1
$\begingroup$

I thought that for momentum integrals in Minkowski space, the Wick rotation to Euclidean space $k_0 \to ik_0$ allows one to write (let's say $f$ comes with an $i\epsilon$ prescription):

$$\int_{\mathbb{M}^4} d^4k \ f(k^2) = i \int_{\mathbb{E}^4} d^4k \ f(-k^2) = 2\pi^2i\int_0^\infty dk \ k^3 f(-k^2).$$

Sorry if the notation is weird, I wasn't sure how to denote the difference between Minkowski space and Euclidean space.

But I've come across a problem (calculating a loop integral) where doing this doesn't give the right answer, namely for

$$f(k^2)= \frac{1}{\left( k^2 - \Delta +i \epsilon \right)^2}.$$

Doing a Wick rotation results in:

$$I = \int_0^\infty \frac{k^3 dk}{\left( k^2 + \Delta \right)^2}$$

But the correct answer should be:

$$I = \int_0^\infty \frac{k^2 dk}{\left( k^2 + \Delta \right)^{3/2}}$$

According to Aitchison, Hey - 'Gauge Theories in Particle Physics', eq. (10.42). I uploaded it here.

Of course, both of these integrals are formally divergent, but suppose there's an energy cutoff. What went wrong here? What conditions must be met for a Wick rotation to work?

One suspicion I have is that maybe these two turn out to be equivalent, up to a choice of the cutoff energy. Especially since there is still a $\int_0^1 dx$ integration to be done to obtain the actual observable quantity, and $\Delta$ is a quadratic function of $x$:

$$\Delta = m_1^2 (1-x) + m_2^2 x -p^2 x(1-x) \equiv Ax^2 +Bx + C.$$

Could this be the reason? Or is the Wick rotation a mistake for some mathematical reason?

EDIT: I checked and the difference between the two integrals exists and equals $\log 2 - 1/2 \approx 0.2$, independent of $\Delta$. So they're not exactly equal even in the limit, but does it matter...?

$\endgroup$
2
$\begingroup$

As noted in the edit, I also believe the two results will be the same. Let's do an example using a convergent integral, taking $I\equiv \int d^4k \, f(k^2),$ where $f(k^2)=(k^2-\Delta+i\epsilon)^{-3}$. Then performing the Wick rotation yields $I = -2\pi^2i\int_0^\infty \frac{k^3dk}{(k^2+\Delta)^3}=-\frac{i}{2}\frac{\pi^2}{\Delta}.$

On the other hand, direct integration over $k^0$ yields $I = -\frac{3\pi i}{8}4\pi\int_0^\infty \frac{k^2dk}{(k^2+\Delta)^{5/2}} = -\frac{i}{2}\frac{\pi^2}{\Delta}.$

$\endgroup$
  • $\begingroup$ In your case the integral of the difference exists and equals zero, since both integrals exist on their own. I tried with this problem and the integral of the difference exists, but is a finite value, see my edit. $\endgroup$ – Spine Feast May 25 '17 at 14:17
  • $\begingroup$ Are you sure you're exactly keeping the integration regions the same when you're imposing the cutoff? The first is spherically symmetric in 4D Euclidean space whereas the second is spherically symmetric in 3D Euclidean space. $\endgroup$ – WAH May 25 '17 at 14:24
  • 1
    $\begingroup$ Apparently that's the problem, found a very similar question here physics.stackexchange.com/q/192643 $\endgroup$ – Spine Feast May 25 '17 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.