# What does it mean for particles to be "far apart" in QFT

I'm currently reading Schwartz's QFT text "Quantum Field Theory and the Standard Model" (although this question does not specifically pertain to this text) and I've got two questions:

1. At the beginning of chapter 6 he talks about how we may consider the system to be non-interacting in asymptotic times, ostensibly because "the particles are far apart". I find this very hand-wavy so my question is can this notion be made rigorous?

Take a massless scalar field with a ## \phi^3## interaction, what's stopping a final momentum eigenstate from splitting into two particles each with half the momenta?I don't see how one could find free asymptotic states in a theory such as this.

2. My second question relates to the first question. Take a massive scalar field with a ## \phi^3## interaction for example. Now even though the momentum operator is unaltered by this term, and hence the ##a_p## at fixed times are still creating/annihilating momentum eigenstates, the Hamiltonian is altered with a ## - \frac g {3!} \phi^3## term which means that the ##a^{\dagger}| 0 \rangle## states are no longer energy eigenstates. However the asymptotic states are said to be on-shell, so does that mean that we are altering the Lagrangian/Hamiltonian at asymptotic times? I'm very confused.

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## Answers and Replies

A. Neumaier
1. This is loose, informal talk. What is really meant is that the preparation and detection events are far apart. This is described by 2-point correlations at equal time and large spatial separation.

2. The asymptotic states are in a different Hilbert space, they are always states of free asymptotic fields, related to the true fields by a limiting process (that is nontrivial to make precise - see Haag-Ruelle theory).

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dextercioby, bhobba and vanhees71
1. This is loose, informal talk. What is really meant is that the detection events are far apart. This is described by 2-point correlations at equal time and large spatial separation.
How are the large spatial separations enforced in practice? I don't see anything in the LSZ formula which does this, all spacetime points seem to be integrated over for the correlation function without any bias (up to a phase). Also aren't the initial and final states meant to be delocalized momentum eigenstates?

A. Neumaier
How are the large spatial separations enforced in practice?
For particles in a beam, the expectation of their fields are tiny except very close to the beam, so 2-point correlations approximately vanish while the particles in the beam are far away from each other. The momenta figuring in the LSZ formula are the momentum of the beams, which gives a state that is a good approximation to a momentum in-state for a collision process taking place in a tiny region.

For particles in a beam, the expectation of their fields are tiny except very close to the beam, so 2-point correlations approximately vanish while the particles in the beam are far away from each other. The momenta figuring in the LSZ formula are the momentum of the beams, which gives a state that is a good approximation to a momentum in-state for a collision process taking place in a tiny region.
Sorry to bring this up again but it's still bothering me. I guess my question is this:

Even if the initial particles are "far apart", what justifies the treatment of them as free particles? Since these states are not eigenstates of the Interacting-Hamiltonian, shouldn't they evolve non-trivially with time?

vanhees71
Gold Member
Of course they evolve "non-trivially with time", as they should since you want to describe collisions. Of course you are right in questioning the notion of "asymptotic free states" since it's one of the difficult mathematical subtleties of the subject.

The physicists' pragmatic answer is as follows: As long as all particles in the theory are massive, there's no general problem (don't worry about Haag's theorem, because you have to regularize the trouble with the infinite volume anyway; the most simple way is to use a finite volume with periodic boundary conditions and then taking the boundary to infinity at the end of the calculation). Then the asymptotic states are indeed free-particle states, as naively assumed in the standard treatment.

Now, of course, the most interesting theories involve massless gauge fields (photons and gluons in QED and QCD), and then you'll ecounter problems with infrared and collinear divergences. The standard treatment is the soft-photon/gluon resummation (see Weinberg, QT of Fields vol 1). Another, very illuminating way to see that the trouble is due to the long-range nature of the interaction "mediated" by massless particles/fields and thus the fact that free-particle states are not the correct asymptotic states. That's already the case for the most simple example of Coulomb scattering in non-relativistic quantum mechanics, but there you can solve the problem exactly since the Coulomb Hamiltonian can be diagonalized analytically, and this is true not only for the bound states (which are treated in any textbook on QM) but also the scattering states. For the more complicated QFT case, one can however still derive the correct asymptotic states approximately in a way to be generalized to QFT. For a very didactical introduction to these ideas, see

P.P. Kulish and L.D. Faddeev. Asymptotic conditions and infrared divergences in quantum electrodynamics. Theor. Math. Phys., 4:745, 1970.
http://dx.doi.org/10.1007/BF01066485

bhobba, eys_physics and Spinnor
DarMM
Gold Member
Sorry to bring this up again but it's still bothering me. I guess my question is this:

Even if the initial particles are "far apart", what justifies the treatment of them as free particles? Since these states are not eigenstates of the Interacting-Hamiltonian, shouldn't they evolve non-trivially with time?
If the quantum field theory obeys certain assumptions (such as the particles being massive) then the Haag-Ruelle framework allows you to prove that the asymptotic time Hilbert Spaces ##\mathcal{H}_{in/out}## have Fock form, in the sense that if ##\phi_{p,out}## is a state which at late times looks like a single free particle with momentum ##p## and ##\psi_{q,out}## is similar for momentum ##q##, then:
$$\psi_{q,out} \otimes \phi_{p,out}$$
will at late times look like a free theory state of two particles.

The full details are given in Haag's book "Local Quantum Physics", but basically the reason is that if the particles have a mass then correlations between them have exponential die off as ##t \rightarrow \pm \infty##

dextercioby, vanhees71 and A. Neumaier
DarMM
Gold Member
To make a small comment on massless states, the situation is much less understood here. The actual asymptotic space of ##QED_4## is very difficult to gain mathematical control over. For example there are states like:
##\psi_{out;H,nml,p}##
which look like at late times like a Hydrogen atom with ##nml## the usual orbital variables and ##p## the atoms momentum. However these states don't come from Tensor producting electron and proton states. They're essentially an independent basis ket for the in/out Hilbert spaces.

Also nonperturbative analysis has shown the electron two-point function has no poles, so the electron doesn't have a sharp mass. Which is funny because the pole mass condition is often used in renormalization for perturbative calculations.

dextercioby and vanhees71
vanhees71
Gold Member
Nevertheless, the on-shell renormalization scheme in QED is not too well-defined either. Since the photon is massless already the one-loop self-energy correction for the electron has a branching point at ##s=m_{\text{e}}^2##, i.e., you have to go at least "an ##\epsilon##" to the left in choosing the renormalization scale.

DarMM
A. Neumaier
To make a small comment on massless states, the situation is much less understood here. The actual asymptotic space of ##QED_4## is very difficult to gain mathematical control over. For example there are states like:
##\psi_{out;H,nml,p}##
which look like at late times like a Hydrogen atom with ##nml## the usual orbital variables and ##p## the atoms momentum. However these states don't come from Tensor producting electron and proton states. They're essentially an independent basis ket for the in/out Hilbert spaces.
This is not textbook QED but QED with an additional proton/antiproton field.

DarMM
DarMM
Gold Member
This is not textbook QED but QED with an additional proton/antiproton field.
Correct. I'm not sure what to call it, as I'm not adding in the full QCD description of a proton, just having it as an extra charged spinor field.

A. Neumaier
Correct. I'm not sure what to call it, as I'm not adding in the full QCD description of a proton, just having it as an extra charged spinor field.
2-flavor QED would perhaps be the right expression. Standard QED with a single spinor field is better behaved as it has no bound states apart from the electrons.
Nevertheless, the on-shell renormalization scheme in QED is not too well-defined either. Since the photon is massless already the one-loop self-energy correction for the electron has a branching point at ##s=m_{\text{e}}^2##, i.e., you have to go at least "an ##\epsilon##" to the left in choosing the renormalization scale.
The branch point means that asymptotic electrons are infraparticles (rather than standard massive particles). Though we don't have a conventional Fock space, the asymptotic structure of QED (with a single spinor field) is reasonably well understood. See Herdegen's work in https://arxiv.org/abs/hep-th/9711066 and https://link.springer.com/article/10.1007/s11005-017-0948-9 and Strominger's work in https://arxiv.org/abs/1705.04311.pdf. Both authors have more papers on the topic.

vanhees71
Standard QED with a single spinor field is better behaved as it has no bound states apart from the electrons.
Could you elaborate on this? Does standard QED predict that hydrogen atoms are unstable? I'm suddenly very interested in how bound states are handled in QED.

PeterDonis
Mentor
2020 Award
Does standard QED predict that hydrogen atoms are unstable?
Standard QED treats systems containing anything other than electrons and photons, like hydrogen atoms, as having a fixed external potential due to the things that aren't electrons or photons, such as the proton in the hydrogen atom. QED therefore does not treat the hydrogen atom as having multiple spinor fields; it has only one, the electron. As the post you quoted notes, standard QED allows electrons to be in bound states, so it predicts that hydrogen atoms are stable.

A. Neumaier
Standard QED treats systems containing anything other than electrons and photons, like hydrogen atoms, as having a fixed external potential due to the things that aren't electrons or photons, such as the proton in the hydrogen atom. QED therefore does not treat the hydrogen atom as having multiple spinor fields; it has only one, the electron. As the post you quoted notes, standard QED allows electrons to be in bound states, so it predicts that hydrogen atoms are stable.
It is not that simple. There are three levels of QED:
1. Standard textbook QED of photons and electrons (without external fields).
2. QED of photons and electrons in an external field (created by stuff not modelled).
3. QED of photons, electrons and nuclei, the latter modelled by additional fields.
In each case, the presence of massless photons implies that there are, strictly speaking, no bound states (defined as poles of the S-matrix), but only infra-bound states (defined as branch points of the S-matrix). The latter become ordinary bound states only in the approximation where a fixed number of photons are accounted for and infrared effects can be ignored. This is usually done in the NRQED framework.

In case 1, the only (infra-)bound objects (in a center of mass frame) are single electrons and positrons in their ground state; positronium, for example, is a resonance only.

In case 2, an external radial field corresponds informally to the presence of an unmodelled nucleus, ignoring recoil effects. In this case, the no-photon approximation produces a spectrum of bound states for as many electrons as the nucleus charge allows and sometimes a few less or more (ions). The 1-photon approximation produces radiative corrections to the spectrum. But in exact QED, the corresponding poles would again turn into branch points (due to an unlimited number of soft photons), and we have no bound states in the conventional sense but only infra-bound states.

Case 3 is needed to take recoil effects into account. Proper modeling requires the nuclei to have non-Dirac form factors in the 0-photon approximation, and everything becomes significantly more complicated. But again, the full theory has to account for soft photons and hence produces branch points instead of poles, and hence only infra-bound states.

vanhees71 and dextercioby
vanhees71
Gold Member
I don't know, what you mean by "standard QED", but if you mean the minimal QED with one matter field should be bound states like positronium. Of course, as with any other non-trivial interacting-fields problem we have only approximations, but usually it's indeed treated by approximately solving the corresponding Bethe-Salpeter equation for the appropriate N-point function (in this case the two-electron-positron 4-point function) and looking for the poles (in the complex energy plane) indicating bound states. Of course, at higher orders this implies soft-photon resummations. For a detailed treatment, see e.g. Landau/Lifshitz vol. 4.

dextercioby
A. Neumaier