What to learn to link EM wave to photon picture?

[Gerenuk]

I know undergrad QM fairly well. There they use the EM potential to introduce EM fields into the momentum.

What should I study to understand the connection between EM fields and actual photon particles?

 

[Manilzin]

Try any textbook in modern quantum mechanics, I recommend J.J. Sakurai - Modern Quantum Mechanics. Basically, a relation between the QM harmonic oscillator and the EM field is established, and the raising and lowering operators of the harmonic oscillators are then interpreted as creation and annihilation operators for photons.

 

[f95toli]

I am not sure Sakurai is the best book for this since it doesn't really cover EM in any great depth.
If you already know the basics (e.g. the first few chapters of Sakurai) of QM you could try e.g. "Introductory Quantum Optics" by Gerry&Knight.
The formalism needed to understand basic quantum optics is actually relatively easy and in most books on quantum optics you will also find discussions about how the formalism/examples relates to real experiments.

 

[borgwal]

Cohen-Tannoudji's book "Photons and Atoms" is all about the fundamentals of QED. Low on applications, though.

 

[0xDEADBEEF]

"Mesoscopic Quantum Optics" by Imamoglu

 

[Gerenuk]

Thanks for all the suggestions. I take a note.

0xDEADBEEF? Is that a common name? Thought I saw it on a Starcraft map...

 

[enotstrebor]

What should I study to understand the connection between EM fields and actual photon particles?

If you want to know the connection between the EM fields and the ``actual photon'' you will not find the answer in textbooks.

The present theory can not tell you if the EM mathematical description is a description of the photon (a property of the particle) or a description of the interaction of the photon (a property of the interaction - an inter-particle property) and thus lacks clearity.

That the EM description is not the photon is also evidenced by the fact that these can not be used when one tries to combine the photon and matter particles (QED), but requires the use of the vector potential. This unification can not be done using the EM description. Lastly the EM photon fields rise and fall together and thus do not continuously conserve total field energy.

To quote Einstein, ``Today every Tom, Dick and Hary thinks he understands the photon but they are wrong.''

 

[Gerenuk]

The present theory can not tell you if the EM mathematical description is a description of the photon (a property of the particle) or a description of the interaction of the photon (a property of the interaction - an inter-particle property) and thus lacks clearity.

Is it possible to explain that difference in a few sentences? For example can you describe what happens when two electrons interaction via the Couloumb interaction? I mean an explanation that fits into a paragraph

That the EM description is not the photon is also evidenced by the fact that these can not be used when one tries to combine the photon and matter particles (QED), but requires the use of the vector potential.

I'm fine using the vector potential for a moment and not understanding what it means.

This unification can not be done using the EM description. Lastly the EM photon fields rise and fall together and thus do not continuously conserve total field energy.

What does that mean? That also reminds me of some momentum problem connected with light beams?! They were argueing about interior and boundary or something... Are there any paradoxes?

To quote Einstein, ``Today every Tom, Dick and Hary thinks he understands the photon but they are wrong.''

Not surprised a clever guys notes that.
That is the most important reason why most people at university fail to achieve exceptional results. One should tell them "Stop believing you know it all, start doubting what you've been told, then make up your own universal complete picture."

 

[jtbell]

One should tell them "Stop believing you know it all, start doubting what you've been told, then make up your own universal complete picture."

"...and don't be surprised if your universal complete picture disagrees with other people's universal complete picture, and don't insist that they accept yours unless you can point to experimental evidence that distinguishes your picture from theirs."

 

[0xDEADBEEF]

Thanks for all the suggestions. I take a note.

0xDEADBEEF? Is that a common name? Thought I saw it on a Starcraft map...

If you allocate memory on a computer it is often filled with whatever another program has put there before. If you read from that memory by accident before putting in values, and your program crashes, it might crash in a different way each time. So on some computers the newly allocated memory is filled with the hex value 0xDEADBEEF. So when you are debugging and looking at variables in hex, and you see DEADBEEFDEADBEEFDEADBEEF in some variable, you instantly know what has happened...
I just needed a stupid name though.

 

[dx]

To understand the relation between photons and the electromagnetic field, you must study what is called "Quantum Field Theory". Particles such as photons, electrons, protons etc., are quanta of the corresponding quantum field. Loosely, quanta are bundles of energy and momentum of the quantum fields.

 

[enotstrebor]

Is it possible to explain that difference in a few sentences? For example can you describe what happens when two electrons interaction via the Couloumb interaction? I mean an explanation that fits into a paragraph

The E,B view of the photon requires the E and B fields to rise an fall together resulting in a violation of conservation of field energy (total energy always includes field energy). But if the E and B are not fields but interaction effects (effective mathematical fields), then a single photon (true) field of a rotational nature (real spin) could effect in the massed particle (simultaneously producing) both an E and B effect. Thus mathematically with a directed rotational field (e.g. a rotating magnetic bar magnet produces a directed rotational field) resulting mathematically with a changing A with time (E= dA/dt) producing the E effect and simultaneously resulting in (B = \nabla A) a B effect where both rise and fall simultaneously as their source is the same (the photon single rotational field)

But the single field produce two effects on the particle making Maxwell's mathematical view of the photon appear to have two fields that rise and fall simultaneously.

Note the single photon field does not change its magnetude and does not violate the conservation of field energy, it is a real (spin) rotational field effect. Note also that now both E and B are velocity dependent effects! E's rotational velocity dependence is hidden by the phenomenological nature (a model of the interaction behaviors, not a model of the interacting particles) of Maxwell's equations.

That the E and B are interaction effects is also indicated by the relationship of the matrix elements of ($F^{uv}$) and the four dimensional gyroscopic field of inertia ($\Omega_{ij}$) which have the same matrix element pattern (G. I. Shipov, ``Theoretical and experimental
research of inertial mass of a four-dimensional gyroscope''). That is to say that the single photon field produces both types of gyroscopic effects in the (real spin) particle's angular momentum, the B field being the traditional gyroscopic reaction at 90 degrees to the spin plane while the E effect is an inplane rotation rate effect.

Not surprised a clever guys notes that.
That is the most important reason why most people at university fail to achieve exceptional results. One should tell them "Stop believing you know it all, start doubting what you've been told, then make up your own universal complete picture."

Yes, and in part believing that mathematics is physics if it produces the correct results.

This along with a lack of fundamental understanding of the phenomenological nature of todays mathematical models which are often only an interaction behavior not the particle behavior, obviates the lack of fundamental clarity, and is an institutionalized problem.