# In $e^+e^-\to e^+e^-$ scattering is the final outgoing state with hard and soft particles together known?

In $$e^+e^-\to e^+e^-$$ scattering there is a production of low energy photons (soft photons). This seems to be connected with infrared divergences. One way to deal with this seems to be the so-called "inclusive formalism", on which one "sums over Feynman diagrams with arbitrarily many soft photons escaping undetected", to get finite results.

I've also read that the hard $$e^+e^-$$ particles in the outgoing state are strongly correlated with the soft photons. I'd like to quantify these correlations, with entanglement measures for instance.

Now, to discuss quantum entanglement we need a state. For instance, if $$\rho$$ is a pure bipartite state (which seems to be the case here), we can quantify entanglement with the Von-Neumman entropy $$S(\rho)=-\operatorname{Tr}\rho \ln \rho.$$

Nevertheless, it is clear that the state is required for us to even start discussing correlations.

But, is the state $$|\psi\rangle$$ of the global outgoing system ($$e^+e^-$$ hard part + soft photons) known? If it is know, how does one finds it?

The closer I've got to an answer was the Faddeev-Kulish approach. But I've also seem people say that "the FK states are highly non-unique", so something is clearly wrong, because then what would be the correct outgoing state?

Another possible answer was at the recent developments on the Large Gauge Transformations at the asymptotic regions of spacetime. In this paper the authors show in the gravitational case, how the conservation of supertranslation charge leads to the soft-particle creation.

But I'm unsure if that is the right way to deal with this.

So, is it known how to describe the full hard + soft outgoing state, and if so how to find it? If not, what is the difficulty and how does one say that "they are strongly correlated" if the global state isn't even known?

• The is no single outgoing state. The distribution of outgoing states can be (and has been) measured to within the limits of various detectors and those can be used to check the theory (which passes). And we write Monte Carlo simulations which match the backgrounds of real experiments pretty well. – dmckee Jan 10 at 14:40
• @dmckee thanks for the comment. I'm sorry, but I'm probably missing something very basic here. Why there is no single outgoing state for the hard + soft system? Does it mean that the final state is described by a density matrix? Isn't that in conflict with unitary evolution? I mean, I can understand that the hard electron part is indeed described by a density operator, because the soft photons are not observed by the limitation of the detectors and we have to trace them out. But I'm failing to see why there should not be a definite hard+soft state. – user1620696 Jan 10 at 14:46
• Perhaps we are talking at cross purposes and I've addressed the wrong question. From the point of view of a greasy-handed experimenter we see the outgoing – dmckee Jan 10 at 14:55