1. Jul 3, 2009

### mgd2009

welcome every body

i have question about the photons generation?

i know that there are two ways to generate what so called photon
1-- the first is by atomic transition and i study this way alot
2-- the second by electrons oscillation due to ac current like in antnna
My question is why when election oscillate it generate photon i know that is a stupid question
but if any one have explanation please say it
second question is these two ways are the only ways to generate photon

2. Jul 3, 2009

### invisigo

Wow, I was about to ask the same question. So it's not a stupid question. I don't think the answer is that available.

If I may repeat the question, there seems to be two theories for photon generation that mgd2009 and myself can't seem to resolve. The first is this atomic transition explanation of photon generation and the second is antenna generation of a photon.

I'm going to think aloud here and maybe help clarify this, or maybe someone else can do it.

I can answer part 2 of mgd2009's question, in that electrons oscillating are accelerating. It is an accelerating charge that generates radiation in classical electrodynamics.

Now how this relates with atomic transitions is a tougher question. I would say that atomic transitions require using quantum mechanical theory. We also have to apply quantum mechanical theory to the antenna. In this case, we are dealing with a metal, which we have to look into the results of solid state physics to treat it quantum mechanically. The metal would have a conduction band and a valence band where the electron could be. A photon is generated when an electron falls from the conduction band into the valence band. Perhaps (I don't know if this is true) an electric current fills the metal with a bunch of electrons in the high energy conduction band, then when the current is reversed, the electrons fall into the valence band emitting a bunch of radiation. However, this explanation says nothing about an "accelerating" charge as is described in classical electrodynamics, which is makes reconciling the two theories confusing.

The other confusing thing is that antenna efficiency is dependent on the length of the antenna to the wavelength of the radiation. But using the solid state explanation of radiation that I just made up in the prior paragraph doesn't describe why one antenna should be any more efficient than another.

I would appreciate if anyone else has a clear explanation.

3. Jul 3, 2009

### mgd2009

you say that ((I don't know if this is true) an electric current fills the metal with a bunch of electrons in the high energy conduction band, then when the current is reversed, the electrons fall into the valence band emitting a bunch of radiation))

i think that is not true because if what you say happen, the photon that emitted will have frequency according to the equation E=hf ,where the E=difference in energy bands and in antenna the frequecy of photon is the frequency of the applied ac signal .

what i think it happen in antenna (iam not sure and any one have different opinion please say it )
what happen in antenna is related in the idea to atomic transition
in atomic transition when election fall on it photon with energy E this election go to higher energy level (difference between to levels =E) and after the electron life time in the higer lever end the electron retern back to the origin energy level and emitted the photon that it take

But in antenna due to the ac signal the electron accelerate so it have amout of energy and when the electron deaccelerate it release the amount of energy it take in form of photon
so the frequency of photon is may be related to the ac applied signal with relatationship

4. Jul 3, 2009

### meopemuk

Yes, these are two basic mechanisms by which photons are emitted. 1. Excited states of bound systems of particles (atoms, molecules, nuclei, etc.) are not stable; they emit photons spontaneously and jump to lower energy states. 2. If a charged particle accelerates (e.g., while interacting with another charge) it emits photons. This is also called the bremsstrahlung radiation.

It is interesting that both these mechanisms have a common description in the "dressed particle" approach to QED, see, e.g.,

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139.

E. V. Stefanovich, "Relativistic quantum dynamics",
http://www.arxiv.org/abs/physics/0504062

and references there.

The idea is that the photon emission results from the presence of interaction terms like $a^{\dag}d^{\dag}c^{\dag}ad$ in the Hamiltonian (in this example I use creation and annihilation operators $a^{\dag}, a$ for electrons, $d^{\dag}, d$ for protons, and $c^{\dag}, c$ for photons). This interaction term tells us that in a 2-particle system (electron+proton) there is always a chance of a third particle (photon) being emitted. This happens either when the electron and the proton form a bound system (an excited state of the hydrogen atom) or when they simply move next to each other (and accelerate).

5. Jul 3, 2009

### ytuab

I doubt quantum field theory (QED).

For example, Lamb shift (energy difference between 2S1/2 and 2P1/2 in hydrogen atom) is really caused by the interaction between the electron and the vacuum as QED says, isn't it?

According to Dirac equation, the 2S1 / 2 and 2P1 / 2 orbitals should have the same energies.

the 2S1/2 electron has higher velocity than 2P1/2, and the 2P1/2 electron has the spin-orbital interaction (2S1/2 electron doesn't have the spin-orbital interaction).
So, the relativistic energy difference (which is caused by the difference of the two electron velocity) must be just equal to the energy difference by the spin-orbital interaction. (up to high orders).

The origin of the upper two energy difference (electron velocity and the spin-orbit interaction) are completely different things. So is it really possible that the two energies just coincide up to high orders?

And Dirac equation of hydrogen atom contains the Coulomb potential which is not Lorentz invariant. (this is only an approximation.)
So we need not consider QED about Lamb shift, don't we? ( The solution of the Dirac equation really shows the 2S1 / 2 and 2P1 / 2 orbitals of hydrogen have the same energies? If not, the two orbitals have different energies before considering QED.)

6. Jul 3, 2009

### Bob S

1) Nuclear decay, like cesium 137 and its 661 KeV gamma ray
2) Annihilation gammas like positron annihilation and its two 511 KeV gammas
3) Bremsstrahlung (photons) caused by electrons scattering off of nuclei
4) Synchrotron radiation (electrons in magnetic field)?

Last edited: Jul 3, 2009
7. Jul 3, 2009

### meopemuk

Fundamentally, this is not different from mechanism 1: a compound system transitions between two energy levels with emission of a photon.

These belong to the category 2: photons emitted from free-moving accelerated charged particles.

This is indeed a new category that has not been considered in this thread yet. Low-order QED is sufficient for its description.

8. Jul 4, 2009

Staff Emeritus
Too bad for you. It's a very powerful theory.

The rest of your message essentially is "maybe someone will someday find a different calculation that gives the same answer." That's surely not evidence.

9. Jul 4, 2009

### ytuab

How do you think about the Lamb shift (energy difference between 2S1/2 and 2P1/2 in hydrogen atom)?

In QED, the Lamb shift is caused by the vacuum polarization.
So the 2S1 / 2 and 2P1 / 2 orbitals of hydrogen must have the "completely" same energies up to high orders if we do not consider the vacuum polarization.

As I said before, the 2S1/2 electron has higher velocity than 2P1/2, and the 2P1/2 electron has the spin-orbital interaction (2S1/2 electron doesn't have the spin-orbital interaction).
So, the relativistic energy difference (which is caused by the difference of the two electron velocity) must be just equal to the energy difference by the spin-orbital interaction. (up to high orders). Is it really possible?

The Dirac equation of hydrogen atom contains the Coulomb potential which is not Lorentz invariant. And it doesn't contain the vector potential (which means the specific reference frame.) So the solution of the equation is only an approximation, isn't it?

Do the solution of the Dirac equation really show the 2S1 / 2 and 2P1 / 2 orbitals of hydrogen have the same energies ( in every reference frame)? If not, the two orbitals have different energies before considering QED (vacuum polarization), don't they?

Last edited: Jul 4, 2009