Contradiction between aymptotically free particles in QFT and unlocalization

Physics Asked by AlanR on November 28, 2020

When studying different interactions in any QFT, one always assumes that the IN and OUT states are asymptotically free particles with definite momenta. For example, one assumes that an electron and a positron come from infinitely separated places at $$tto-infty$$, they interact, and another $$e^-e^+$$ pair leaves. When infinitely separated, one assumes they are "free", i.e., one particle does not influence the other one at all because they are "far away".

BUT, we always consider these particles as states with definite momenta, which means that they are completely unlocalized in space, and (I think) there would be no reason to believe there is no room for any interaction whatsoever. So the statement "infinitely separated" wouldn’t have any sensible meaning.

How is then this apparent contradiction solved? What am I missing?

2 Answers

In and out states are not localized and do include the interactions between particles. The idea is that looking at the full interacting Poincare group (which includes the Hamiltonian) there are certain states that transform irreducibly. These are called the one-particle states. Any reducible state can be expanded as a sum of tensor products of irreducible one-particle states. This is just group theory (in a non-rigorous physicist sense).

But the energy of a tensor product of one particle states is just the sum of the individual energies. There is no potential energy correction. So it looks like the particles are not interacting, even though we are using the full interacting Hamiltonian. The resolution is that these tensor product states, which we identified on pretty abstract grounds, should be considered 'asymptotically free' in and out states.

To see the difference between asymptotically free states, which are really eigenstates of the full Hamiltonian, and actual free states, let's split up the Hamiltonian into a free and interacting part $$H=H_0+V$$. This $$H_0$$ isn't what you would naively choose as the free part, because we need to have the same mass spectrum in the free and interacting theories. In other words $$V$$ should contain counterterms related to mass renormalization. But this is something let's not worry about at this point.

Following Weinberg chapter 3 which d_b mentions, let's call eigenstates of the full Hamiltonian $$Psi_a$$, and eigenstates of $$H_0$$ $$Phi_a$$. The $$a$$ index refers to the particle content of the theory (it tells you which one-particle states it is a tensor product of).

Now the important thing is we can expand the $$Psi$$ basis in terms of the $$Phi$$ basis (there are mathematical subtleties about this but let's not worry about them). Schematically, $$Psi_a=Phi_a + int db M_{ab}Phi_b.$$ $$M$$ is just a collection of expansion coefficients I won't bother to write here (the detailed form is in Weinberg's book). This is called the Lippmann-Schwinger equation. The important thing is that interacting $$Psi_a$$ has the same energy and momentum as free $$Phi_a$$, but it is not the same since it also has components from every $$Phi_b$$ that has a non-vanishing $$M$$. These extra components are telling you everything about the interaction. Also you will find that there are two ways to define $$M$$ appropriately. Depending on this choice $$Psi$$ will be either an in or an out state.

Correct answer by octonion on November 28, 2020

This is treated in Vol. 1, Chapter 3 of Weinberg's QFT book. The solution is that instead of eigenstates $$|mathbf{p},sigma,nrangle$$ one should consider wavepackets that are superpositions of eigenstates, begin{align}|Psirangle = sum_{n,sigma}int dmathbf{p} ,g(mathbf{p}, sigma, n), |mathbf{p},sigma,nrangle.end{align} These can be localized in space and time. (The indices and momentum here should be understood as composite labels for multiple particle states.)

Answered by d_b on November 28, 2020

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