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Divergence of forward Bhabha/Moller scattering

  1. Sep 24, 2010 #1

    I have a question about the divergence of forward Bhabha/Moller scattering.

    I guess the classical analog of it is the Rutherford cross-section divergence, but that can be explained by the infinite impact parameter.

    In the QED version - the Bhabha/Moller scattering, it is the matrix element for given states that diverges - not only the cross-section, and I can't see how two plane-wave particles can have an impact parameter that could resolve this divergence.

    Also, it seems that the divergence stems from the zero-momentum divergent photon propagator here. I saw explanations that this is typical for any infinite range interaction.

    Could somebody please explain what is the solution for this divergence. Is this an unphysical one? What was the wrong assumption that caused it?

    Thank you.
  2. jcsd
  3. Sep 26, 2010 #2
    Hi again.

    Well, I'm a bit surprised I didn't get any answers.

    If the subject doesn't look that interesting - I think it is interesting. Infinite probability without explanations of the simplest electron-electron scattering looks like a serious problem in QED, which is the most accurate physical theory we have.

    If it is because nobody has an answer - this makes it even more interesting. But my guess is that I just misunderstood something here, and I'd appreciate it very much if some one could point me to the issue.

    Thank you.
  4. Sep 27, 2010 #3

    can you be more specific as exactly which quantity diverges ? Please keep in mind that differential cross-section is not probability : it is probability density. There is nothing wrong for the density to go to infinity as long as the integral remains finite. Often asking a well-posed question amounts to defining a domain of integration.

    There are quite a few divergences, AFAIK all of them are "innocuous" in the sense that they are either "fake" or if they are real, they have a physical interpretation. Unfortunately, they are mostly non-trivial.

    For the long range Coulomb divergence, they are indeed due to infinite range of the interaction. If one is asking "what is the cross-section for arbitrarily low angle scattering ?" one is faced with arbitrarily large impact parameter because the Coulomb potential has "infinite range". So if you think carefully about what cross-section means, in terms of effective scattering surface, the impact parameter going to infinity means that the surface also goes to infinity.

    The question is still well-posed for a finite angle and checked with experiment.

    However, I am not clear that this is precisely your question. There are other collinear divergences which have another interpretation. For instance, if you try to take into account radiative correction (next order in coupling constant) with yet another (low energy) photon in the final state, one is faced with infinities due to the fact that a propagating charged lepton has unit probability to have an additional arbitrarily low energy collinear photon.
    Last edited: Sep 27, 2010
  5. Sep 27, 2010 #4
    Thanks for the response. I also assumed that the divergence is false, but I want to understand why it is false.

    To be more specific: In QED, electron-electron scattering matrix there is a photon propagator with 1/(k1-k2), where k1 is the initial electron momentum and k2 is the final electron momentum for one of the particles. For zero-deflection case k1=k2, and the photon propagator diverges. This makes the matrix element diverge.

    I agree that the cross-section can be infinite for infinite impact parameter and that in experiment there are no ideal plane wave etc, but theoretically, as I mentioned before, two things are not clear to me here: 1) how can an impact parameter be defined for two plane-wave electrons? 2) It is the matrix element that diverges - before the calculation of the cross section.

    Maybe this is my main problem. Is matrix element squared for two electrons having definite initial and final states (e.g. plane waves) a probability density, and not probability? If I were calculating the probability density of scattering into an infinitesimal angle, it would indeed be OK to get an infinity that would become finite after integration over a finite angle range. Doesn't a matrix element <k2|S|k1> represent probability and not probability density for normalized plane waves?

    I know that some radiative corrections yield divergences that are fixed by renormalization, but I don't think this is the case here.

  6. Sep 27, 2010 #5
    My understanding is that a matrix element such as <k2|S|k1> is neither a probability nor a probability density. A probability (per unit time per unit luminosity) would be an integrated cross section. A probability density would be a differential cross section. What the usual Feynman rules allow one to compute is not S but the matrix elements of S-1. To go from those matrix elements to the differential cross section requires (of course taking the square modulus and) a phase space density factor taking into account various normalization conventions (including factors of 2 and pi) for the scalar product of bras and kets. So the link between QFT and experiments is usually at the level of differential cross sections. You are now asking what happens in a singular soft limit which is ill defined experimentally. Although it may formally be possible to absorb normalization conventions for the density of states into the normalization of plane waves, I am unsure this is good idea to approach the problem. The normalization of plane waves for instance usually includes the volume of the box, which will also go to infinity in the soft limit you are worried about.

    As a consequence, I do not think I managed to be much helpful !
  7. Sep 27, 2010 #6
    I guess I'm not sure about one thing: Doesn't a theory have to be free of divergences regardless of the experiment. Is it a valid theoretical argument to say something like "this will take for ever to record" or "this requires infinite precision of equipment" so we don't have to worry about the singularity?
  8. Sep 28, 2010 #7

    Vanadium 50

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    Well, yes. The theory has to match observation, and observation means you are looking in some region of time, space and minimum energy.
  9. Sep 28, 2010 #8
    Then why do QFT people worry so much about eliminating infrared and ultra-violate divergences in the theory? As much as going to string theory. If the theory has to be consistent only as far as experiment is concerned, and if no one can measure infinitesimal and/or infinite momenta - why bother?
  10. Sep 28, 2010 #9
    The question you ask has an infinite answer not because of a theory. It already has an infinite answer in classical mechanics. Unless I am mistaken, you are also doing a classical calculation using the QFT mechanisms (although it is difficult to tell, because despite several requests, you have not bothered writing down the formula for what goes to infinity, apart from 1/virtuality of the exchanged photon). In this case, the infinity is the question : you are asking the value for an effective surface of infinite radius.
  11. Sep 28, 2010 #10

    Vanadium 50

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    Among other reasons, this has a divergence because it is fixed order.
  12. Sep 28, 2010 #11
    I appreciate your efforts to help very much, and it is not a matter of me "bothering writing down the formula". I'm just new here, and I have no idea how to display equations in posts. Here's a page in Mandl nad Shaw where the matrix element (Eq. 8.48a) diverges for p1=p1':


    In both quantum and classical mechanics I can understand the divergence of the cross-section. But in quantum version I don't understand the divergence of a step before the calculation of the cross section - the divergence of the matrix element. In the classical version one step before the cross-section is the impact parameter which comes out infinite. I can't think of a quantum analog of an impact parameter for plane waves (p1 and p1').

    Again - this is simple in the classical version, but what is the "radius" from the "center" for two plane waves?
  13. Sep 28, 2010 #12
    Do you mean that this can be probably fixed by higher-order IR divergent terms? Similar to the solution of Bremsstrahlung IR divergence? I couldn't find anything like that in the literature
  14. Sep 28, 2010 #13
    Actually, I don't. The classical calculation is done simply by calculating the trajectory of a point charge in a Coulomb potential - e.g. Rutherford scattering:


    or in more detail here:

    http://www.worldscibooks.com/etextbook/5460/5460_chap1_2.pdf [Broken]
    http://www.worldscibooks.com/etextbook/5460/5460_chap1_3.pdf [Broken]
    Last edited by a moderator: May 4, 2017
  15. Sep 28, 2010 #14

    Vanadium 50

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    You don't expect a fixed order prediction to necessarily be finite. Only the all-orders prediction needs to be finite.
  16. Sep 28, 2010 #15
    In fact I wanted to suggest to look at Rutherford scattering, because this is the same 1/sin(theta)^4 divergence. Unless you take into account more than one photon exchange, in which case you do need to renormalize you result, then by taking into account only one photon exchange you get the classical result (no quantum correction).
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