How can scattering with interaction at all time be resolved using QFT?

  • Thread starter Thread starter ndung200790
  • Start date Start date
  • Tags Tags
    Interaction Time
ndung200790
Messages
519
Reaction score
0
S-Matrix is calculated in that the interaction at infinite past and future time is zero.Then how can we solve the problem of ''scattering'' where the interaction presents at all time(e.g considering interaction of electron in an atom) with QFT?
 
Physics news on Phys.org


I think that the coupling constant(of an interaction) is the probability of a process(create and destroy particles) that the interaction Hamintonien describes.So I do not understand why the strong coupling constant at low energy is greater than 1(the probability that greater than 1).
 


I think that the coupling constant(of an interaction) is the probability of a process(create and destroy particles) that the interaction Hamiltonian describes.
where did you find that?
 


OK,now I could understand that all things related with ''creating and annihilation processes'' can be referred to S-Matrix,but not to the coupling constant(?).
 


you can say that.
 


In bound state(e.g hydrogen atom), if we use QED Feynman diagrams purturbation then the series can not be truncated because the terms in the series are power of \alpha/v where velocity v of electron is same order to \alpha.But I do not understand why in the higher order term in the series then the more vertex in the diagram but all terms in the series still have the same magnitude.
 


In treating bound states,you will have to resort for more convenient situation such as bethe salpeter eqn.I am not sure what you mean by velocity here.
 


v is characteristic velocity of electron(I think it is of wave packet).If we ''use'' Feynman diagrams,electron exchanges virtue photons with immovable proton,so I do not understand as they say all diagrams have the same magnitude.
 


v is characteristic velocity of electron(I think it is of wave packet).
this is of no use here.
Feynman diagrams,electron exchanges virtual photons with immovable proton
electrons exchange virtual particles with other electrons also.One has to take care with protons who are not point particle.A simple treatment of electron proton interaction is given by rosenbluth formula in terms of modifying the vertex factor at proton which involves form factors.
they say all diagrams have the same magnitude.
who says.
 
  • #10


The book:''Concepts of Particle Physics'' of Gottfried Vol 2,section 4.Bound states Page 243-244,Footnote 7,says that.
 
  • #11


there are two types of diagrams.One which can be built up from other irreducible diagrams will constitute a reducible one.All the reducible diagrams will contain same order of interaction because a single irreducible diagram is used to built it namely 31 there which is irreducible and makes infinity of diagrams from this single ones.There are other irreducible diagrams which of other higher order when used will constitute different reducible diagrams which will not have same order as the other one.A bound state can be represented by an infinity of such irreducible diagrams where a single one of irreducible will make infinite of reducible diagrams.These reducible diagrams can be avoided by using only irreducible one's by using bethe-salpeter formalism.That's why they say that all in 32 represents same order.
 
  • #12


Please explain what is irreducible diagrams and how to build up from them to reducible diagrams.
 
  • #13


A reducible diagram is built up from irreducible diagrams.A diagram is reducible if you can cut it into two diagrams by a horizontal cut which does not cut any boson lines(photons here).You can see you can not cut 31 to make two other because it will cut photon line.While in 32 you can cut the diagrams horizontally to get diagrams which are just 31.
 
  • #14


Is it correct if I say that ''probability amplitude'' of whole reducible diagram is the sum of ''probability amplitudes'' of irreducible diagrams cut off from the reducible diagram?.So that you say all the reducible diagrams have same order(?).
 
  • #15


No,one irreducible diagram represents represents infinity of reducible diagram.In case there is interaction present at all times as in bound state.One single diagram in course of time represents interaction over a long time by those ladder diagrams(infinity of).In standard formalism there are only irreducible diagrams and you can see that a single diagram represents infinity of diagrams so in lowest order approximation also when you consider one diagram(irreducible),you count infinity of diagrams.So they are called of same order not because they are same order feynman diagrams but they represent same interaction over the time.
 
  • #16


But in Gottfried Vol 2 they say the series of diagrams is power of anpha/v~1 so all diagrams have same order?
 
  • #17


Could I say that the reducible diagrams produced from the ladder diagram are not a ''normal'' Feynman diagrams?
 
  • #18
ndung200790 said:
But in Gottfried Vol 2 they say the series of diagrams is power of anpha/v~1 so all diagrams have same order?
they give same interaction because of a bound state in which one photon exchange for example occurs in due time.It is not a simple scattering.
Could I say that the reducible diagrams produced from the ladder diagram are not a ''normal'' Feynman diagrams?
What do you mean by normal.They are all feynman diagrams.But it is only the irreducible one which is considered.I think it will be better to give a reference here
http://prola.aps.org/abstract/PR/v84/i6/p1232_1
it is available freely from
http://fafnir.phyast.pitt.edu/py3765/BetheSalpeter.pdf
 
Back
Top