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I Analytically continuing 2,3,4pt integrals

  1. Sep 13, 2017 #1

    CAF123

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    I was reading a paper that gives a nice collection of all scalar integrals that crop up in QCD loop calculations. Such integrals are computed in some kinematic region and then the authors say the results may be analytically continued if so desired. I just wonder how is this analytic continuation done in practice? It's a relatively short paper and I've pasted it here https://arxiv.org/pdf/0712.1851.pdf

    Thanks!
     
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  3. Sep 13, 2017 #2

    mathman

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  4. Sep 14, 2017 #3

    CAF123

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    I was wondering how it applies to the loop integrals with the feynman +iepsilon prescription, e.g

    The authors state that a particular kinematic region allows for the ##i\epsilon## to be dropped and then one can analytically continue results via the prescription $$p_i^2 \rightarrow p_i^2 + i\varepsilon, s_{ij} \rightarrow s_{ij} + i\varepsilon, m_i \rightarrow m_i - i\varepsilon.$$ I just wonder why this is the case and if the sign choices here are significant?

    As a simple example, the analytic continuation of the massive tadpole is given as $$I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\varepsilon)^{1-\epsilon}$$ but what should I do with this result as it contains an explicit ##\varepsilon## in ##m^2-i\varepsilon##?
     
  5. Sep 14, 2017 #4

    vanhees71

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    Since you are interested in the limit ##\epsilon \rightarrow 0## the ##\mathrm{i} \epsilon## is unproblematic.
     
  6. Sep 14, 2017 #5

    CAF123

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    Could you elaborate? There is an ##\epsilon## (epsilon) from dim reg and there is a ##i\varepsilon## (varepsilon) from the feynman prescription.
     
  7. Sep 14, 2017 #6

    vanhees71

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    Argh, yes. I meant the ##\mathrm{i} \varepsilon## of course. Don't use ##\epsilon## for both quantities :-).
     
  8. Sep 14, 2017 #7

    CAF123

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    Ok. I'm just wondering though if I take the ##\varepsilon \rightarrow 0## limit don't I just get the result before analytic continuation? E.g for the massive tadpole given earlier, I was presuming the result before analytic continuation is with ##m^2-i\varepsilon## replaced with ##m^2##?
     
  9. Sep 14, 2017 #8

    vanhees71

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    Let's write your expression as
    $$I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\eta)^{1-\epsilon}.$$
    What I meant is that you are interested in the limit ##\epsilon \rightarrow 0##. Then the ##\mathrm{i} \eta## is irrelevant for ##\eta \rightarrow 0^+##.

    The Laurent expansion is
    $$I_1^D=\frac{m^2 - \mathrm{i} \eta}{\epsilon} - (m^2 - \mathrm{i} \eta) \left [ -1 +\gamma_{\text{E}} + \ln \left (\frac{m^2-\mathrm{i} \eta}{\mu^2} \right) \right ] + \mathcal{O}(\epsilon).$$
    As you see, you can just take ##\eta=0## without problems.
     
  10. Sep 14, 2017 #9

    CAF123

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    I see thanks, a few follow up questions

    1)I guess the idea of analytic cont is the process of taking a result computed in some kinematic region and looking for the result in some other kinematic region. So for the tadpole example, is the analytic cont a means of finding the result for ##m^2<0## if the original result was computed in ##m^2>0##?

    2) What does the fact that one can take ##\eta \rightarrow 0## without problems mean? I suppose the original result before analytic cont. was computed in the region ##m^2>0## but if I take ##m^2<0## and send ##\eta \rightarrow 0## the log has a negative argument. Perhaps I'm thinking about this in the wrong way.

    2) why is it necessary that the correct analytic cont for ##m^2## is ##m^2-i\eta## and not, say, ##m^2+i\eta##? I suppose this is why you made ##\eta \rightarrow 0^+##.

    Thanks!
     
  11. Sep 15, 2017 #10

    vanhees71

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    The ##\mathrm{i} \eta## comes into the game, because in vacuum-QFT pertubation theory you need the time-ordered propagator, i.e., you have a pole structure (in Minkowski space with the (+---) signature)
    $$G(p) \propto \frac{1}{p^2-m^2+\mathrm{i} \eta}.$$
    Alternatively you can use this pole structure of the free propagators to Wick rotate the energy component ##p^0## to make it purely imaginary. Then you deal with Euclidean QFT since the Minkowski product becomes (up to a cumbersome sign) the positive definite Euclidean scalar product. The prize to pay is that you have to Wick rotate back after you've done your integrals, which is not as simple as it looks at the first glance, particularly if you have vertex functions with ##\geq 3## legs, which depend on more than one independent four-momentum.

    In the case of the tadpole loop you get of course a function that does not depend on any external momentum, because it's effectively a one-point function and the vacuum is translation invariant. With ##m^2>0## the tadpole loop with a scalar propagator has superficial (and since there are no subdivergences also true) degree of UV divergence 2 and no infra-red/collinear singularities. The UV divergence has to be regularized, and in your case this has been done using dimensional regularization. As long as ##m^2>0## and because the degree of divergence is positive, there are no poles or other singularities, and thus the ##\mathrm{i} \eta## has no function anymore, and you can make ##\eta \rightarrow 0## without trouble.

    Since the superficial degree of divergence is positive, there's also no problem in making ##m^2 \rightarrow 0^+##. This is because there are no IR problems with the diagram and the fact that dimensional regularization is a socalled mass-independent, when the (modified) minimal-subatraction scheme is applied, i.e., you don't subtract any log terms containing the mass. Thus both the regularized as well as the MS renormalized tadpole diagram vanishes for ##m^2 \rightarrow 0^+##.

    For more on renormalization, see my QFT manuscript:

    http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
     
  12. Sep 15, 2017 #11

    CAF123

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    If I put ##m^2## to zero in the integral representation for the tadpole then there is now an IR singularity ( to be cancelled with the UV divergence to render the result zero in dim reg because integral is scaleless). How does this agree with what you said about being able to take ##m^2 \rightarrow 0^+## ?

    Also , regarding the analytical continuation , in that paper I referenced they mention one can correctly analytically continue expressions using some relations between logs and dilogs which crop up in massive triangles. Can I ask if you understand what is being said in this section of the paper (sect. 2.2) so I can ask a few things ?

    Thanks!
     
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