenari said:
I think of this as the EM wavefront but whose energy is distributed over the wavefront in a granular way.

This is wrong and not how the classical limit of quantum electrodynamics works. The classical limit is more akin to coherent quantum states, i.e., states which do not have a definite number of photons, but are superpositions of states with any number of photons.

davidbenari said:
The electron gained momentum, and it accelerated. Why didn't it produce an ordinary Maxwellian wave?

Because all of this happened at a level where quantum effects are dominating. At this level, an electron does not accelerate as a classical particle, and it does not emit a classical field in the same way a classically accelerating particle would.
 
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  • #5
Orodruin said:
Because all of this happened at a level where quantum effects are dominating. At this level, an electron does not accelerate as a classical particle, and it does not emit a classical field in the same way a classically accelerating particle would.

Just out of curiosity, when is the picture of the Maxwellian wave created by an accelerated charged particle adequate? When is the phenomenon not too quantum?

Orodruin said:
This is wrong and not how the classical limit of quantum electrodynamics works. The classical limit is more akin to coherent quantum states, i.e., states which do not have a definite number of photons, but are superpositions of states with any number of photons.

In class we did a calculation of the number of photons crossing an area ##A## due to a laser light. It assumed a constant intensity, and the energy of the photons as ##E=hf##. Was this calculation wrong because it supposes a definite number of photons then?

Thanks.
 
  • #6
davidbenari said:
In class we did a calculation of the number of photons crossing an area ##A## due to a laser light. It assumed a constant intensity, and the energy of the photons as ##E=hf##. Was this calculation wrong because it supposes a definite number of photons then?

Thanks.

While you cannot count photons in this way (just as little as you can know the slit the particle went through in the double slit experiment), you can compute the expectation value of the number of photons.
 
  • #7
Also, for a very very very very large number of photons, such as you get in a typical laser, the variations from the expectation value are very very very very tiny, in percentage terms.
 

What is a photon?

A photon is a basic unit of light or electromagnetic radiation. It is a particle that carries energy and has zero mass and zero charge.

What is an electron?

An electron is a subatomic particle with a negative charge. It is one of the fundamental particles that make up atoms.

How do photons and electrons interact?

Photons and electrons can interact through a process called the photoelectric effect, where a photon transfers its energy to an electron, causing it to be ejected from an atom. They can also interact through scattering, where a photon is deflected by an electron's electric field.

What are the applications of studying photon/electron interaction?

Studying photon/electron interaction is important for understanding the behavior of light and matter, as well as for developing technologies such as solar cells, photodetectors, and lasers. It also has applications in fields such as medical imaging, telecommunications, and quantum computing.

What is the relationship between the energy of a photon and the energy level of an electron?

The energy of a photon is directly proportional to the energy level of an electron. When a photon is absorbed by an electron, it can cause the electron to jump to a higher energy level. Conversely, when an electron emits a photon, it loses energy and moves to a lower energy level.

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