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What is an effective field theory

  1. Feb 16, 2009 #1
    Can anyone tell me if this is the basic concept behind effective field theories:

    1) You begin with your original Lagrangian, and you want to construct an effective Lagrangian out of it.

    2) You redefine your coupling constants in your original Lagrangian so that they include the integral from a cutoff to infinity (i.e., you combine the higher order corrections into the coupling constants, but only from the cutoff to infinity). This way, when you make an effective Lagrangian, you no longer need to integrate to infinity, since this was already done in the coupling constants.

    3) Step 2 isn't perfect for some reason (why?), so your effective Lagrangian needs to include not just a redefinition of the coupling constants in your original Lagrangian, but new terms (an infinite amount of them) whose coupling can have negative mass dimensions.

    4) As of yet all your parameters are undefined - you didn't specify a mass scale or anything. You decide that the mass scale is the cutoff. At this point you could determine the value of all your parameters evaluated at the cutoff by comparing to experiment?

    5) Now you integrate below your cutoff, and in doing so you derive a renormalization group equation for all your coupling constants.

    6) Now you're done and can see how parameters in your effective Lagrangian changes as you vary the cutoff.

    What's the point of all this? What's this used for that regular renormalization can't do?
  2. jcsd
  3. Feb 17, 2009 #2


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    I don't know this stuff very well, but I'll give it a try. As I understand it (and maybe I don't), step 2 should be more like this: Choose a cutoff and split the integrals into two parts, integration up to the cutoff, and integration above the cutoff. Then do the integrals above the cutoff so that only the integrals up to the cutoff remain. The result of the integration that was performed shows up as an adjustment to all of the coupling constants in the Lagrangian that appears in the integrals up to the cutoff. Note that it's not just the coupling constants in the renormalizable terms that get adjusted. The new Lagrangian contains all the terms that are consistent with the symmetries of the original Lagrangian, including an infinite number of non-renormalizable terms.

    The fact that any adjustment of the cutoff changes all the coupling constants suggests that it doesn't really make sense to start with a renormalizable Lagrangian, but it can also be shown that non-renormalizable terms have a negligible effect in experiments performed at low energies. I don't remember the details of that argument, but it was based on order-of-magnitude estimates that you can make once you have figured out the dimensions (units) of the coupling constants.
  4. Feb 18, 2009 #3
    The infinite number of additional terms that you have to add in an effective Lagrangian do seem to be negligible at low energies.

    Does this mean that the Standard Model Lagrangian might really be an effective field theory, and we just don't know it because all the experiments are at low energy? If this is true, how do we know if our theory is a complete theory, or an effective field theory? Take phi^4 theory. This theory is perfectly renormalizeable. But if you construct an effective field theory from it, you can find through a renormalization group equation that the cutoff can't be taken to infinity. Isn't that an inconsistency? Before performing an effective field theory analysis on it, you were integrating out to infinity perfectly fine, with no worries about a cutoff or anything. Anyways, does this finite cutoff that you find equal the symmetry breaking scale? Is this how unification is done?
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