In Quantum Field Theory, one has to deal with loop integrals which are divergent. The way out is to regularise your integral. For example one can introduce a cutoff, like in the Pauli-Villlars regularisation scheme.

I was trying to learn this method, and I found some useful material in Zee's book: Quantum Field Theory in a Nutshell. But I'm having some math-related difficulty, and I hope someone could offer some help.

In short words, once we find a divergent integral (in QFT they appear in a common form), we can exploit the fact that the following integral, which sometimes is called a master formula, is convergent,

$$\int \frac{d^4 k}{(2 \pi)^4} \frac{1}{(k^2 - c^2 + i \epsilon)^3} = \frac{- i}{32 \pi^2 c^2}$$

Where, k is a 4-momentum, and c is usually a mass.

Then, if we face a divergent integral, like (we can see it by power counting),

$$\int \frac{d^4 k}{(2 \pi)^4} \frac{1}{(k^2 - c^2 + i \epsilon)^2}$$

We first introduce a cutoff in the Pauli-Villars way, which is to replace the integrand by:

$$\int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}]$$

Where, $\Lambda$ is the cutoff. And we are taking very large k and $\Lambda^2 >> c^2$.

In order to calculate this integral with this cutoff, I first differentiate the "divergent" integral wrt $c^2$ until I find a convergent integral. So,

$$ \frac{d}{dc^2} \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \int \frac{d^4 k}{(2 \pi)^4} \frac{2}{(k^2 - c^2 + i \epsilon)^3} = \frac{- i}{16 \pi^2 c^2} $$

Then, I integrate back,

$$ \int d \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \int dc^2 \frac{- i}{16 \pi^2 c^2} $$

Which gives (according to me :p and here where I think I'm missing something),

$$ \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{- i}{16 \pi^2} log(c^2)$$

But the correct answer is,

$$ \int \frac{d^4 k}{(2 \pi)^4} [\frac{1}{(k^2 - c^2 + i \epsilon)^2} - \frac{1}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{i}{16 \pi^2} log(\frac{\Lambda^2}{c^2})$$

Similarly for other integrals, such as

$$ \int \frac{d^4 k}{(2 \pi)^4} [\frac{k^2}{(k^2 - c^2 + i \epsilon)^2} - \frac{k^2}{(k^2 - \Lambda^2 + i \epsilon)^2}] = \frac{-i}{16 \pi^2} [\Lambda^2 - 2c^2 log(\frac{\Lambda^2}{c^2}) + c^2 + ...] $$

Which I don't know how to reach with my calculation. Is it that when we integrate back, we have the freedom to add a suitable constant (like $+ log(\Lambda^2)$ for the first case, and $+\Lambda^2 + log(\Lambda^2)$ for the second)? Or something else?

I hope the question is clear and someone will help me out here.

  • $\begingroup$ you should probably fix the error in your first integral... as it stands now it diverges quadratically.... $\endgroup$ – Kyle Nov 28 '11 at 1:18
  • $\begingroup$ Oh! Thanks for mentioning the typo. It's fixed now. $\endgroup$ – stupidity Nov 28 '11 at 1:23

I think the answer is that you can split up your integral into two identical integrals (up to a sign), you know the answer to both, the one with $c^2$ comes with a $-$ and the one with $\Lambda^2$ comes with a $+$ sign. Just make sure when you are integrating back after taking the derivative that you take the same bottom limit i.e.

$\int_{Const.}^{c^2} f'(s)ds$ and $\int_{Const.}^{\Lambda^2} f'(s) ds$ where $f(x)$ is the integral

$f(x) = \int \frac{d^4 k}{(2 \pi)^4} \; \frac{1}{(k^2-x^2 + i \epsilon)^2}$

hope that makes sense

  • 2
    $\begingroup$ it's worth commenting that although this works, like many quick and dirty ways of computing in QFT the maths is a little suspect! $\endgroup$ – Kyle Nov 28 '11 at 1:39
  • $\begingroup$ Thanks for your answer. That's what I did first, and I got a correct result for the log divergent integral. But for the quadratically divergent one, I get something like $~ \Lambda^2 - \frac{\Lambda^2}{c^2}log{\Lambda^2}{c^2} + c^2$ But this expression has the wrong dimension for the middle term (it's dimensionless), whereas $\Lambda^2$ and $c^2$ are of $(mass)^2$ dimension. The correct answer for the middle term should be something like $~c^2log\frac{\Lambda^2}{c^2}$ $\endgroup$ – stupidity Nov 28 '11 at 14:06
  • $\begingroup$ There's a typo in the previous comment. the middle term in the first expression reads, $\frac{\Lambda^2}{c^2} log(\frac{\Lambda^2}{c^2})$ $\endgroup$ – stupidity Nov 28 '11 at 14:13

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