Consider the following integral:
∫d3k(2π)31k3
In dim-reg, such integrals evaluate to 0. However, if we instead consider
∫d3k(2π)3(p+k)2k3(p+k)2=∫dx∫d3k(2π)3k2+(1−x)2p2(k2+x(1−x)p2)5/2Γ(52)Γ(32)(1−x)1/2=∫dxμ2ϵ(4π)3/2−ϵ[Γ(ϵ)Γ(52)Γ(32)Γ(52)(1x(1−x)p2)ϵ+finite]32(1−x)1/2=14π2Γ(ϵ)μ2ϵp−2ϵ+const
Even though the integrals are the same (after canceling out (p+k)2), they give different results when evaluated under dim-reg. My understanding of this is that when one did the Feynman parameterization, an IR regulator was secretly introduced when interchanging the dx and the d3k.
- How exactly is the IR being regulated in the second case?
- The fact that the first case evaluates to 0 is due to a cancellation between UV and IR divergences. Is there an easy way to evaluate only the UV divergence of the first integral within dim-reg?
Answer
You have to be careful when performing certain operations in dimensional regularization, and you always have to check the convergence.
What meaning does an integral like ∫dDk(2π)D1k2+m2
This happens in this case. The integrand goes to kD−1 when k is near zero (the presence of a non null m2 is crucial for this fact), so you have convergence around zero when Re(D)−1>−1, while near ∞ you have that the integrand goess to kD−3, so you have convergence when Re(D)−3<−1. In total, the integral is convergent in the strip 0<Re(D)<2, and analytic in that strip. So, you can perform analytical continuation in the complex plane and obtain a formal expression for each complex value of D, to use in calculations.
If, however, m=0 you can run in some problems. The integral ∫dDk(2π)D1k3=12D−1πD2Γ(D2)∫∞0kD−4dk
Now, let's compute this sum by using the Feynmann trick, to see if we get zero. For the first integral, you have Γ(32)Γ(12)∫10dx∫dDk(2π)D1√x(k2+δ2(1−x))32=√π2Γ(3−D2)(4π)D2Γ(32)δD−3Γ(D−12)Γ(D2).
I will omit calculations for the second. At the end, you just have 3√π4Γ(5−D2)(4π)D2Γ(52)δD−3Γ(D−32)Γ(D2).
Now we sum. Note that here you have to use the properties of Γ. The sum is √π21(4π)D2Γ(D2)(Γ(5−D2)Γ(D−32)+Γ(3−D2)Γ(D−12))δD−3.
To summarize, your error lies in the exchange of the dx and dk integrals. To do this exchange, you have to find an open set in the complex plane where the integral makes sense and is convergent. In your case, you could try by expanding the numerator and "splitting" (improper terminology, as you are defining, not splitting) the integral: each term of the sum has a strip of convergence, so you can use the Feynmann trick. Try to do that and see if you can obtain zero.
My reference for this answer is the book by Damiano Anselmi, "Renormalization", freely downloadable from his site, http://renormalization.com/ (under Documents->Books. Yeah, he owns that domain!). Chapter 2's beginning introduces dimensional regularization.
No comments:
Post a Comment