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A "Quantum chromodynamics is a zero-parameter theory"

  1. Sep 28, 2016 #1

    ShayanJ

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    I'm trying to understand the large N expansion scheme and one of the resources that I glanced is Zee's "Quantum Field Theory in a Nutshell". The quote in the title is in the first sentence of the "Large N Expansion" chapter of the book.
    I don't understand this sentence. Of course QCD have some parameters. Putting aside the masses of the quarks, we have coupling constant. I know its a running coupling and depends on energy but in QED we use the fine structure constant(which is a running coupling) as the expansion parameter!
    So what's the meaning of the quote?
    And why do we need the large N expansion scheme? Is it because in low energy where the coupling constant is large, we have no small expansion parameter? Or is there any other reason?
    I will also appreciate it if anyone can suggest a pedagogical introduction to the large N expansion scheme in the context of high energy physics(and not statistical physics).

    Thanks
     
  2. jcsd
  3. Sep 28, 2016 #2
    As far as I am aware, the statement applies to massless, pure QCD. The only free choice seems to be the coupling constant, or ##\alpha_S(\mu)## at some scale ##\mu## as you noted. Equivalently, we may choose the energy scale ##\mu## at which ##\alpha_S(\mu)## equals some particular value, say ##\alpha_S(\mu)=0.1##, as the running is then fully predicted. However, as we don't have a meaningful mass scale, changing the value of ##\mu## is just the same as changing unit we measure energy in. This of course can't have physical consequences, as our choice of system of units is arbitrary.
     
  4. Sep 28, 2016 #3

    ShayanJ

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    But how is that different from QED? If we don't consider charged fermions, pure QED is massless too!
     
  5. Sep 29, 2016 #4
    Well, but in QED the case ##\mu \rightarrow 0## is meaningful, ##\alpha_{em}(\mu=0)## is a dimensionless observable.
     
  6. Sep 29, 2016 #5

    ShayanJ

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    Yeah, that's right. But why can't we use ## \alpha_s(\mu \to \infty) ## for QCD?
    QCD has asymptotic freedom so that quantity should be well defined, right?
     
  7. Sep 29, 2016 #6

    vanhees71

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  8. Sep 29, 2016 #7

    ShayanJ

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    Thanks, but actually I asked for a pedagogical review of large N expansion, as applied to high energy physics. PDG doesn't seem to have any!
    I myself found a few, so that was actually the optional part of this thread!
     
  9. Sep 29, 2016 #8

    vanhees71

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  10. Oct 3, 2016 #9

    ShayanJ

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    I found some other quotes about the same issue that I don't understand:

    From the introduction of this paper.

    From Shifman's Advanced topics in Quantum Field Theory.

    what do they mean exactly? How does it prevent g from being used as a expansion parameter? How does it differ from QED? Why it still doesn't prevent us from doing perturbation theory using the coupling constant as the expansion parameter, in the weakly coupled regime of QCD?

    Thanks
     
  11. Oct 3, 2016 #10

    Haelfix

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    Both of your questions were answered already in the thread, and the answer is basically given by what Shifman writes. The QED coupling constant stays dimensionless, the QCD one gains a mass dimension via dimensional transmutation. The former is thus a good expansion parameter (it is a small number), the latter is not (in the pure QCD case).. Suppose you were to redefine the QCD g^2 as a large number.. How would you do perturbation series?
     
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