Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon as electrical force carrier

  1. Mar 27, 2008 #1
    I'm reading Feynman's book QED, and while I'll need at least another couple of read-throughs to grasp what I can, there are already a few fundamental questions I have.

    Feynman explains that electrons and protons within an atom exchange photons (p. 113 of my copy). I'm interpreting this as the photon is used to transmit the electrical force between protons and electrons, and thus it is what keeps the electron attracted to the nucleus. Is this basic idea correct?

    If so, how do photons exert the electrical force? I've seen Feynman diagrams of electron-electron interactions using photons, and in these diagrams the electrons repulse each other via what appears to be the conservation of momentum in emitting/absorbing photons. In other words, e1 emits a photon towards e2, and in doing so, will change direction to away from e2 to conserve momentum from emitting the photon. e2 does a similar thing upon absorbing the photon, and thus the two particles have repulsed each other. Is this the mechanism of electrical force, carried out by photons, or am I reading too much into this?

    If so, how does this picture change with a proton and electron such that the force is attractive?

    Furthermore, even if this is explained, it leads to the issue of when photons are emitted. I was under the impression that electrons in high-energy states randomly emit photons, and return to lower-energy states. If photon exchange is fundamental to proton-electron attraction, it would seem that much more regular exchange of photons would be necessary, even if the electron is already in the lowest energy state -- and thus, how can it emit a photon towards the proton? And what if the photon "misses" the proton? How is that energy recovered/regained? Is it the case that in some circumstances particles "shoot" photons at other nearby particles, and this is different from an excited electron randomly emitting radiation?

    Explanations, or a pointer to something that explains this to a physics newbie are much appreciated!
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2
    Welcome to PF !
    Excellent book.
    Superficially, yes.
    This matter is not trivial. I strongly suggest you begin by reading John Baez's FAQ About Virtual Particles

    You must realize that we are talking about virtual processes here. Quantum mechanics is mandatory to make sens.

    I would also advise you to check "The Quantum Vacuum, an intro to QED" by P.W. Milonni. There is in 3.3 entitled "Atomic stability" a very enlightening (but short, and semi-classical although justified) discussion.

    I would actually be very happy if a theoretician could give his opinion on this book, or more precisely this very paragraph. I use to think very differently until I read that. I know that we can pretty much calculate everything about the hydrogen ground state except the Lamb shift ignoring virtual photons. I was therefore under the feeling that QFT (canonical) language is not appropriate to bound state problems. In fact, it is suited to calculate scattering problems. If you take the electron/proton Feynman diagrams, perturbation expansion, none of them will generate a pole at [tex]\sqrt{s}=m_{p}+m_{e}-13.6[/tex] eV. We are dealing with a non-perturbative situation. Usually people perturbe well defined wave-functions calculated by other approximate means, not the free asymptotic states, so it basically amounts to working in a different basis. However, as shown by Milonni's argument, there is a very interesting separation of contributions to spontaneous emission rates. You can separate vacuum fluctuations and radiation reaction in a unique (meaningful) manner. This had been shown rigourously and generally by Cohen-Tannoudji et al long ago (1982). Both terms contribute equally. In an excited state, they give the same contribution. In a ground state however, they cancel exactly (of course, no spontaneous emission in the ground state. But both contributions are non-zero and cancel). Now comes the punchline, if you write this cancellation you find Bohr's quantization condition (for the ground state) [tex]L=\hbar[/tex]. Assuming a Coulomb potential, you find the 13.6 eV of course. This deeply fascinates me, I mean the physical interpretation. If a someone is interested to comment, I would be glad to reproduce Milonni's argument here in details.
    Last edited: Mar 27, 2008
  4. Mar 27, 2008 #3
    Thanks! I look forward to lurking and occasionally posting some silly questions. :)

    Excellent. This is exactly the kind of thing I was hoping for. I did a first pass, and am looking forward to rereading it at a more leisurely pace.

    Fair point. While I "realize" that at a high level, it's all too easy to slip back into comfortable real-world analogies. In this case, the fact that momentum seemingly explains electrical repulsion so intuitively made me try to think of attraction in an intuitive way as well.

    I took a peek on Amazon, and the book appears to be fairly math-intensive. While I'd like to eventually (re) learn the math necessary to do some non-trivial physics, I'm not at all there now. Let me know if my perception of the book is incorrect.

    Thanks for your answer!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook