In quantum field theory, the concept of a force is not explicitly present, and we speak of interactions. I guess we could say that a force is an emergent phenomenon.

Interactions manifest themselves in (at least) three important ways: particle decay, scattering and forces.

In (introductory) QFT, typically the first two, particle decay and scattering processes, are treated extensively and computations are made, but forces, e.g. the repulsion between two fermions, or bound systems, are usually not studied. My questions:

  • What is the reason for that? One possible reason I could think of is that for forces we need the concept of a localized particle, whereas in QFT it seems that we mostly work in the momentum representation. Maybe it is just much more complex (for this reason or others) to study forces than decay and scattering.

  • Is it possible to see, not necessarily very formally, how an attractive or repulsive force could emerge from an interaction Lagrangian?

  • $\begingroup$ Sure, you can calculate something like a force in QFT. For example, if $\phi$ is a force carrier field, you can add a source $J(\mathbf{x}) = e \delta(\mathbf{x}) + e \delta(\mathbf{x} - \mathbf{x}_0)$ and compute the energy of the resulting field configuration, which can be done with standard Feynman diagrams. Then differentiate with respect to $\mathbf{x}_0$. (This is sort of an artificial example, though.) $\endgroup$ – knzhou Oct 20 '16 at 23:59
  • 1
    $\begingroup$ Potentials, for example, the Coulomb potential (and therefore, the force), is evaluated using Born approximation in almost all QFT texts. @doetoe $\endgroup$ – SRS Jul 10 '17 at 12:22
  • $\begingroup$ @SRS Thanks. If I understand correctly, the Born approximation perturbatively computes the scattering given a potential. In QFT we can compute the scattering, and turning this approximation around, we obtain a potential that would give rise to that scattering, is that a correct way to understand it? $\endgroup$ – doetoe Jul 10 '17 at 12:33
  • 1
    $\begingroup$ @doetoe Exactly! $\endgroup$ – SRS Jul 10 '17 at 13:31

In quantum mechanics forces are emergent. They can arise when you try to keep track on the correlations through time of your quantum system. You may try to measure over and over a position of a quantum particle and see if the correlations obey some dynamical law (Newton's law).

In quantum field theories, because locality, forces are always produced by interactions, i.e. correlations made by virtual particle exchange, or field fluctuations if you prefer. So, a repulsion between particles are actually a probabilistic phenomena. The probability of two electrons be close together decreases as they get close. This is so because they have a probability to exchange virtual photons that affects the probability of each particle, producing this kind of correlation. Everything in quantum mechanics are probabilistic phenomena.

You can see all this applied to the Coulomb force here

Also, you can define force in quantum mechanics through:

$$ F\leftarrow i\left[H,\,P\right]/\hbar $$ that is, how momentum operator behaves under infinitesimal evolution on time. You can see more here. But all this is yet a probabilistic phenomena.

  • $\begingroup$ I think force is a out-dated concept in modern physics. We rarely use force and Newton mechanism, instead we use action, lagrangian, etc. $\endgroup$ – Turgon Oct 21 '16 at 2:42
  • $\begingroup$ A force is nothing more than a gradiant of a virtual work ;) You change a little bit your configuration and see where the energy are going. $\endgroup$ – Nogueira Oct 21 '16 at 4:38
  • 1
    $\begingroup$ I think OP would be happy to see a draft of the calculation of the Coulomb potential, for example, from first principles of QFT, that is from the form of the E/M propagator. $\endgroup$ – Prof. Legolasov Oct 21 '16 at 11:53
  • $\begingroup$ @Paradoxus, good suggestion. I gonna do this when I found some time. Tree-level approximation giving the Coulumb potential. $\endgroup$ – Nogueira Oct 21 '16 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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