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Minimal subtraction vs Momentum subtraction schemes

  1. Mar 6, 2014 #1
    Hi everyone,
    I have a question about renormalization schemes. Consider for example the self energy of the photon (i.e. renormalization of the electric charge in QED) and take two famous schemes: the Minimal Subtraction (MS) and the Momentum Subtraction (MO). In the MS case we choose the counter terms of the renormalized Lagrangian in order to just cancel the divergent part of the self energy loop:
    where [itex]\epsilon\to 0[/itex] when we restore the original 4-dimensional space-time.
    This lead to a beta-function given by:

    The MO scheme, on the other hand, requires the two-point function to be zero at a particular momentum [itex]-p^2=M^2[/itex]. This lead to a more complicated counter term and to a beta-function given by:
    \beta_{MO}(e)=\frac{e^3}{2\pi^2}\int_0^1dx x(1-x)\frac{M^2x(1-x)}{m^2+M^2x(1-x)}.

    My question is: these two beta-functions will lead to different behaviors of the running electric charge. How is that possibile? The running of the electric charge has been experimentally measured and the physical predictions of the theory should be independent on the renormalization scheme used. How do we reconcile these aspects?

    Thank you
  2. jcsd
  3. Mar 6, 2014 #2
    I'm not familiar with the second renormalisation scheme.

    The coupling constant is only meaningful when computed with the corresponding amplitude for the process.

    Does the difference cancel when the amplitude is also included? The observed quantity has to be renormalisation scheme independent.
  4. Mar 6, 2014 #3
    No, it doesn't (at least I don't think so). What I know is that the MS scheme becomes meaningless when the renormalization scale [itex]\mu[/itex] crosses the natural scale of the theory (say for example the mass of the electron in a pure QED case) . I would like to know if this is true and how can we explain that.
  5. Mar 6, 2014 #4
    It does not matter whether you choose MS scheme or Off-shell subtraction scheme (momentum subtraction scheme),the observables like S-matrix elements or cross-sections will be independent of renormalization scheme.It is a virtue of MS scheme that renormalized constant are in a simple form and propagators are not,while the reverse is there for Off-shell subtraction scheme.The physical predictions are independent of renormalization scheme because of the renormalization group,which tells you how the coupling strength along with other things should vary with renormalization group scale to give same physics.(I cannot expand on the answer right now because of a broken keyboard but I think this gives you the idea)
  6. Mar 6, 2014 #5
    My point is that, from what I know, the renormalization group tells you, once you choose one particular renormalization scheme, how the parameters of your theory should vary with the renormalization scale in order to keep the Green's functions the same. It doesn't tell you how to change from one scheme to another.

    Now, take for example the two schemes I listed before. Starting from their beta-functions the renormalization group tells you how the coupling runs. In our case this means [itex]e(\mu)[/itex] for the MS and [itex]e(M)[/itex] for the MO. My point is that, since the two beta-functions are different then the two runnings are different as well. Still, the running of the coupling is a perfectly measurable quantity and it shouldn't be scheme-dependent.
  7. Mar 6, 2014 #6
    The beta function certainly is scheme dependent, though the first few terms in its perturbative expansion are scheme-independent. Googling gave me this link, which might be useful: http://bolvan.ph.utexas.edu/~vadim/classes/2012f/ms.pdf

    When you say "the running of the coupling is a perfectly measurable quantity and it shouldn't be scheme-dependent," I think you need to be more careful. When you measure the coupling, you do it by picking a renormalization scheme, computing the theoretical prediction for a scattering amplitude (say), comparing the theoretical prediction to experiment, and figuring out what value of the coupling makes theory agree with experiment. This gives you the coupling *in a certain renormalization scheme*. You measure the running of that coupling by repeating the same procedure at several energy scales. This gives an experimental measurement of the beta function for a particular renormalization scheme. To do the same for another renormalization scheme, you have to redo the theoretical calculations in that renormalization scheme, and you will end up with a different beta function.
  8. Mar 6, 2014 #7
    That sounds reasonable! So when for example we see an experimental plot of [itex]\alpha(\mu)[/itex] with the varying scale, are we looking at the running in one particular scheme? Because I alway encountered just one plot and I deduced it was the only one. Maybe it's computed in the MS which is the most famous one.
  9. Mar 6, 2014 #8
    I would guess most plots you see are probably in the MS-bar scheme. But it shouldn't matter a whole lot: when the coupling is small the differences between renormalization schemes are also small.
  10. Mar 7, 2014 #9
    and I said that whatever renormalization scheme you choose,you will get different dependence of β function but renormlization group assures that your physical prediction will be unaffected by this.
    It is not a perfectly measurable quantity to all orders but when you are doing perturbation expansion and trying to get the physics out of it,there is problem.Say you have a physical quantity f(g),which will be renormalization scheme independent but it's expression in term of g and m vary depending on renormalization scheme employed.
    $$f(g)=f_0+f_1(g^2)+f_2(g^4)+..$$(I have dropped the dependence on m)
    the f's are renormalization scheme dependent,since g is too because it is green function at specific value of arguments.However we demand that this perturbation series summed up to all orders is renormalization scheme independent.However in perturbation theory we truncate the series at a certain order and thus make the quantity scheme dependent in neglected orders.Now if you adopt two different renormalization scheme then $$f(g)=f'(g')$$,for the other one you can do again the perturbative expansion in g',where the coupling constant itself are related by a finite renormalization.
    $$g=(1+z_0g'^2+z_1g'^4+...)g'$$,if you use it in $$f(g)$$ and compare terms with $$f'(g')$$,you will find that $$f'_0=f_0$$,$$f'_1=f_1$$,$$f'_2=f_2+2z_0f_1$$,..,so $$f_0$$ and $$f_1$$ are scheme independent but next higher orders are not.The perturbation series as a whole is scheme independent but to a finite order it is not.
    Last edited: Mar 7, 2014
  11. Mar 7, 2014 #10
    I don't think this is what the renormalization group does. The renormalization group shows you how to deal with a change in the renormalization scale (basically thanks to the Callan-Symanzik equation), not how to deal with a change in the renormalization scheme. In fact, as you correctly said, the beta-function, which is produced by the renormalization group, is scheme dependent (although weakly and just in perturbation theory).

    Looking around I found that this perturbative scheme dependence is, in fact, a pretty annoying source of theoretical uncertainty. However, it seems that the problem could have been recently solved. Take a look at:


    However, thank you very much for your proof!
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