Take the 2-minute tour ×

Suppose particle A interacts with particle B by means of an exchange particle E. Presumably, since this is quantum physics, the exchange particle emission and receipt are just one of many possible paths by which the two particles can interact.

Does that mean that A is in a continual state of superposition of having emitted an exchange particle in every direction? And that B then is in a continual state of superposition of having absorbed it (and thus moved if there was a force involved) and not?

Somehow that seems like the most sensible option, because the alternative is that A knows where B is and sends it E at some frequency determined by the strength of the fields.

The process above would also entangle A and B as they cannot disagree over whether they exchanged E.

share|improve this question

3 Answers 3

If we are in QM, we have to admit that Particle A is "everywhere" and particle B is "everywhere" so their "exchanges" also happen everywhere. Their "exchanges" lead to correlation of relative position of these particles. Both particles are in mixed states but the system quasi-particles (relative motion and center of inertia motion) may be in pure states, as if they belonged to non interacting and "independent" systems.

share|improve this answer

You are correct in your reasoning but I'd want to warn about thinking of A as being in a superposition of having sent a messenger particle and not, and B in a similar superposition. Like you write, the process "entangles" A and B, so the entire process is in a superposition - A+B and A+B+E, for every moment of E which is compatible with your measurements momenta of the incoming and outgoing A and B's.

Note that usually a spacetime (4D) analysis is done so you don't have to think conceptually about A having sent out an E into "thin air".. either there is an exchange (completely described in the 4D volume) or not.

share|improve this answer

the only interesting sublety here that i think its worth mentioning is that, indeed, the state is in superposition of (many) states of the two charges and a carrier, but the carrier momentum 4-vector does not need to be on-shell. You can say that in general, this is what makes different real field bosons from their virtual exchange carrier counterparts

As a quick reminder, being off-shell basically means that $p_{\mu}p^{\mu}$ can be anything, rather than the usual value of zero for photons

share|improve this answer

Your Answer


