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I am learning renormalization in Quantum field theory and following mainly Schwartz (Quantum field theory and standard model) for it. While explaining Renormalization group equations it says it mainly $\textbf{resum the large logarithmic depedence}$. Please explain :

  1. What is the meaning of large logarithms here (is it leading order in log type terms)?

  2. What is so special about log terms? I have observed that in QED all infinities appear in form of log.

  3. What if there appears quadratic divergence, RG equation wont help I suppose. What is the solution for that case?

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    $\begingroup$ I found answer to first part already. Perturbation theory breaks if argument of log is sufficient large and $e_R^2 log(..)$ becomes 1. This is what they call "large logarithms" $\endgroup$ Commented May 11, 2016 at 6:56
  • $\begingroup$ Not all infinities appear as logs. In particular, corrections to the Higgs mass diverge quadratically. $\endgroup$
    – jwimberley
    Commented Sep 16, 2016 at 18:32
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    $\begingroup$ Only log divergences are renormalizable. $\endgroup$ Commented Sep 17, 2016 at 4:17
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    $\begingroup$ @jwimberley then how do you tackle quadratic divergences because as much as I understand RG equations tackle logarithmic divergences only. $\endgroup$ Commented Sep 19, 2016 at 5:46
  • $\begingroup$ @seeking_infinity I think the answer below addresses this. Basically, for the Higgs field you get lucky enough that it doesn't break things completely, but it's still not renormalizable in the same sense as log divergences, and it requires fine tuning (i.e. a counter term that agrees that matches a UV term to a large number of decimal places). $\endgroup$
    – jwimberley
    Commented Sep 19, 2016 at 13:31

1 Answer 1

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Requirement of logarithmic divergence basically is result of local nature of counterterms and renormalizability of the Lagrangian of the theory however there are some exceptions. If you can have a local counterterm (product of fields or their derivatives) which has already a form similar to one already present in the Lagrangian then it is possible to redefine some parameters to absorb the divergence.

In general, a divergence higher than logaritmic can destroy renormalizability of the theory and to make a better sense of it one has to use an explicit cut-off making it an effective theory upto that cut-off.

In case of QED, suppose you encounter a quadratic divergence for self-energy of the electron; this would require to introduce a term like $C_1(\partial_\mu\psi)^2$ as $\partial_\mu$ gives a mass dimension one and does not introduce an explicit cut-off scale of energy. In QED, we don't have a term like $(\partial_\mu\psi)^2$ as electron field introduced only comes with first order derivative term in Lagrangian. As a result, it is not possible to absorb the counterterm with any other existing term. This process is basically an indication of non-renormalizability of the term which can be seen simply also from counting the mass dimension of the term. A logarithmic divergence on the other hand does not introduce any extra mass dimension to a counterterm so these problems do not arise.

Another case of Higgs field is interesting as the quadratic divergences do not spoil renormalizability. A quadratic divergence requires a counterterm of the same form $B_1(\partial_\mu\phi)^2$ but now this has the same form as the kinetic term of the scalar field in the Lagrangian. So we have a term of the same form in the Lagrangian already and hence it's not problematic on the general ground . Power divergences however cannot be handled by renormalizable theories (without using a cut-off obviously); these theories are termed unnatural and other extensions are needed to treat it properly (like SUSY, Technicolor). Without these extensions one have to fine-tune the theory suffering from power divergence.

This discussion is parallel to Collins 'Renormalization' chapter 3.

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  • $\begingroup$ Are you saying SUSY and Technicolor require fine tuning? I thought they were introduced precisely to solve the problem of fine tuning in standard model. $\endgroup$ Commented Sep 19, 2016 at 4:30

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