1. Nov 14, 2005

### DaTario

Hi All,

Does any one have a way to estimate the average electric field amplitude of one atomic photon with a given frequency and spectral width (in case it matters) ?

Does any one have a way to estimate the average volume of an atomic photon with a given frequency and spectral width (in case it matters) ?

Thank you so much,

DaTario

2. Nov 15, 2005

### Allday

hi Da Tario,

the way i understand it, the intensity of light is proportional to the photon flux, and the energy of light is proportional to the frequency of light. The electric field amplitude (squared) is proportional to the intesity of light (power/area or W/m^2) therefore it seems that the amplitude would be most closely related with the photon flux and therefore the E-field amplitude for one photon isnt a very well defined quantity.

the volume of a single photon is a pretty tricky idea as well. i think the best way to think about this quantum particle is with a wavefunction. you could then choose a volume and integrate the probability distribution to find the odds that the photon is in whatever volume you choose. take some cut off (say 90% chance) and consider that volume to be the volume of the photon if you like, but remember that the cut off is arbitrary.

gabe

3. Nov 15, 2005

### Kruger

"average volume" [of a photon]

Something like that don't exist. In QFT a photon is just the quantum of the electromagnetic field and the em-field is described by the quantum mechanical oscillator and this oscillator is spread over the whole space. So a photon is spread over the whole space. And therefor you must be a little bit more specifically in what you mean with volumina.

And there isn't such a thing as a "photons wave function".

4. Nov 15, 2005

### Ratzinger

Is that really so? (I'm asking.) Why?

Is that because photons are energy concepts? But aren't there (quantum) wave packets of photons in space??

5. Nov 15, 2005

### Kruger

""" But aren't there (quantum) wave packets of photons in space??"""¨
<-- No there aren't. We can make such packets with quantum harmonic oscillators. And remember the energy states of such oscillators are quantizised (photon)

6. Nov 15, 2005

### X-43D

7. Nov 15, 2005

### Ratzinger

But why can't you just simple take a the EM wave as your probability amplitude for photons?

8. Nov 15, 2005

### Kruger

This would make no sense because we normally connect the amplitude of the photon with the strength of the electromagnetic field, i.e. electric field. And we can measure the same electric field, i.e. classical electromagnetic wave, at two points in space at the same time. But from quantum mechanics we know that if we measure a particle we won't measure it at another space point the same time. And a field cannot be described via particles. We need quantum field theory. And a photon is basically spread over everywhere in space. (We can measure the em-field at several positions the same time and will get results (if we would measure an electron at several positions at the same time we would only measure it at one position)).

Do you see a little bit what I mean?

9. Nov 16, 2005

### reilly

Actually, the so-called semi-classical approach to QED uses the classical energy density, 1/2(E*E + B*B) as an effective photon density. (This is discussed in Jackson's E&M text, and elsewhere I'm sure.

The electric field associated with a photon, with vector potential, A(k,x)=a(k)exp(ikx-kt) + HC, where a(k) is a photon creation operator, is E = - dA(k,x,t)/dt, and is an operator. A great reference for all the gory details of photon fields and their associated operators and their interactions is Cohen-Tannaoudji,et al, Photons and Atoms; Introduction to QED.
Regards,
Reilly Atkinson

10. Nov 16, 2005

### DaTario

I believe there are some possible, in principle, superpositions which yield localized photon states. The last book by Mandel and Wolf on quantum optics and coherence talks about this.

Let me ask the thing positively:
would it be reasonable to produce the volume of an atomic photon (quantum of light produced by an atomic decay) considering it as a cillinder having radius equal to the atomic diameter (10^-10 meters) and height equal to the coherence length of a photon ( 3 meters ) ?

volume = PI x (10 ^-10) ^2 x 3 which is approx. = 10 ^-19 m^3.

Best Regards

DaTario

11. Nov 16, 2005

### DaTario

I have this book, but it seems not to contain any estimative on the mean field amplitude. Perhaps it catains the methodology to be used to find this value.

Thanks

DaTario

12. Nov 16, 2005

### DaTario

Any of you have typical valued of electric field amplitudes of ordinary light fields, laser light 1 mW/cm^2 and things like this ?

I would appreciate so much finding it. I have tried to look up Google, but it has failed to return me any appropriate result to my list of key words.

Best Regards

DaTario

13. Nov 17, 2005

### Staff: Mentor

The time-averaged irradiance (energy flux density) in W/m^2 of a classical electromagnetic wave is $\frac{1}{2} \epsilon_0 c E_0^2$ where $E_0$ is the amplitude of the electric field.