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A Understanding renormalization / regularization basics

  1. Aug 8, 2016 #1
    I've been trying to get a rough understanding of what renormalization involves (in a particle physics context; I'm aware it has many other applications eg. condensed matter) but it hasn't quite clicked yet. The things I have in my head so far are as follows:

    - A particle will be surrounded by a cloud of virtual particles, which affect the measurement of the particle's properties, ie. if you collide two electrons, the energy of the collision determines how deep into this cloud you penetrate, and thus the charge of the electron you measure is not actually a constant, but rather depends on your experiment. So it is said that the electron has some intrinsic 'bare' charge, but this is not what is being measured.

    - In Feynman diagrams, you have your basic "tree-level" diagrams, but also you can have additional diagrams that contain virtual particle loops.

    - If you want to calculate an amplitude for, say, a scattering process, you need to take these additional loop diagrams into account.

    - The energies/momenta of the incoming/outgoing final state particles do not determine what the energy/momenta of a virtual particle must be, so instead one must integrate over all energies/momenta to determine the contribution that some loop diagram makes to the overall scattering amplitude.

    - However, sometimes these integrals blow up to infinity (and I think it's that for high momentum this is called an "ultraviolet divergence", conversely an "infrared divergence" is an infinity corresponding to momentum becoming very small?). So, these infinite contributions ruin your amplitude calculation.

    ==> Solution required, "renormalization".

    - First of all, instead of integrating all the way to infinite momentum, one imposes some cut-off upper limit to the integral, then your orginal integral can be split into a finite part and a divergent part. The imposition of a cut-off is referred to as "regularization".

    *** (I'm aware of there being different ways to do this regularization - it may not be a momentum cut-off that is used, but rather other things such as a dimension cut-off that are more useful/appropriate [for reasons like preserving gauge invariance I think?] but they sound like more complicated things and I'm trying to understand in the simplest way possible. And regardless of the specific details of the cut-off, they should always produce only one result). ***

    Anyway, assuming nothing I've said so far is incorrect, here's where I start getting shaky.

    At this point the material I've read starts talking about of some "counter-terms" (are these also called "radiative corrections?") that get added to the Lagrangian to cancel out the divergent parts of these integrals, and the bare parameters (charges, couplings, fields) in the Lagrangian get redefined... somehow... by some sort of subtraction? And somewhere in there, you require an experimentally measured value made at one specific collision energy? (or rather, one measurement per parameter involved?) And it is said that your theory is "renormalizable" if this can be done with a finite number of counter-terms? The infinities get absorbed into the bare parameters? (which is fine, there's supposedly nothing wrong with them being infinite?) And you end up with a Lagrangian in terms of "renormalized" parameters, with no dependence on the arbitrary, user-selected cut-off/"regularization".

    But I don't fully understand what I just said there. What I'm hoping for here is if someone can talk about what I'm getting at in that previous paragraph. Also I'm not sure if perturbation theory / power series expansions have any relevance to the scattering example I've mentioned here, but I have seen them in some of the texts I've looked at, just not quite sure if / how they fit into this picture.

    Thanks.
     
    Last edited: Aug 8, 2016
  2. jcsd
  3. Aug 8, 2016 #2
    Hi Anchovy,

    renormalization is a tricky business, and it is difficult to find a clear description of this technique in textbooks. So far, your understanding is pretty good, though not complete.

    I think, the crucial thing that needs to be understood is that when we do renormalization we are basically saying: "Our original Hamiltonian (or Lagrangian) was not correct, because it resulted in divergent scattering amplitudes. So, we are going to "fix" this Hamiltonian by adding to it divergent counterterms. If we choose the counterterms wisely, then their divergences would cancel divergences in loop diagrams and the S-matrix calculated with our new Hamiltonian will become finite and it will agree with experiment."

    The next question is: which conditions should be used to derive specific forms of the counterterms? Roughly speaking, there are two physical conditions. They are the "mass renormalization condition" and the "charge renormalization condition". The first one demands that with the new Hamiltonian, single particles (e.g., electrons) do not experience "self-scattering" or, in other words, the interaction has no effect on the particle's mass. The second one renormalizes the interaction strength, by fitting it to known experiments. For example, in QED we can demand that electron-electron scattering at low energies and long distances is the same as the classical Coulomb scattering of two charges "-e".

    After the renormalization is done, we end up with a Hamiltonian containing divergent counterterms in each perturbation order. One can find some satisfaction in the fact that these counterterms can be combined into corrections to the mass and coupling constant parameters of the Hamiltonian. But still, it is important to realize that the Hamiltonian of the renormalized theory is an ill-defined divergent operator. For example, it cannot be used to define the time evolution, and we cannot simply diagonalize the Hamiltonian to get eigenvalues and eigenvectors of bound states, as we usually do in ordinary quantum mechanics.

    Nevertheless, our new "Hamiltonian with counterterms" is very good for one purpose -- calculation of the S-matrix (scattering amplitudes). Since 99% of high energy physics concerns S-matrix and related properties, the renormalization is considered a success and bad divergences in the Hamiltonian are simply ignored.

    Eugene.
     
  4. Aug 8, 2016 #3
    Hi Anchovy,

    It's true that when you compute a 'bare amplitude' at one-loop, you will likely experience UV divergences. Which, like you say are associated with the large momentum configurations.

    *** aside ***
    The infrared divergences can be soft or collinear. The soft divergences certainly appear from low energy configuration (like a soft gluon or something where there is an emission of a gluon with Egluon -> 0), but a collinear divergence need not necessarily be from Egluon->0. Anyway, thats not that relevant here.
    *** end aside ***

    When you calculate physical quantities, the UV divergences present in the 'bare amplitude' gets cancelled by those present in the definition of parameter/fields. The renormalisation procedure/technique allows you to separate out the bare parameters and fields which appear in the lagrangian and eases the calculation of processes. For the counter-term approach, this would proceed as follows

    Typically, one would choose an independent set of parameters:
    e.g. M_W, M_Z, m_f, m_H, e
    so masses and the electric charge of the broken SM (after higgs mechanism)

    Then, separate out the bare parameters and fields into the 'renormalised ones' and the UV counterterms (which contain the UV divergent parts)
    e.g. m_f^0 (0 for bare) -> m_f + delta m_f, q_L^0 -> (1 + delta Z_L^q) q_L
    so a mass counter term (delta m_f) and a wave function renormalisation constant delta Z_L^q. By expanding out these terms in the lagrangian, when you compute the feynman rules, you will literally arrive with the same rules as before for renormalised fields and in addition the new pieces which are the counterterm contributions.

    The form of these terms depends on the renormalisation scheme. For example, in the on-shell scheme these constants
    have a specific form and are calculated from two-point functions.

    Having computed all the renormalisation constants, you can then compute the predictions for observables. Making sure
    to express observables in terms of the chosen set of input parameters (M_W, M_Z, m_f, m_H, e).

    Of course, you don't know what any of these values are, unless you can measure them. So, after you have measured all of them and you know
    what they are, you can use them as inputs for your new calculations.

    The tricky point is the extraction of these parameters is scheme dependent. So if you use the on-shell scheme, it's the on-shell masses you want. You may alternatively use a scheme where the masses are running masses (mu-dependent). In this case, the extraction of the parameters is performed at a particular scale (typical of the process). For example, extracting the value of alpha_s in hadronic tau-decays occurs at ~ mtau. This is why often parameters are quoted in a number of different schemes.

    I strongly recommend the document of Ansgar Denner - https://arxiv.org/pdf/0709.1075v1.pdf. It's very informative.

    Since I see renormalisation as a calculational technique, I don't see any intuitive problem with isolating all these divergences dimensionally. So doing the loop integrals in d-dimensions and isolating the divergences as 1/epsilon poles.

    Hope this helps a bit.
     
  5. Aug 8, 2016 #4
    This brings up something that has been confusing me, and affects how I interpret the rest of your response. Having a scattering amplitude that diverges sounds straightforward enough. However, I'm not sure what it means for a (counter) term in a Lagrangian/Hamiltonian to 'diverge'. Say we're talking about the electron scattering process, with the tree-level diagram and additional loop diagrams constructed using vertices corresponding to the QED Lagrangian's interaction term, [itex]-ie \overline{\psi} \gamma^{\mu} A_{\mu} \psi [/itex]. What form does a "divergent" counter term take? Is it something like [itex] -1^{*} (-ie' \overline{\psi} \gamma^{\mu} A_{\mu} \psi )[/itex], where [itex] e'[/itex] has the same sign as, but a different value to, the original electric charge [itex]e[/itex] from the original interaction term?
     
  6. Aug 8, 2016 #5
    I believe I understand how the divergence in the scattering amplitude could arise (ie. if you have a loop factor in your amplitude calculation that blows up through integrating up to high momentum), but where does a divergence in the definition of parameters/fields come from / what does such a 'definition' look like mathematically?
     
  7. Aug 8, 2016 #6
    What you wrote is exactly one of the counterterms. It is called the "charge renormalization" counterterm. The value of the parameter [itex] e'[/itex] is infinite there (I am not sure about the sign.) More precisely: this value is a function of the ultraviolet cutoff [itex] \Lambda [/itex] and it tends to infinity as the cutoff goes to infinity. The job of renormalization is to choose the correct dependence [itex] e' (\Lambda)[/itex], so that scattering amplitudes become [itex] \Lambda [/itex]-independent and converge to desired values in the limit [itex] \Lambda \to \infty[/itex].

    In addition to the above "charge renormalization" counterterm, in QED there are few other counterterms ("mass renormalization", "vacuum polarization", etc.) of different operator structure. All of them diverge in the limit [itex] \Lambda \to \infty[/itex]. Their exact [itex] \Lambda [/itex]-dependence in each perturbation order is figured out by the renormalization theory.

    Eugene.
     
  8. Aug 9, 2016 #7

    vanhees71

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  9. Aug 9, 2016 #8
    OK, that's a relief. Usually my guesses are wrong and the right answer involves going down a rabbit hole.

    Hmm, so in that charge renormalization counterterm, this [itex]e'[/itex] can be equated to some function of the cut-off [itex]\Lambda[/itex]. Is this where perturbation theory comes into the picture, ie. is it that we'll have some power series [itex]e' = C + a\Lambda + b\Lambda^{2} + O(\Lambda^{3}) [/itex] ?

    Also, regarding:
    Are you able to show me a quick example of this?
     
    Last edited: Aug 9, 2016
  10. Aug 9, 2016 #9
    No, your formula for [itex]e' [/itex] doesn't make sense. Usually, renormalization constants (like [itex]e' [/itex]) have logarithmical dependence on the cutoff momentum [itex]e' \propto \ln(\Lambda/m) [/itex], where m is the electron mass. These expressions should be evaluated in each order of the perturbation theory independently. Here are step-by-step instructions for renormalization that you could find useful:

    1. Start from your original Hamiltonian with normal masses of particles and usual interaction [itex]-ie \overline{\psi} \gamma^{\mu} A_{\mu} \psi [/itex].
    2. Write the Feynman-Dyson series for the S-matrix based on this Hamiltonian
    3. Choose the desired perturbation order and draw all diagrams for your scattering process in this order.
    4. Isolate loop diagrams and classify them into types of counterterms that can be used for their suppression. For example, the counterterm you wrote earlier in the thread is good for cancelation of the so-called "vertex loop diagram".
    5. Pick one loop diagram and evaluate its contribution to the S-matrix element. This calculation of the loop integral is the most difficult part of the renormalization theory. Now there are lot of techniques for loop integrations, but I still prefer the original Feynman's paper "Space-time approach to quantum electrodynamics". Phys. Rev., 76 (1949), 769. As a result you'll get a (usually logarithmic) function [itex] Loop(\Lambda) [/itex] that goes to infinity as [itex]\Lambda \to \infty [/itex]. This is your loop divergence that you are trying to cancel.
    6. To cancel the above divergence you add a counterterm to the original Hamiltonian. The coefficient in front of this counterterm is a divergent function of [itex] \Lambda [/itex], which is chosen such that this additional interaction cancels the loop contribution [itex] Loop(\Lambda) [/itex].
    7. Repeat steps 5-6 for each loop diagram, thus producing a number of counterterms that ensure the finiteness of the S-matrix in your selected perturbation order.
    8. Go to the next perturbation order and treat all loop diagrams there in a similar way. You may wonder: the number of different loop graphs in all orders is infinite; does this mean that we need an infinite number of counterterms for renormalization? The answer to this question is given by the property of renormalizability. It appears that in different orders you meet only few types of counterterms. Only their coefficients are different in different orders. The counterterms have the same operator structures as pieces of the original Hamiltonian (or Lagrangian). So, you can regard counterterms as "corrections" to various constants (masses and charges) in the original Hamiltonian. That's why renormalization can be regarded as a replacement of masses and charges with divergent [itex] \Lambda [/itex]-dependent expressions.

    Eugene.
     
  11. Aug 10, 2016 #10

    haushofer

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