Energy Distribution of EM Radiation ~ One Photon

  • Thread starter kcodon
  • Start date

For a single photon emission, the r=ct sphere represents:


  • Total voters
    7
  • #1
81
0
Hi All,

I've been recently reading a thread, and a question keeps popping up there. Its not very difficult, it just requires a simple yes or no answer...in a single photon emission, is the energy of the photon distributed evenly on the sphere r=ct, or does this sphere represent the probability of the photon being there, and in fact all the energy of the photon is at one point...i.e. a detector would get all of the energy. Ok not quite yes/no but you get the picture.

I believe in the latter, however I would like to know what others think, thus the poll.

Hopefully this will be short and sweet,

Kcodon
 

Answers and Replies

  • #2
chroot
Staff Emeritus
Science Advisor
Gold Member
10,226
34
The photon cannot be said to be anywhere until it interacts with something. When individual photons interact with matter (like a screen), they always act at discrete locations.

- Warren
 
  • #3
81
0
So you're in favour of probability?

Darn, I should've posted this in QM I think...

Kcodon
 
  • #4
mda
117
0
The second option is obviously correct.
The first is the average result for a large number of emissions.
 
  • #5
Claude Bile
Science Advisor
1,471
19
In QM a photon with a spherically symmetric wavefunction, and a similar photon that has been observed are two different scenarios.

By my understanding, for a photon that hasn't been observed, the energy will be distributed symmetrically. The wavefunction that describes the photon represents the probability of the photon being detected at that point.

For a photon that has been observed, the energy will be localised at a point, and as such the detector making the measurement will receive all the photons energy. The wavefunction of the photon in this case is obviously very different to the previous case.

Claude.
 
  • #6
81
0
Thanks again Claude,

Your explanation makes sense - I never considered how under QM they are considered as different events so to speak. I suppose I need to clarify then - I am reffering to what energy the detector would observe, and it seems we agree it would get the full energy of the photon.

Kcodon
 
  • #7
119
0
I could've sworn I posted in this thread.

The second one is correct, the 'wave' can be understood as a wave until it is detected, in which the wave function collapses and it retains particle-like characteristics.

Look up: The Copenhagen Interpretation
 

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