# Electron vs. Photon's electrical potential

1. Jul 26, 2013

### zheng89120

Hi guys,

After reading a paper about electron scattering, I became very interested. So if an electron is travelling in empty space, the electron's potential (according to the paper) is supposedly:

http://www.flickr.com/photos/31694672@N07/
(I used flickr and Glui to post the particular Electron equation)

In the equation, the high-lighted part is an associated Legendre's function, μ is cos θ, and I was not sure what ω would be.

Anyhow, I was wondering what the potential would look like for a PHOTON, travelling in empty space (instead of an electron).

(Un-necessary background information: I was reading a paper about how an Electron is scattered by a dielectric hemisphere. The theoretical space was divided into 3 parts: 1. before the electron hits the hemisphere, 2. when the electron is within the hemisphere or passing right by the hemisphere, and 3. after the electron passes by the hemisphere. The paper presented the potential for 1, 2, and 3. I wanted to focus on part 1 for now, being the topic of the paragraphs above.)

Z. Zheng

Last edited: Jul 27, 2013
2. Jul 26, 2013

### Simon Bridge

That equation was: $$\varphi_1(\vec{r},\omega)=\sum_{l=0}^{\infty}\sum_{m=-1}^{l}A_{lm}\frac{a}{r^{l+1}}P_m(\cos\theta)e^{im\varphi}$$

I think a single photon belongs to a different model, so you wouldn't represent it that way.

... citation please: important for context.
Sounds like you are reading around plasmons etc. There are equivalent papers for photons to the electron scattering ones.

Last edited: Jul 26, 2013
3. Jul 27, 2013

### zheng89120

Right, so the paper that I was referring to was called "Electron-energy losses in hemispherical targets":

http://dipc.ehu.es/aizpurua/publications/PRB54-2901.pdf

So I was not sure where to start in order to calculate the electrical potential of a Photon, travelling in empty space, being massless and charge-less.

Last edited: Jul 27, 2013
4. Jul 27, 2013

### Simon Bridge

Yep - a photon is the EM field - so you'd be pressed to figure how "the electric potential of a photon" makes sense. We also don't normally think of them as interacting with potentials either.

Off the paper - the potential is associated with the hemisphere rather than the electron, and w is the plasmon frequency. The paper is explicitly written in terms of classical dielectric theory - in this framework, light is an electromagnetic wave: the photon model is not applicable.

If you go to photons, then the electron is handled differently too and you need a QED-style approach.

5. Jul 30, 2013

### zheng89120

Hello,

Thank you for your informative replies. Anyhow, I was just wondering if there is any possibility of approximating a photon with classical dielectric theories, and without quantum electrodynamic theories. My 'photon as scattered by hemisphere' project was suppose to be due April/13, hence has been very past due already. I think only an approximation of how the photon gets scattered by a (dielectric) hemisphere w'd be suffice. Could there already be a calculation of a photon as a dielectric function, similar to:

φ1(r⃗ ,ω)=∑l=0∞∑m=−1lAlmarl+1Pm(cosθ)eimφ

?

6. Jul 30, 2013

### zheng89120

Hello,

Thank you the replies. Unfortunately, my 'photon as scattered by hemisphere' project was suppose to be due April/13. So I was wondering if there is any possibility of representing a photon as some classical dielectric function, without using quantum electrodynamic theories. Could a photon be represented in a way similar to:

$$\varphi_1(\vec{r},\omega)=\sum_{l=0}^{\infty}\sum_{m=-1}^{l}A_{lm}\frac{a}{r^{l+1}}P_m(\cos\theta)e^{im\varphi}$$

(thank you for the Latex)

7. Jul 30, 2013

### Simon Bridge

A photon is a QM object from the QM models.
The wave-optics is what happens to the photons on average.

A scattering experiment would normally be modelled by posing plane-wave states for the incoming beam - which may not be monochromatic, so you get a sum, maybe a continuous sum, of modes.

There are a lot of papers which parallel the one you cited, but using photons. You should go look.