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In a certain lecture of Witten's about some QFT in $1+1$ dimensions, I came across these two statements of regularization and renormalization, which I could not prove,

(1) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + q_i ^2 \vert \sigma \vert ^2} = - \frac{1}{2\pi} ln \vert q _ i \vert - \frac{1}{2\pi}ln \frac{\vert \sigma\vert}{\mu}$

(..there was an overall $\sum _i q_i$ in the above but I don't think that is germane to the point..)

(2) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + \vert \sigma \vert ^2} = \frac{1}{2\pi} (ln \frac{\Lambda}{\mu} - ln \frac{\vert \sigma \vert }{\mu} )$

I tried doing dimensional regularization and Pauli-Villar's (motivated by seeing that $\mu$ which looks like an IR cut-off) but nothing helped me reproduce the above equations.

I would glad if someone can help prove these above two equations.

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The source is apparently pages 24-25 of lecture 12 here math.ias.edu/QFT/spring –  Mitchell Porter May 12 '12 at 5:12
@MitchellPorter That surely it is. I didn't link it because I thought the larger context was unnecessary and might even be diversionary from what is probably a "basic" calculation trick that I am unable to see. (very sadly!) May be you can help? –  user6818 May 12 '12 at 20:15

2 Answers 2

up vote 3 down vote accepted

Let's just look at the integral $$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2}.$$ The other integrals should follow from this one. Introduce the Pauli-Villars regulator, $$\begin{eqnarray*} \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} &\rightarrow& \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} - \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\Lambda^2} \\ &=& (\Lambda^2-\alpha^2)\int \frac{d^2k}{(2\pi)^2} \frac{1}{(k^2+\alpha^2)(k^2+\Lambda^2)} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \int\frac{d^2k}{(2\pi)^2} \frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \frac{1}{2} \frac{2\pi}{(2\pi)^2} \int_0^\infty dk^2\,\frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\beta^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\Lambda^2 - x(\Lambda^2-\alpha^2)} \\ &=& -\frac{1}{2\pi} \ln \frac{|\alpha|}{\Lambda} \end{eqnarray*}$$ Where we have combined denominators with the Feynman parameter $x$, with the intermediate variable $\beta^2 = \Lambda^2 - x(\Lambda^2-\alpha^2)$. Of course, this could also be approached with dimensional regularization with the same result.

Addendum: After regularization we must renormalize. Using the minimal subtraction prescription we find $$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} \rightarrow -\frac{1}{2\pi} \ln \frac{|\alpha|}{\mu},$$ as required.

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Thanks for the reply. That does explain much of the claimed equations. But can you explain the appearance of the ``\mu" on the RHS. (..how is that related to the UV cut-off of $\Lambda$ on the LHS?..) –  user6818 May 18 '12 at 20:17
@user6818: Glad to help. The $\mu$ comes from the renormalization prescription, minimal subtraction. We have $\ln |\alpha|/\Lambda \rightarrow \ln |\alpha|/\mu$. Roughly, the divergent part (and only the divergent part) is taken care of by the counterterm and what is left over is some arbitrary mass scale $\mu$. –  user26872 May 18 '12 at 20:54
I am familiar with the minimal subtraction scheme but in this case its not clear as to in what sense is $ln \frac{\vert \alpha \vert }{\mu}$ non-singular compared to $ln \frac{\vert \alpha \vert }{\Lambda}$ ? (..here I would think that you are subtracting $ln \frac {\mu} {\Lambda}$..) But isn't it a bit odd one had to have an UV cut-off ($\Lambda$) and also do a minimal substraction? (..I would typically think that one would do a minimal subtraction to remove the $\epsilon$ in the dimensional regularization..) –  user6818 May 18 '12 at 21:21
@user6818: The term with the UV cutoff, $\ln\Lambda$, is totally analogous to the term $1/\epsilon$ in dimensional regularization. We regularize in Pauli-Villars by adding the Pauli-Villars regulator. In dimensional regularization we allow the dimension of spacetime to vary. In either case, after regularization we have divergences that must be renormalized. There are many ways to do this, but the gist of it is that the counterterms are used to absorb the divergences. For the result to make any sense the term $\ln\Lambda$ must be absorbed by the counterterms. –  user26872 May 18 '12 at 21:39
@user6818: Note that $\mu$ is an arbitrary mass scale so $\ln\frac{|\alpha|}{\mu}$ is not singular for any reasonable choice of $\mu$. For example, we could let $\mu = 100$ GeV. –  user26872 May 18 '12 at 21:39

perhaps since your itnegral is logarithmic divergent you could do the following

$$ \int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}\to \int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}- \int_{0}^{\infty}\frac{dx}{x+b}+\int_{0}^{\infty}\frac{dx}{x+b} $$

then the integral $$ A=\int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}- \int_{0}^{\infty}\frac{dk}{k+b}$$ is convergent so we must now regularize

$ \int_{0}^{\infty}\frac{dx}{x+b} $ which can be made by using Ramanujan's resummation to get $ \sum_{n=0}^{\infty} \frac{1}{n+b}= -\Psi (b) $ now use the Euler-Maclaurin summation formula

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