I was reading in Griffiths, Quantum mechanics 2nd Edition about the Exchange forces, so it says that identical bosons attract each other, like the case of Einstein Bose condensates, identical fermions repel, and when we incorporate the spin to the wave function it can have the opposite effect.

So I have been thinking, what about photons, they just disperse, not attract nor repel because they don't have electrical charge, while gluons which have color charge tend to form loops. My main question is why don't they affect other photons (unless colliding), shouldn't there be an attraction or repulsion by the exchange force? Or is it because I would need a QFT treatment in that case?

Also I was wondering something related. The book says that exchange force is not a real force but a geometric consequence, what is the difference, and in the case of electrons, they repel each other because of their electric charge, is there any link of charges to exchange forces. If any argument is incorrect I would be grateful if you correct me.

  • $\begingroup$ Well, you answered a bit yourself. You need QFT to describe the geometrical consequences. $\endgroup$ – Tamoghna Chowdhury Feb 27 '16 at 4:15
  • $\begingroup$ We are indeed taught at a basic level that bosons exhibit gregarious behavior, I'm not sure, maybe photons do too, as they are bosons. $\endgroup$ – Tamoghna Chowdhury Feb 27 '16 at 4:17
  • $\begingroup$ Related: physics.stackexchange.com/q/1361/2451 and links therein. $\endgroup$ – Qmechanic Jan 27 '17 at 6:47

My main question is why don't they affect other photons (unless colliding), shouldn't there be an attraction or repulsion by the exchange force? Or is it because I would need a QFT treatment in that case?

Photons are quantum mechanical entities, so yes, it is a QFT case. Any interaction between two photons goes through exchange diagrams. Two photon interactions occur, with very low probability because the diagrams are box diagrams with at least four 1/137 couplings depressing the probability . For light frequencies this is a very small number . This probability grows with energy so even gamma gamma colliders are envisaged.


Firstly the case why fermions 'repel' each other is simple, it is due to Pauli exclusion principle which states that no two fermions can have the same set of quantum numbers. Therefore if you try to confine them to a single state, then they exert a Fermi pressure and try to 'repel' each other. Bosons don't obey the Pauli exclusion principle by the vitue of the spin statistics theorem and hence don't repel each other. Since they don't repel each other, therefore you can clump a large number of bosons into a single state. The same happens for photons as they are bosons.

Why photons don't attract and the case for gluons The Lagrangian for QED doesn't have any terms that cause photon-photon interaction and as a result, photons don't attract/repel each other (in the sense that they don't scatter off each other, although at higher loops there are diagrams in which photon photon scatter but they are indirectly mediated by other charged particles). However the Lagrangian for QCD has gluon-gluon interaction terms along with the fact that gluons carry color charge which results in color confinement, due to which they exhibit different features from photons. If photon would have been electrically charged the situations would have been different.

Exchange forces Now, Exchange forces are descibed by a theory with interacting mediator particles. Let us take the case of intermediate (carrier) bosons described by a QFT. In an example of such a theory, with Yukawa like interactions, that is, where interactions are mediated by bosons, if we have a spin-0 mediator boson, then the interaction is attractive. This can be easily seen by writing down the Feynman amplitude and taking the Born approximation and Fourier transforming it back to the position space. Similarly the interaction mediated by a spin -1 vector boson is repulsive. Photon is a spin -1 boson and that is why like charges repel. The corresponding case for a spin - 2 boson (graviton may be an example) is again attractive and that is why gravity is attractive.

  • 1
    $\begingroup$ "The Lagrangian for QED doesn't have any terms that cause photon-photon interaction" You're putting the cart before the horse. The mathematics is a model of reality; it doesn't cause the properties we observe. $\endgroup$ – David Richerby Feb 27 '16 at 17:37

Photons do not interact with each other, in QED, because there is no such interaction term in the Lagrangian or Hamiltonian. The only interaction term is the electromagnetic field A (4 vector) with charged particles. There are also deep reasons for that. An interaction of two photons would be depicted in QFT as nonlinear in A. The square term would require the photon to have mass, which we know pretty accurately it does not. Other nonlinear terms in A, such as A contracted with some derivative terms cause other problems. The most general form is derived from gauge invariance, and the imposition of other symmetries (like P or T, parity or time reversal) and that the theory be renormalizable (so no infinities), eliminates any other nonlinear terms.

The strong force also obeys symmetries, but the gauge symmetry is non Abelian which allows for the self interactions of gluons. The weak force similarly allows for the self interactions, with some dependence on parity.

The bottom line is that it is not possible to device a spin 1 quantum field (like the photon which is spin 1) that has no mass like the photon, and obeys the symmetries we've observed, and including special relativity, with no infinities, that is self interacting.

The exchange force is a real force, what they mean is that in some cases the photon exchanged between two particles are not seen at the end of the interaction and are labeled virtual photons, as opposed to when we see scattering of charged particles with photons, when you see a photon going in and coming out. They are real forces. In QED and QFT forces are all derived from interactions between particles, and represented by terms that must obey the symmetries we know exist (time, translation, rotation, and in most cases T, C (charge) and P), and special relativity (i.e., the speed of light is the same in all inertial frames of reference). And for electromagnetism an Abelian gauge symmetry which is what causes the field to exist and not interact with itself, and be zero mass. And for all of them no infinities.

When one tries to include gravity quantum theory needs to incorporate General Relativity, and a consistent theory has still to be achieved. String Theory is one contender. So generally one deals with non quantum gravity, with some interactions with quantum fields -- that has been able to be done. I'm just mentioning this to clarify that special relativity is the way relativity fits into quantum theory (quantum theory of fields, QFT), and we'll see what happens in trying to integrate gravity I the future. So, for now, we have a QFT for the 3 forces (electromagnetism, strong, weak), and a non quantum theory for the 4th, gravity, in General Relativity. Each is separately consistent but so far not consistent with each other when gravity is very strong such as in black holes or the Big Bang.


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.