Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Running Coupling in MS scheme

  1. Aug 5, 2016 #1
    Hi everyone. I've got some doubts about the interpretation of the running coupling. I'll expose to you some of the results and comments i read in the lectures notes of my QFT professor. As an example, we are studying ##\phi^4## and the 4-point amplitude accounting for the next to leading order correction, the 1-loop diagram. In order to deal with the divergence we apply renormalization and we choose the dimensional regularization and the on-shell prescription. This choice has the advantage of canceling the dipendence on ##\mu##, the renormalization scale. So we find that the effective coupling is:
    $$\lambda_{eff} = \frac{\lambda}{1-\frac{3\lambda}{16\pi^2}\ln{\frac{k}{m}}}$$

    This result has a quite straightforward interpretation. All the parameters have a direct physical meaning and this encodes the running of the constant when ##k## increases.

    After that he decides to do again the calculation for the running constant using the group renormalization equation. To do so, he changes the renormalization scheme to MS. In this scheme ##\mu## does not go away and we find the running:
    $$\lambda_{eff}(\mu) = \frac{\lambda(\mu_0)}{1-\frac{3\lambda(\mu_0)}{16\pi^2}\ln{\frac{\mu}{\mu_0}}}$$

    Now, ##\mu## has not any physical meaning and neither has ##\lambda##, in this scheme the parameters of the lagrangian are not the physical ones. Nevertheless, the professor states: "If we choose ##\mu = k## and ##\mu_0 = m## then we get exactly the same result".

    I don't understand why we want to compare the 2 coupling, one is physical and the other is not. In addition, ##\mu## is arbitrary, the observables don't depend on it, so why do we have to choose ##\mu = k##? What am i missing here?
     
  2. jcsd
  3. Aug 5, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Are you allowed to link to the lecture notes? It seems to me that ##\lambda_{\text{eff}}## has a different meaning in the two formulae. In the 2nd one it's a running coupling in the usual sense, depending on the renormalization scale ##\mu##.

    In the 1st case it's a RG resummed vertex function. This example shows that you can use the RG to resum large logarithms, i.e., for ##k^2>m^2##. I'm also not sure what ##k/m## means since ##k## should be a four-momentum, and the dependence in the log should be of the form ##\ln(-k^2/m^2)##, where ##k^2=\eta_{\mu \nu} k^{\mu} k^{\nu}##.
     
  4. Aug 5, 2016 #3
    Of course, here's the link: http://www.robertosoldati.com/archivio/news/114/advqft.pdf
    The first result is at the beginning of page 61, the second one is in page 72.

    I'm not sure i have understand the link between these two formulae, they are different object but at the same time one gives information about the other?
    What is the real meaning of the second one?

    PS k is the module of the spatial momentum, there is an approximation for large values of mandelstam variables.
     
    Last edited: Aug 5, 2016
  5. Aug 5, 2016 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I'll have a look at it. Of course, if you are in the spacelike region, you are in fact dealing with Euclidean QFT. Then ##k=|\vec{k}|## makes sense.

    The argument on page 61 is only a teaser for the more systematic RG study. I think for a systematic approach with RG methods you have to make an analysis in the spirit of Sect. 5.12 in my manuscript. I have to think about the RG in the on-shell scheme myself first, but I've the feeling that should work. In ##\phi^4## the on-shell scheme is restricted to ##m^2>0##, where ##m## is the renormalized and at the same time the physical mass in the OS schem, i.e., your renormalization conditions read something like
    $$\sigma(p^2=m^2,m^2)=0, \quad \partial_{p^2} \Sigma(p^2,m^2)|_{p^2=m^2}=0, \quad \Gamma^{(4)}(s=t=u=0,m^2)=-\lambda.$$
    Then you can analyze the behavior of ##\Gamma^{(4)}## under scaling momenta and mass as described in Sect. 5.12 of my manuscript.
     
  6. Aug 5, 2016 #5
    I think that my problem is that i don't understand the meaning of the running coupling that depends on ##\mu##. It seems like something artificial, but sometimes is presented as something with physical meaning. For example, how do i choose ##\mu## and ##\mu_0##? Aren't they arbitrary?
     
  7. Aug 5, 2016 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I have the opposite problem. I understand the lecture notes very well concerning the running coupling in the usual meaning, i.e., as its dependence on the renormalization scale. It's the standard treatment of the modern perturbative approach to the RG using dim. reg. and choosing the MS renormalization scheme (a la t'Hooft and Weinberg 1973). Of course the physical quantities should be independent of the choice of the arbitrary renormalization scale. That independence is used to derive the renormalization group equations which tell you how the renormalized parameters (wave-function normalization, mass, and coupling constant for, e.g., ##\phi^4## or QED). The change of these renormalized parameters is such that the observables like S-matrix elements don't change when you change the renormalization scale. The parameters have to be fixed at one arbitrary renormalization scale. Then the RG tells you how they run when changing this scale. In perturbation theory the RG provides a resummation of leading logs in such a way that the perturbative S-matrix element become renormalization-scale independent. Of course perturbation theory can only be valid in the realm, where the coupling constant (as the formal expansion parameter of the corresponding asymptotic series) is small.

    The argument on p. 61 of the manuscript is handwaving in the step from the perturbative result to simply giving a kind of resummed result, but I guess you can derive it with a renormalization-group equation for the on-shell renormalization scheme, where you define these parameters at certain arbitrary momenta, for which purpose you have to introduce a momentum scale, which then serves as the renormalization scale, and again the RG tells you how the parameters change with a change of this renormalization scale. This is what's called a momentum-subtraction (MOM) scheme, which I also treat in my manuscript.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Running Coupling in MS scheme
  1. L-S coupling scheme (Replies: 3)

  2. Running couplings (Replies: 1)

  3. Running coupling (Replies: 1)

Loading...