I am reading about scattering and S-matrix in the context of quantum field theory and although I understand the math and the physical interpretation of the final results, I am confused about some initial assumptions. 1) They assume that the particles are free, at $T=\pm \infty$ and hence they have definite value of the momentum (i.e. they are on shell). Does this mean that they are also located spatially at $\pm \infty$? As they travel with momentum p, which is finite, and we collide them at a measurable point in space and time $t_0,x_0$ it would mean that they also travel an infinite distance in space, but I am not sure. 2) They assume that the field vanishes at spatial $\pm \infty$ and I am not sure I understand why. In a free theory, say $$L=(\partial_\mu \phi)^2$$ the solutions are, classically, $\phi(x)=e^{ipx}$ and in QFT they are a superposition of creation and annihilation operators i.e. something like $\phi(x)=\int(a_pe^{ipx}+a_p^\dagger e^{-ipx})$. In both cases, $\phi(x)$ doesn't vanish at infinity, so why would the field in the interaction theory would vanish at infinity. I understand that we can't (usually) calculate it exactly, but why would the solution to that interaction Lagrangian would vanish if the solution to the free one doesn't vanish? What exactly do they mean by the field vanishing at infinity? 3) They say that at $\pm \infty$ (when we create our initial momentum states) the vacuum state is $|\Omega>$ which is (usually) different from the normal free vacuum state $|0>$. If the field vanishes at infinity, why would the vacuum be different there? I think I am confused about considerations of spatial versus temporal infinities. Any answer would be greatly appreciated.

  • $\begingroup$ The asymptotic states are just as in QM.. the momentum eigenstates are a superposition or position eigenstates and the states can decay with position as well $\endgroup$ – InertialObserver Sep 21 '18 at 8:02

Think more physically than mathematically then you will understand everything better.

particles do not have infinite separation before and after the event because we considered only finite range interaction potentials (& for potential like coulumb, we need to have a different approach to account for infinite range or we could just make it a finite range potential by invoking shielding). e.g., Consider H atom in ground state. the ground state energy of e is -13.6eV. that means if we supply 13.6eV, the e will become free. Similarly, mathematically when we say that T tends to + or - infinities the particle are free we only physically mean that our scattering event time range is very small and before and after this event time, all the particles are free. We also mean that they are widely separated(not infinitely). Also, particles are just localized excitations of respective fields. And for our purpose*, fields do exist all over the spacetime but the interaction hamiltonian or lagrangian do not. This is gurunteed by finite or shielded potential.

This answered your 1) and 2) parts.

3) Can have similar explaination, when T tends to +-infinities, vacumm is just |0>,because particles are free. But during the event vacuum will change due to interactions. Therefore, the vacuum state will change with time.

{EXTRA }Now, i'm explaining your doubts regarding edge of the universe(in comments bellow):

By Noether theorem, in large scale Total energy of the universe is not conserved becuase of violation of time translation symmetry: Our universe is expanding(accelerated), the reason is dark energy. Dark energy density is constant(or may be varying very very slowly) but universe is expanding very very fast(volume is incresing rapidly), therefore total energy is not conserved.

Also, if our universe do have a boundary i.e, if it have a closed shape or it is not infinitely flat then the space translational symmetry is also broken at this edge.(since there is no spacetime beyond that) hence from Noether theoren, momentum will also not conserved.

In our scattering processes, we do not care for such a boundary or expanding space, because our processes happens only locally in a very very very small region, where we have spacetime translational symmetries and hence consrvation of energy and momentum.

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  • $\begingroup$ Thank you for the reply! I am still a bit confused about the vanishing of the field. If I think of the Higgs field (which is the only scalar field we actually observed), it has a non-zero value everywhere. There is no point in the universe where you can place a particle, such that it can be considered free from the Higgs field. Same for the vacuum, in any point of the Universe, the Higgs field affects the vacuum, so it will always be different than the free (no Higgs) vacuum, no matter where you are. I know I am missing something, but I am not sure what. $\endgroup$ – Alex Marshall Sep 20 '18 at 21:55
  • $\begingroup$ May be the boundary conditions of the universe are such that all field vanishes! We can't even think of such a bizzar boundary of the universe. $\endgroup$ – Aman pawar Sep 20 '18 at 22:55
  • $\begingroup$ No but I mean, if the theory is invariant under Poincare group, and my scalar field is not zero here, or in the Andromeda galaxy, why would it be zero at the edge of the Universe? If you think of electron-positron pair production in a vacuum, that definitely doesn't happen in a free theory, but it happens in QED. Why would you assume that far away (even if I am not sure what that means if the physics is invariant under translations) the vacuum would be free and you wouldn't have these pair productions? $\endgroup$ – Alex Marshall Sep 21 '18 at 0:01
  • $\begingroup$ I'm editing my answer to give you a more suitable explaination about your question and about the universe edge fuzziness. $\endgroup$ – Aman pawar Sep 21 '18 at 15:11
  • $\begingroup$ @Alex if you satisfied with this answer then please accept it. $\endgroup$ – Aman pawar Oct 4 '18 at 11:05

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