\phi^{4} theory rigorously

  • #26
strangerep
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QED's infra-red problem is mainly related to the asymptotic space of states not being a Fock space. Essentially because photons are massless, all electrons pick up a soft photon cloud. These electron + cloud and photon + cloud states are the appropriate in/out states. However getting enough information on these states to show a unitary S-matrix exists between them is very difficult. So QED's infrared problem is a massless force carrier problem which makes a treatment of the S-matrix difficult.
I was under the impression that the old work by Chung, Kibble, Kulish+Faddeev,
and more recently Lavelle+McMullan+et al, had already found the appropriate
asymptotic states -- in which asymptotic electrons are accompanied by a Coulomb
and radiation field -- and that this solves the QED IR problems to all orders.

Kibble constructs a very large nonseparable space, but Kulish+Faddeev just
concentrate on the asymptotic spaces, deriving an S-matrix that maps
between them. (This is probably only at the level of rigor of theoretical physics,
borrowing Arnold's phrase. :-)

In these treatments of IR, they don't bother to dress the asymptotic photons since
no residual part of the interaction remains at light-like infinity. Could you give some
details/references about the "photon + cloud" you mentioned, as relevant in this
case, please?
 
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  • #27
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The [itex] g\phi^4 [/itex] is a change in the potential. From a dynamical perspective, it can be seen as a perturbation to the initial Hamiltonian, or from an interaction perspective, it is a self-coupling term.
And this is taken from the total energy of the system [tex]1/2 \nabla^2 \phi^2 + 1/2 M^2 \phi^2[/tex]?

So if M^2 \phi^2 is the potential energy on of the system and g\phi^4 is the the change in the potential, are they effectively describing the same potential, or is it like describing V(x) and V(\alpha) as two potential depending on different functions?
 
  • #28
dextercioby
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No, it's just one single potential for the whole system. [itex] V(x) = \mu \varphi^2 + \lambda \varphi^4 [/itex].
 
  • #29
DarMM
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I was under the impression that the old work by Chung, Kibble, Kulish+Faddeev,
and more recently Lavelle+McMullan+et al, had already found the appropriate
asymptotic states -- in which asymptotic electrons are accompanied by a Coulomb
and radiation field -- and that this solves the QED IR problems to all orders.
Oh yes, it solves it to all orders in perturbation theory. I'm only talking about its solution nonperturbatively at a rigorous level. Like most things however the solution is basically what those authors describe, but it is difficult to show this is the case in constructive field theory. For instance it is obvious that [tex]\phi^{4}_{2}[/tex] with a non-zero mass term has a mass gap. However showing this nonperturbatively at a rigorous level is very difficult.

In these treatments of IR, they don't bother to dress the asymptotic photons since
no residual part of the interaction remains at light-like infinity. Could you give some
details/references about the "photon + cloud" you mentioned, as relevant in this
case, please?
Cloud is a very loose way of referring to the coherent photon states. Technically every state has to have this non-Fock coherent state structure tagged on. The book by Strocchi "Symmetry breaking" and Othmar Steinmann "Perturbative quantum electrodynamics and axiomatic field theory". Of course the basic picture is as you have stated, all this is simply making it mathematically rigorous.
 
  • #30
strangerep
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strangerep said:
In these treatments of IR, they don't bother to dress the asymptotic photons since
no residual part of the interaction remains at light-like infinity. Could you give some
details/references about the "photon + cloud" you mentioned, as relevant in this
case, please?
Cloud is a very loose way of referring to the coherent photon states.
Technically every state has to have this non-Fock coherent state structure tagged on.
Let's see if I understand. The physical asymptotic electron states are constructed
by acting on the bare electron states with an exponential operator whose exponent
involves the bare photon operator. I think you're saying that the bare photon
states must be acted on by this same operator to get physical asymptotic photon
states? I.e., the bare photon is essentially dressed with itself, but not with (e.g.,) an
expression involving electron-positron pairs?

The book by Strocchi "Symmetry breaking" and Othmar Steinmann "Perturbative quantum electrodynamics and axiomatic field theory". Of course the basic picture is as you have stated, all this is simply making it mathematically rigorous.
I have Steinmann, but had trouble extracting a comprehensible picture from it.
I'll take a look at Strocchi. Thanks.
 

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