Are photons the wave sources of Huygens-Fresnel principle?

Click For Summary
SUMMARY

The discussion centers on the relationship between photons and the Huygens-Fresnel principle, asserting that while photons are quantum mechanical objects with wavelengths, they do not act as point sources of waves. The Huygens-Fresnel principle states that every point in a wavefront can be considered a source of secondary waves, which is conceptually distinct from the nature of photons. In the context of quantum electrodynamics (QED), photons can be related to the electromagnetic field, but they are not the sources of that field. The principle serves as a heuristic for solving wave equations, particularly in three-dimensional electrodynamics.

PREREQUISITES
  • Understanding of the Huygens-Fresnel principle
  • Familiarity with quantum electrodynamics (QED)
  • Knowledge of electromagnetic theory and fields
  • Basic grasp of wave equations and Green's functions
NEXT STEPS
  • Study the mathematical formulation of the Huygens-Fresnel principle in wave theory
  • Explore quantum electrodynamics and its implications for photon behavior
  • Learn about Green's functions and their application in solving wave equations
  • Investigate the differences between classical and quantum descriptions of electromagnetic fields
USEFUL FOR

Physicists, particularly those specializing in quantum mechanics and electromagnetism, as well as students and researchers interested in wave theory and quantum field theory.

DoobleD
Messages
259
Reaction score
20
1 - Huygens-Fresnel principle states that every point in a light wave is itself a wave source,
2 - light is made of photons,
3 - photons have a wavelength, they are QM objects.

Hence my question : are photons the actual wave sources in Huygens-Fresnel principle ? Or are those two interpretations totally different things ?
 
  • Like
Likes   Reactions: member659127
Physics news on Phys.org
They are completely different things.
 
Nugatory said:
They are completely different things.

Thank you for answering. Is there a physical explanation for the wave sources of Huygens-Fresnel principle ? I think I read once it had something to do with QM ?
 
Historically yes, totally different things, but the modern view in the context of quantum electrodynamics or qft two pictures are not completely irrelevant I would argue, since the "photon" is explained in terms of the field.

Edit: good question by the way
 
erbahar said:
Historically yes, totally different things, but the modern view in the context of quantum electrodynamics or qft two pictures are not completely irrelevant I would argue, since the "photon" is explained in terms of the field.

Edit: good question by the way
But a photon is not a point source and is not of defined size.
In addition, we know that Huyghens gives the right answer, but is it for the correct reason? We only see diffractiion where there is an obstruction i.e. charges. Maybe radiation from the currents on the diffracting object give the same result?
 
tech99 said:
but is it for the correct reason?
For mechanical waves, yes! Absolutely... Just imagine every point of the wavefront of a water wave trying to "push down" and acting like a point source... Similar sort of argument can also worked out for EM waves like the "local" disturbance in the EM field. And since QFT is based on fields I believe the two pictures are not totally irrelevant...
 
On the fundamental level the sources of the electromagnetic field are electric charges and magnetic moments of the elementary particles (quantum fields). Since electromagnetic theory is an Abelian gauge theory the photons themselves are uncharged and thus not sources of the electromagnetic field.
 
Hyugens principle is one of the family of mathematical theorems that allow the "change of basis".

An example is Fourier transform that allows any periodic function to be re-expressed as weighted sum of sines.
Another example is Taylor series that allows any smooth function to be re-expressed as weighted sum of polynomials.
A very simple example is a change of basis of a vector space. You can take any linearly independent set of vectors and then for any vector you can re-express it as a weighted sum of such vectors.
There are many more such theorems.

Hyugens principle is of similar kind. It says that any function satisfying a wave equation may be re-expressed as weighted sum of spherical waves originating at every point of space.

So yes, photons in quantum dynamics do support such reexpression, so they satisfy Hyugens principle.

Moreso, this is exatly the rationale behind the Feynman quantization, with an important constraint. Feynman said that any quantum field may be re-expressed as a weighted sum of a special basis of fields, where every field has the same amplitude and they differ only in phase.

To say it once again:
You want to have a certain sophisticated field of unit amplitude and some phase.
You select a basis of simple fields, each originating at every point of space, which have unit amplitude and zero phase (by some convention).
You can modify the above fields by changing their amplitude or shifting their phase, and you can sum them up.
It turns out that you don't need to change their amplitude to derive any field you need. You only need to shift their phases. Quantum field theory can pass under this important restriction.

Once more, photons (or any other quantum field) satisfy the Hyugens principle and this is an important fact.
 
  • Like
Likes   Reactions: kurt101 and member659127
Huygens's principle is a heuristic principle, guiding you to find an ansatz to solve wave equations under certain conditions (e.g., the problem of refraction on openings, lits or gratings etc.). It's in some approximative sense quite well satisfied for electrodynamics in 3 spatial dimensions (Kirchhoff's approximation). It's not at all valid for 2D wave problems.

The much more precise and powerful mathematical concept is that of Green's functions of the corresponding linear differential operators relevant for the equation in question. For the wave equation the paradigmatic example is the d'Alembert operator ##\Box=1/c^2 \partial_t^2-\vec{\nabla}^2##.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 64 ·
3
Replies
64
Views
5K