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Qft and bound states

  1. Dec 26, 2005 #1
    I read somewhere that quantum field theory does not allow calculations and predictions of bound states in a satisfactory way. Is that true and how much is that a problem given that qft claims to be so fundamental?
  2. jcsd
  3. Dec 27, 2005 #2
    As far as i know, QFT does a good job in some problems concerning bound states, e.g. the Lamb shift in light atoms, which has been calculated using QFT in Weinberg's book. But in bound states with strong interaction, as a bound state is a low energy state(~MeV), the strong interaction becomes so strong that perturbation theory breaks down in any sense. So we can't use common QFT method to calculate hadron mass, etc. But we do have some approaches to settle this problem, e.g. lattice QCD, but the solution is far from exact.
    Last edited: Dec 28, 2005
  4. Dec 27, 2005 #3
    Well this problem is more general. The entire QCD field theory suffers from it.

    Lattice QCD ? You mean non perturbative QFT or infrared QCD, right ?

  5. Dec 27, 2005 #4
    I hear Lattice QCD can do some non perturbative calculations, for example, it can give hadron mass up to a few percent in most sence. In lattice QCD, we can begin with the Euclidean but entire action, take the action as a probable denisty of all field configurations, place space on lattice and do sampling according to the action.

    It's my understanding. I hope it is helpful.

    Last edited: Dec 28, 2005
  6. Dec 27, 2005 #5


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    wangyi, I believe what you say is true. BTW the particle name you want is hadron, with the d coming before the the r. It is important to observe this silly detail because your original spelling has an embarassing slang meaning.
  7. Dec 28, 2005 #6
    Sorry, My mother language is not English, but I will pay more attention on these details. Thank you for your correction.
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