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Wick's Rotation contour integration

  1. Dec 23, 2012 #1

    I have Wick's rotated a contour integration of the form

    [tex]\frac{i\lambda}{2}\int d^{3}p\int \frac{dE}{(2\pi)^{4}}\frac{1}{E^{2}-p^{2}-m^{2}}[/tex]

    this is the form where we integrate along the real line, we rotate this to 'Euclidean' Space such that we make the changes

    [tex]E\rightarrow iE\: ,\: p^{2}\rightarrow-p_{E}^{2}[/tex]

    Where I think the change on the right of the momentum is the same as

    [tex]p\rightarrow ip_{E}[/tex]

    I'm not sure how to impose these changes on the integration measures dE and d^{3}p? I think we just get an 'i' factor from the change to the dE but not sure how to do the p integration measure.

    Thanks for any help guys,
  2. jcsd
  3. Dec 24, 2012 #2
    wick rotation is performed so as to make calculation easy.It transfoms it from minkowski space to euclidean one.it corresponds to a rotation by ∏/2 of interation path of E here.so only
    E--->iE and nothing for P2.It amounts to write the propagator term as
    E2+P2+m2.after it you don't need any iε prescription.There is no change in p.
  4. Dec 24, 2012 #3


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    The integral as given in the original post is simply not defined, since there's no rule given, how to integrate along the real [itex]E[/itex]-axis, because of the singularities of the integrand.

    I guess, it's about the tadpole integral in [itex]\phi^4[/itex] theory. If it's also for vacuum quantum field theory than it's very likely that the propagator at hand is the time-ordered one, which in this case is identical with the Feynman propagator. Thus, what you want to calculate is the integral
    [tex]\Pi=\frac{\mathrm{i}\lambda}{2} \int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \frac{1}{p^2-m^2 +\mathrm{i} 0^+}.[/tex]
    Of course, this integral is divergent as it stands, and one has to regularize it.

    One possibility is to use Wick rotation for the [itex]p^0[/itex] integration. This is an allowed procedure for this integral due to the residue theorem of function theory. As a function of [itex]p^0[/itex], which variable we extend to the entire complex plane now, the integrand has two poles
    [tex]p_{1,2}^{0}=\pm \sqrt{\omega^2+ \mathrm{i} 0^+}=\pm (\omega + \mathrm{i} 0^+)[/tex]
    with [itex]\omega=\sqrt{\vec{p}^2+m^2}[/itex].

    Now take a path consisting of the real axis a large quartercircle at infinity in the right-lower quadrant of the [itex]p^0[/itex] plane then up along the imaginary axis and another quarter circle at infinity in the upper left quadrant back to the real axis. The quarter circles obviously don't contribute anything, and the whole integral is 0 since we have chosen the contour such that it doesn't contain any of the two poles of the integrand in its interior. Thus you can just integrate along the imaginary axis. Thus we set
    [tex]p^0=\mathrm{i} p_4.[/tex]
    The Minkowski product then goes over to [itex]p^2=-p_4^2-\vec{p}^2=-p_E^2[/itex], where I defined the Euclidean scalar product as positive definite and wrote [itex]p_E=(\vec{p},p_4)[/itex] for convenience. From this we get
    [tex]\Pi=\frac{\lambda}{2} \int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p_E}{(2 \pi)^4} \frac{1}{p_E^2+\omega^2}.[/tex]
    The imaginary factor is gone, because there is a factor [itex]\mathrm{i}[/itex] from the switch from the [itex]p^0[/itex] to [itex]p_4[/itex] as an integration variable and a factor [itex]-1[/itex] from the use of the positive definite Euclidean metric. In this Euclidean integral we can forget about the regulator in the denominator, because (at least for the case of massive particles) there are no singularities of the integrand along the integrations. Of course the integral is still UV divergent, and you have to regularize it to make sense. A brute-force regularization is to just cut off the integral and integrate over a four-dimensional sphere [itex]p_E^2<\Lambda^2[/itex] or you use dimensional regularization or whatever else you like.
  5. Dec 24, 2012 #4
    yes,the integral is quadratically divergent in high energy limit.He has already posted that integral somewhere for evaluation where I have told him that it is divergent.But this one was about wick rotation so i just pointed out what is it.
  6. Dec 24, 2012 #5
    Hey guys thanks for getting back to me, thanks for going in to detail vanhees71 - I understand this much better.

    Indeed I have posted this before and the integral itself is undefined and I need to introduce some UV limit I believe to get any result.

    I believe the function is the first order contribution of the loop to the two point function or the first order contribution to the mass of the two point function, I really have studied any Quantum Field Theory at all so this is all quite confusing for me.

    Also I can't use 4 vector notation which, ironically, complicates things for myself.

    Nonetheless, I can see what I need to do now... I think.

    Cheers guys!
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