Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

For electromagnetic waves we have the photon association, one imagines light as particles "flying around".

What is the analogy for a constant electrical field, one which doesn't change in time and maybe not even in space? What shold I "imagine" the photons (or probably some non-particle like superposition of these photons) doing, if the situation is static.

How I came up with this is really through the following question:

Let's say I have two parallel capacitor plates and between these is just vacuum, except for one single atom. Can I "turn on" the capacitor in a way which ionizes the atom? Can I "tackle" the atom from the left by turning on a homogenous electrical field and would that depend on the turning on-velocity?

share|improve this question

5 Answers 5

Sadly but the analogy is same, also "particles flying around" but now we just call them virtual. Which means they dont obey energy momentum relation for the real photon. This is just "talking" on the side of theoretical physics. There is no way to detect anything there. Its just electrical field.

Shortest answer is - there is no clear picture up to date in this area. Conception of a FIELD invented by Michael Faraday is very fundamental and still the most profound one in physics, so you should look at it as a FIELD.

share|improve this answer

Photon Bosons of electromagnetic field don't have enough energy to knock electrons out. So, photoelectric effect isn't possible with electromagnetic field only.

In the photoemission process, if an electron within some material absorbs the energy of one photon and acquires more energy than the work function (the electron binding energy) of the material, it is ejected. If the photon energy is too low, the electron is unable to escape the material. So, photoelectric effect can't be seen even with low frequency light rays. Don't expect it with photons of electromagnetic field.

share|improve this answer
What is bounding the photon energy from above? –  NikolajK Aug 6 '12 at 19:07
The ionization is possible without real photons, namely tunnel ionization. –  twistor59 Aug 6 '12 at 20:35
@Nick Unable to understand your question. Please, clarify. –  S S Aug 6 '12 at 20:44
@twistor59 True, but still its not Photoelectric effect. Its mandatory to knock out one electron per one photon. –  S S Aug 6 '12 at 20:51
You said "Photon Bosons of electromagnetic field don't have enough energy to knock electrons out." and I wondered why the photon energy could not be high enough. @twistor59: Yeah, seems the Landau reference at the end of the article "proves" that this is possible, thanks for that. –  NikolajK Aug 6 '12 at 22:03

The photon is the particle that carries the electromagnetic force i.e. charges exert a force on each other by exchanging virtual photons.

In your example of a capacitor one plate has a positive charge and the other has a negative charge, and the two plates are continuously exchanging virtual photons, which causes the attractive force between the two plates. Your atom in between the plates can interact with the virtual photons, and indeed if you ramp up the field strength there will come a point where the atom ionises. The electron will whizz off towards the positive plate and the positively charged ion will whizz off towards the negatively charged plate.

But whether this is really analogous to the photoelectric effect I'm not sure. In your capacitor the photons are virtual and you can't simply claim a virtual photon hits the atom and ionises it like a real photon hitting a metal surface. Hopefully someone who knows more than me about quantum electrodynamics can comment.

share|improve this answer
Rougthly I'd expect I have to compute the amplitude ${}_{t_0+T}\langle \text{freed electron + lots of photons}|e^{i\int \dots+jA+\dots}|\text{bound orbital electron + lots of photons}\rangle_{t_0}$, but that's quite sketchy. –  NikolajK Aug 6 '12 at 19:14
You shouldn't say "virtual" photon. According to attributes, Its real photon found in light rays. But, it doesn't follow symmetries to violate conservation laws. –  S S Aug 6 '12 at 21:03
Its not photoelectric effect. For photoelectric effect, its mandatory to knock one electron out per one photon. In case of high intensity electric field, density of photons would be higher involving many photons to pull out electron. –  S S Aug 6 '12 at 21:07
@SachinShekhar: It's really about if ionization is possible, you don't have to call it photoelectric effect, if this is bugging you. This was only the analogy. –  NikolajK Aug 6 '12 at 21:56
@Nick If you didn't have special interest in photoelectric effect, then this is the answer. –  S S Aug 7 '12 at 9:46

First of all you have a profound issue with accepting the idea of "photon the quantized excitation of the electromagnetic field".

Light waves, you're good with that. Spherical, flat, standing light waves and even impulses. Just as Planck liked to think about them. But a light .. like a particle? "Naah I don't buy it!" Am I right?

I had this problem for a very, very long time, and it stems from thinking that distortions/waves in the electromagnetic field (or any other field, say gravitational) should at least resemble waves in water. And Maxwellian spherical light waves do that very well. Photons - unintuitive, dreadfully, in this manner.

Until you see this - http://www.youtube.com/watch?v=mHyTOcfF99o - you've probably never seen kind of waves in water, but they're legitimate, as smoke rings from cigarettes. A vortex ring is a stable, quantized directional wave, strongly localized, with a given energy, velocity, and angular momentum.

Since I'm freshly registered, I can't post pics thumbnails, but this is what I would point my finger to - vortex ring

We all need to observe nature more often.

Just imagine those flying around in the electromagnetic field between two plates of the capacitor (more precisely - distortions in the electric and magnetic field intensities, and thus also distortions in the total local energy density of the electromagnetic field, bouncing between the two plates). Just like with a gas of particles, say air, in case of isotropic pressure (constant field strength), the temperature (energy density) is constant between the plates - this is a thermodynamical line of thinking with electromagnetic analogies in the brackets.

Hope that helps you conceptually to get your head around the idea of the photon. Yep, it's unintuitive. Yep, the truth is within the math and the equations, and not in the interpretations and analogies.. but nevertheless, they are helpful. Sometimes.

Sometimes not :P

share|improve this answer
Dophins! Well, I'm not too concerned with the photon discription actually. I guess one has to plug in the \vec E operator between a multi (multi multi) photon state and it will look more and more like the classical electromagnetic field. I'm really asking for the that state, I guess. Might well be that there are a bunch of established restriktions, you know like $\vec k \vec E\overset{!}{=}0$ for each individual momentum eigenstate or things like that. –  NikolajK Aug 6 '12 at 19:21
"Am I right?" No. You're not. QED is one of the best tested theories in existence. The $g-2$ of an electron agreement between theory and data is good to 11 digits. I believe that a vortex ring is an example of a soliton, and that is a different critter than virtual particle exchange. –  dmckee Aug 6 '12 at 19:25
@dmckee: He's not right with what here? Me having a profound issue with accepting the idea of photons? ^^ –  NikolajK Aug 6 '12 at 22:04

If it is ionising an atom with a constant field that you want, that is certainly possible in principle and it works in much the same way that ionization in an intense IR field does, via tunnel ionization.

The way this works is that the constant field adds a linear potential $V=eE_0 x$ to the atomic Coulomb attraction, which means that in some scale the potential looks like $1/r-x$. However deep the atomic ground state is, it is facing a barrier before there is a classically allowed region at the same energy. Thus, it will tunnel out with some nonzero rate.

For laser fields at frequencies much slower than any resonance, the fields appear frozen and therefore ionize using precisely this mechanism. Unfortunately, to get any measurable results, you need to get the barrier appropriately low, and that means producing electric fields of the magnitude of the atomic fields - i.e. (electron-)volts per armstong, or $\sim10^{10}\textrm{V m}^{-1}$, which is waaay outside of anything you can do with DC. With ultrashort laser pulses, though, you can compress a whole mJ or even more into a single few-cycle pulse at IR frequencies, and that can make fields strong enough even for over-the-barrier ionization.

The Wikipedia article on tunnel ionization is a bit short but it has some DC results!

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.