# B Photon shape

1. Apr 3, 2017

### Daniel Petka

What is the best way to imagine a photon?

Sometimes I hear that they are simply transverse waves with oscillating EM fields, which just move forward. Other times, though, the photon is said to move along the wave.
The third image reminds me of a probability wave.

So which one is correct?

In fact, none of them really explains, why a photon doesn't interact with a particle that is smaller than its wavelength. I am not able to imagine it well and that confuses me.

2. Apr 3, 2017

### woody stanford

How about a guess? The answer to your first question: a particle (I think I'm on firm physics ground with this one).

But, we do know about wave/particle duality, so I suppose the visualization of it in your first diagram (a classic in 20th century physics and true) is the best, most accurate way to conceive of its "shape". I have to admit a certain skepticism of a particle travelling in the wobbly path conjecture, seems somehow...inelegant.

I personally visualize it in terms of W/P duality that there is a particle there (like in your second diagram of the little quantas shooting along) but its a wave as well (like the little quanta waves there). But I don't think the physics community as a whole collectively has anything to say past what they have said.

3. Apr 3, 2017

4. Apr 4, 2017

### Daniel Petka

Anyways it's always a sine function, right?

5. Apr 4, 2017

### Staff: Mentor

The best is not to try an imagine photons at all!

For the vast majority of situations, simply consider a classical EM wave, as in the first picture you posted. Even then, you have to be careful to remember that the axes perpendicular to the direction of propagation represent the electric and magnetic fields, and don't have a physical extension.

6. Apr 4, 2017

### Jilang

If the photon has a wavelength larger than the particle there will be interference effects, which tend to cancel each other.

7. Apr 4, 2017

### woody stanford

The electrical component (I think). Interestingly you can model the magnetic component with cos(X) which places it at the right phase for the model.

Here is a very cool equation I found:

f(x) = y = A sin (kx+d)

To use it you realize that A affects amplitude, k inversely affects wave length and d provides a phase shifter. I thought it was very cool when I came across it. For example for red visible light of wavelength 500nm, you just set k=(2*PI)/5E-7 in your computer.

If you use g(x) = A cos (kx+d) you can get a really interesting model of the above diagram (the EM diagram of light). A simple way to do it.

AND if you like the little wiggly quanta things, to do that you take two sine waves, slowly adjust the phase of one, add the two and all of a sudden you have a wonderful little gadget that produces them...btw.

8. Apr 4, 2017

### Staff: Mentor

That will indeed give you little wiggly wave packets, and they are indeed fascinating (I don't think it's possible to tire of the things that can be done by superimposing waves).... But all of this is part of the classical wave model of electromagnetic radiation, of which @DrClaude spoke above. It has nothing to do with photons, and the wave packets that appear when you play with sine waves are not photons.

9. Apr 5, 2017

### edguy99

I like this picture as it reflects many aspects of the photon, including reinforcement, cancellation and wave nature that one can visualize diffraction around corners.

Landau & Lifshitz Vol. II. On page 108, the wave equation section talks about electromagnetic waves “in which the field depends only on one coordinate, say x (and on the time). Such waves are said to be plane”. Electromagnetic waves are ever changing “plane waves moving in the positive direction along the X axis”.

Last edited by a moderator: Apr 30, 2017
10. Apr 5, 2017

### Staff: Mentor

You forgot something there...

11. Apr 5, 2017

### calinvass

I think the animation might be misleading, at least it is to me. What is the spot that changes size? In a normal wave, the maximum amplitude is always present.
I don't see how we can draw a photon based on QM description.
The wave and the particle don't exist at the same time. It is either one or another.

Last edited: Apr 5, 2017
12. Apr 5, 2017

### edguy99

The spot is representing the Landau & Lifshitz plane wave moving along the X-axis.

13. Apr 5, 2017

### vanhees71

A photon is, by definition, described as a normalizable one-quantum Fock state, and in this very specific sense you deal indeed with "wave packets".

14. Apr 5, 2017

### Daniel Petka

The photon doesn't change in "size" as some would suppose, the electric an magnetic fields don't have any influence on the position, do they?

15. Apr 5, 2017

### Daniel Petka

Btw I highly doubt the second picture (oscillating particle) I posted to be true, since it would mean that the amplitude has an influence on the position of the photon, which it doesn't, right?

16. Apr 5, 2017

### vanhees71

A photon is a state of the quantized electromagnetic field. It is highly misleading to think of it as a little billiard ball. Since it is a massless quantum of spin 1, you cannot even define what "position of a photon" should mean. There is also no wave-particle duality, as you can often read in popular-science books (and, horribile dictu, even in textbooks at the highschool or university level; there are way too many bad textbooks on quantum theory around ;-)). This was a bunch of ideas floating around between 1900-1925 when the physicists where forced to rethink the very fundamentals of physics from scratch by observations and experiments. The upshot is that the theory found to describe matter until today is modern quantum theory, developed by Heisenberg, Born, Jordan and Schrödinger as well as by Dirac.

In the non-relativistic case (applicable to massive particles which are much slower than the speed of light) you can describe a particle's state with a wave function (in the formulation of modern QT as "wave mechanis" a la Schrödinger). Although Schrödinger first thought, he could think of the particle as being just this wave function, i.e., an extended object smeared over all space, this idea is not confirmed by observations. Whenever you measure, e.g., an electron, you find a single spot on the screen, not a smeared cloud of spots! On the other hand, using many electrons (e.g., running through a double slit) hitting the screen under certain circumstances leads to a pattern, which is well described by the wave function calculated from Schrödinger's equation. The up to today accepted interpretation of the wave function is that it's modulus squared is the probability density to find an electron at the given spot, i.e., $|\psi(x)|^2 \mathrm{d} x$ is the probability to find the electron in a little intervall of length $\mathrm{d} x$ around $x$.

The relativistic case is more involved. There you don't have a fixed number of particles since in collisions at relativistic energies you may always create and destroy particles. So you need a formalism that deals with such reactions, changing the particle number, and the physicists figured out early that the most elegant way to describe it is Quantum Field Theory. Also in relativistic physics there is the additional possibility to have particles (or better said particle-like excitations of fields) that are massless (you cannot define such a thing in non-relativistic physics; at least if you formally do so it doesn't lead to anything observed in nature), and these massless quanta have quite weird properties which are not easily described in any everyday language.

17. Apr 5, 2017

### edguy99

It is common to see the photons as little wiggles as in your first post. Sometimes "size" is used to signify wavelength (say 300nm) , ie. a little photon would signify high energy and fast wavelength, while splitting that photon into two with a crystal and getting two 600nm photons would be shown larger (physically) and slower to cycle in an animation.

Size in this kind of animation is trying to present visually, the equation E for energy = h / λ, that tells us a photon with low energy will take much longer to complete one cycle of the wave then a photon with high energy.

18. Apr 6, 2017

### Daniel Petka

So the amplitude doesn't matter, just the wavelength...it's really hard to imagine

19. Apr 6, 2017

### woody stanford

OK, question, if photons are particles of light, then can you have an RF photon? I mean its "light" (EM radiation) right? I'm pretty sure there are gamma and x-ray photons, so why not RF ones? And if there isn't where is the cut off point (and I want it down to one decimal place of precision....Mhz)...kidding. lol :D

20. Apr 6, 2017

### Staff: Mentor

You can have photons of any frequency/wavelength.