Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Quantum electrodynamics (QED) is the relativistic quantum field theory of electrodynamics that is the first theory where full agreement between quantum mechanics, special relativity and electrodynamics is achieved.

$$F=\frac {q^2}{4\pi \epsilon_o r^2}$$

  • What is, it's 'relativistic-developed' equation?
share|cite|improve this question
A good exposition of how classical fields are the limiting case of the underlying quantum mechanical equations is given by Lubos at… . If one is serious to understand this, one should spend the requisite effort to acquire the mathematics that would help understand the problem and its solution. – anna v Nov 20 '12 at 12:57
up vote 1 down vote accepted

My advice is that you wouldn't take what the wikipedia says too seriously. In the first place, QED is a relativistic quantum field theory (QFT) and, as stated in standard textbooks --e.g. Mandl & Shaw, Landau & Lifshitz--, quantum field theory (QFT) and quantum mechanics are two different theories. In the second place, there is not "full agreement between quantum mechanics and special relativity" --a point already known by the father of relativistic QM and of QED: Paul Dirac-- [*]

Regarding your questions, the Hamiltonian equation of motion

$$\frac{d\mathbf{p}}{dt} = \mathbf{F}_\mathrm{E}$$

where $\mathbf{F}_\mathrm{E}$ is the electrostatic classical force given by $\mathbf{F}_\mathrm{E}={e^2}/{4\pi \epsilon_0 \mathbf{r}^2}$. This equation is generalized in quantum mechanics to

$$\frac{d\hat{\mathbf{p}}}{dt} = \hat{\mathbf{F}}_\mathrm{E}$$

where the hats denote quantum mechanical operators. The above is one of the Heisenberg equations, which are the quantum analogue of the Hamilton equations of classical mechanics. The electrostatic quantum mechanical force operator is given by $\hat{\mathbf{F}}_\mathrm{E}={e^2}/{4\pi \epsilon_0 \hat{\mathbf{r}}^2}$. The explicit expression for the quantum mechanical position operator depends on the representation used.

The relativistic classical equation given by classical electrodynamics is

$$\frac{d}{dt}\left( \mathbf{p} -e\mathbf{A} \right) = \mathbf{F}_\mathrm{L}$$

where $\mathbf{A}$ is the vector potential and $\mathbf{F}_\mathrm{L}$ is the Lorentz force given by $\mathbf{F}_\mathrm{L} = e(\mathbf{E} + \mathbf{v} \times \mathbf{B})$ for given electric $\mathbf{E}$ and magnetic $\mathbf{B}$ fields.

The corresponding relativistic quantum equation proposed by QED is

$$\frac{d}{dt}\left( \hat{\mathbf{p}} - e \hat{\mathbf{A}} \right) = \hat{\mathbf{F}}_\mathrm{L} = e(\hat{\mathbf{E}} + \hat{\mathbf{v}} \times \hat{\mathbf{B}}) \;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$

with $\hat{\mathbf{v}} = c \alpha$ with $\mathbf{\alpha}$ being the vector of Dirac matrices [#]. Here $\hat{\mathbf{F}}_\mathrm{L}$ is the quantum Lorentz force proposed by QED.

You can find the equation (1) in the Eleventh Lecture in Feynman's QED textbook but with a slightly different notation --Feynman uses c=1 units, do not use hats to denote operators, and uses the dot notation for the time derivatives e.g.; $\dot{\mathbf{p}}=d\mathbf{p}/dt$--.

As Feynman correctly notices the QED equation (1) is "not completely acceptable", because there are some mathematical and physical issues with the QED expression for the quantum Lorentz force $\hat{\mathbf{F}}_\mathrm{L}$.

[*] You would also check the sections 7 and 8 --specially section 8.3 "Does QFT solve the problems of relativistic QM?"-- of this recent Foundations of Physics paper. I completely agree with his conclusion:

Thus, instead of saying that QFT solves the problems of relativistic QM, it is more honest to say that it merely sweeps them under the carpet.

[#] The MathJax engine used here does not render correctly bold Greek letters, but in Feynman's textbook you will find a beautiful alpha symbol in bold font.

share|cite|improve this answer

At long distances and low energies the equation you give for the electrostatic force remains unchanged. At short distances and high energies the electrostatic force gets stronger, however there is no simple equation to describe how it varies.

The change in strength of the force is usually described by a change in the coupling constant known as running. This will be described in any book on QED. A quick Google found this popular science level description, and I'm sure a more determined Google would find many similar articles.

share|cite|improve this answer
Is it that the running of the coupling is inherently related to short distances, or does the running appear at big distances too, if you decide to make more elaborate computations (higher order diagrams)? – NikolajK Oct 5 '12 at 7:52
I suppose the running exists at long distances but it's immeasurably small. The running starts becoming important when the energy is high enough to create virtual particles, and the energy only gets this high at short distances. – John Rennie Oct 5 '12 at 8:17
The running essentially stops at distances longer than the Compton wavelength of the electron, i.e. the energy scale associated with the lightest charged particle. – Luboš Motl Oct 5 '12 at 11:30
@JohnRennie, it is possible to increase the energy density over a macroscopic field such that we'll see the Schwinger limit where nonlinear electromagnetic effects in vacuum will be observable. The next generation of high energy lasers might make this possible. read: – lurscher Oct 7 '12 at 3:20
@lurscher: to be honest I don't know. Why not ask that as a question? – John Rennie Oct 7 '12 at 6:10

It is not only the "electrostatic force" that is developed in QED, but also the other notions. For example, the number of particles is not conserved, but the total charge is. Strong electromagnetic fields can create particle-antiparticle pairs so the strong EMFs are "unstable".

As the number of particles is not conserved, one cannot write a meaningful equation of motion for a given charge $q$ - there is no guarantee that it will not suffer a "transfomration" in course of interaction. Instead, one delas with probabilities to observe this or that in experiment.

As well, there is such a thing as identical particles in QM. That prevents us from "following" a given particle evolution and requires taking into account impossibility of distinguishing identical particles.

Thus, not only the electrostatic force is modified, but the "mechanical equations" too.

share|cite|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.