Energy Of A "Single" Photon In Em Radiation?


by Robin*
Tags: energy, photon, radiation, single
jfizzix
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#19
Aug30-13, 05:56 PM
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A photon is a hard thing to pin down (I'm a grad student in quantum optics, so it's something I think about a lot).

A photon follows most of the same quantum rules that other quantum particles do, and in that sense, it's just as hard to talk about a single photon as it is a single electron, or any other particle.

We can only infer properties about photons through the experiments we perform (like everything else). Photons are especially hard to discuss because most of the time, by measuring a photon, you destroy it (i.e. it gets absorbed by a detector).

The full-blown mathematical treatment of a photon (in my field) describes it as a particular kind of quantum state of the electromagnetic field. It's a very low energy state with only one photon in the field, but it is more than the vacuum. You can have 2-photon states, 3-photon states, coherent states, thermal states, entangled states, and many more yet to be thought of.

You can of course make approximations, which is why we don't need quantum field theory to design mirrors and lenses and such (yet). Almost all of the time (when you see quadrillions of photons) Classical Electromagnetism will describe everything just fine (easier said than done).
jfizzix
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#20
Aug30-13, 06:03 PM
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The photoelectric effect is a simple experiment explaining how photons can have an energy proportional to their frequency, provided one is okay with light being divided up into chunks known as photons.

That being the case, scientists have shown that with high frequency light, you can knock electrons off pieces of metal. The brighter that light (the more photons), the more electrons get knocked off, and we can measure these electrons hitting detectors.

However, at low frequencies, no electrons get knocked off the metal, no matter how bright the light is.

We know that if the photon has enough energy, it can knock an electron off, and it can't if it doesn't.

More importantly, we can measure the kinetic energy of the electrons that get knocked off and see that the higher the frequency of the light, the higher the kinetic energy of the ejected electrons.

This relationship between electron energy and frequency of the incident light is precisely what we expect if we say the energy of a photon is proportional to the frequency of the light it's a part of.

It's hard to think of photons on their own, but if a photon is just a very tiny part of a beam of light, then it makes sense to speak of the frequency of a photon as the frequency of the light it is part of.
Robin*
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#21
Aug30-13, 11:21 PM
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Quote Quote by DaleSpam View Post
You could do a frequency sweep, but this would typically be done with RF where the quantum mechanical description (photons) is not terribly helpful and you are generally better sticking with a classical description.

My personal recommendation is that it seems like you should learn classical EM before attempting to learn quantum electrodynamics.

It might help if you identified the system you are interested in. The mechanism for producing photons is different if you are talking about a radio antenna, an incandesent light, an x-ray tube, or a flourescent substance.
I am talking in the context of sun
Drakkith
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#22
Aug31-13, 02:44 AM
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Quote Quote by Robin* View Post
So there is only one stream of photons and no perpendicular streams ?
Ignore photons for now. I'll try to give you a very brief idea of what an EM wave is.

Classically, an EM wave is an oscillation of the electric and magnetic field vectors. A vector is a direction. This means that as an EM wave passes over an antenna, it will cause the electrons in that antenna to move one way and then the other as the phase of the wave changes. If you were to draw an arrow to represent the force exerted on the electrons, it would alternate between "up" and "down". (Or whatever two directions the wave was oscillating in)

Note that this oscillation is NOT in the direction of propagation. Imagine a water wave. The wave moves up and down, but it propagates outward, in a direction perpendicular to the oscillations. EM waves behave the same. They move outwards in a direction that is always 90 degrees, aka perpendicular, to the directions of oscillations. So you can have the electric field vector oscillating "up and down", the magnetic field vector "left and right", and the wave would move forwards.

Well, that's the gist at least. It's not 100% accurate, but it should help you understand what an EM wave is.
sophiecentaur
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#23
Aug31-13, 04:38 AM
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Quote Quote by Robin* View Post
I am talking in the context of sun
I would recommend that you read around a lot more about EM waves before you try to tie them in with the idea of Photons. Your questions and comments are all very random and disjointed and I have a feeling you are jumping into this subject far too deep, to start with. Your present 'picture' of photons is really not leading you anywhere useful and asking questions based on this picture seems just to be leading you further into confusion.
You will not be able to 'bend' Physics to fit your ideas - it has to work the other way round.
Do some reading and then come back to the thread.
DaleSpam
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#24
Aug31-13, 05:52 AM
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Quote Quote by Robin* View Post
I am talking in the context of sun
The sun is a plasma, meaning that it consists of a bunch of charged particles. Because the particles are charged they are constantly emitting and absorbing EM radiation as they accelerate.


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