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

A Anomalous field dimension in the on-shell scheme

  1. Aug 12, 2017 #1
    In learning QFT, I've found that the renormalization group is often introduced with the MS scheme. I noticed that if one uses the on-shell scheme instead and calculates the anomalous field dimension using
    [itex]\gamma_{\phi} = \mu \frac{\partial \ln Z_{\phi}}{\partial \mu} [/itex]
    one finds that [itex]\gamma_{\phi} = 0[/itex]. Is this correct or am I making a mistake? What is the physical interpretation?

    I've found this to be the case for both [itex]\phi^3[/itex] and [itex]\phi^4[/itex] theory. I haven't learned QED or anything about QFTs in particle physics so I'm not sure how it plays out there.

    I wasn't sure if this is the place to post, or the homework or high-energy section; if this is not the right place, apologies and please feel free to move this thread.
  2. jcsd
  3. Aug 12, 2017 #2

    king vitamin

    User Avatar
    Gold Member

    In massless [itex]\phi^4[/itex] theory in 4 dimensions,

    \gamma_{\phi} = O(g^2)

    That is, there's no correction until 2-loop, so if you just do the one-loop calculation, you should get zero.

    In massless [itex]\phi^3[/itex] theory in 6 dimensions (where the interaction is renormalizable), Srednicki does the calculation in his book and he does find a one-loop correction. (Note that the one-loop contribution is proportional to the square of the coupling still, unlike the phi4 case).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted