I have followed two courses on QFT, which both involved renormalization by dimensional regularization. My confusion is that one of the professors claimed that dimensional regularization can only be used to regularize logarithmically divergent integrals, whereas the other professor claimed that the scheme can renormalize higher order divergencies. Let me make my question more precise.

Consider the standard integral which is computed in a dimensional regularization scheme: $$ I(n,\alpha) = \int \frac{d^np}{(p^2 + m^2)^{\alpha}} = i\pi^{n/2}\frac{\Gamma(\alpha-n/2)}{\Gamma(\alpha)}(m^2)^{n/2 - \alpha}. $$ For $\alpha=2$ and $n=4\pm\epsilon$ (depends on convention) the integral is logarithmically divergent and we can continue standardly by expanding the gamma, obtaining a $1/\epsilon$ pole and adding a counterterm to the Lagrangian. Generally, the integral is only convergent for $\alpha > n/2$. In the case of $\alpha = 1$ for example my first professor claimed that one has to introduce a Pauli-Villards cut-off in order to renormalize correctly. My second professor claims that one may use: $$ \int d^n q \frac{\partial}{\partial q^{\mu}}(q^{\mu}f(q)) = \int d^n q q^{\mu}\left(\frac{\partial}{\partial q^{\mu}} f(q)\right) + n\int d^n q f(q), $$ such that one throws away the boundary term as it is a surface integral by Gauss' theorem. He states explicitly that this may be done if $f(q)$ vanishes quickly enough at infinity, but also that "analytic continuation will be implemented by ignoring the surface term irrespective of the asymptotic behaviour of the integral." If this is true/legit, one may write: $$ \int d^n q q^{\mu}\left(\frac{\partial}{\partial q^{\mu}} f(q)\right) = - n\int d^n q f(q), $$ such that one can express divergent integrals in terms of finite ones.

Is the last approach correct? Is the first approach wrong? Could you comment on these two approaches?


1 Answer 1


You may check that for $\alpha\leq 0$ when the integral is UV-divergent, the result for $I(n,\alpha)$ you wrote down – which is an analytic function i.e. a result of analytic continuation (and yes, the ignoring of the surface terms for all values of the exponents and dimensions is a way to easily cancel the seemingly divergent but morally vanishing terms which is enough for the analytic continuation) – is still perfectly finite because the pole from $\Gamma(\alpha-n/2)$ (which exists assuming that the argument is a non-positive integer) is cancelled against the pole in the denominator $\Gamma(\alpha)$.

For $n/2=2$, the only values of $\alpha$ for which the pole in the numerator exists and isn't cancelled are $\alpha=2$ which is the logarithmic divergence; and $\alpha=1$ which is the quadratic divergence. Moreover, the pole-like result in this $\alpha=1$ case may be attributed to the "log divergence" part of the integral while the "quadratic divergence" part of the integral, when the integral is properly divided, is equal to zero in dim reg. Dimensional regularization automatically "annihilates" higher-order divergences such as the quartic one (where the admixture of the log divergence is already zero in your integral); a quartic divergence appears in vacuum energy (cosmological constant) diagrams.

Even when the quartic divergences (or the purified quadratic divergences) are rendered as finite by dim reg, it is still true that the theory is sensitive on the parameters of the theory we use at high energies so the elimination of these divergences doesn't mean that we eliminated the finite parameters that still have to be specified and whose values aren't uniquely dictated by any principle if we interpret our QFT as an effective one only.

Maybe the first professor doesn't like the fact that dim reg completely annihilates the infinite term for higher-order power-law divergences – that's a situation he doesn't call a "successful regularization" because in the physical limit, he still wants the result to look infinite. But there's nothing wrong if a regularization assigns a finite value to a naively divergent integral. And indeed, dim reg likes to assign a finite (or even vanishing) result to many diagrams that are divergent in other regularizations. For example, dim reg may automatically preserve the gauge symmetry which means that it cancels the infinite corrections to the mass of the photon.

Analogously, the zeta-function regularization used in 2-dimensional CFTs (and string theory, among other places) automatically preserves the conformal symmetry. The conformal symmetry implies, among other things, that the only symmetry-preserving finite value of the sum of positive integers is $1+2+3+\dots = -1/12$ so similar sums are automatically rendered finite and the value is completely unambiguous in the zeta-function regularization.

  • $\begingroup$ I don't really get what you mean by "Dimensional regularization automatically "annihilates" higher-order divergences such as the quartic one". I have seen zeta-function regularization in string theory. I was also wondering why this is consistent. One regularizes the sum $\sum_{i=1}^{\infty}n = lim_{\epsilon\rightarrow 0}\sum_{i=1}^{\infty}n e^{-\epsilon n} = 1/\epsilon^2-1/12$, but this seems a bit arbitrary to me. Why not divide it by n! for example? I guess this would yield a different finite part. $\endgroup$
    – Funzies
    Jul 7, 2013 at 22:52
  • $\begingroup$ @Erik: if you use the analytic continuation of Riemann zeta $ \sum_{n=1}^{\infty}n = -1/12 $ this is also similar for divergent integrals :) vixra.org/pdf/1305.0171v3.pdf $\endgroup$ Jul 13, 2013 at 8:39
  • $\begingroup$ If you don't mind, could you elaborate a bit on the last paragraph, in particular "the only symmetry-preserving finite value of the sum of positive integers is −1/12"? I'm curious about where the "specialness" of -1/12 comes from after reading the answers to this question. $\endgroup$
    – user76284
    Jul 21, 2019 at 7:12
  • $\begingroup$ It's a half a semester course to explain the things really clearly. But the value -1/12 is the only one that respects conformal symmetry and modular invariance in various partition sums and derived quantities. There are also many diverse ways to calculate the only correct result. See eg math.stackexchange.com/questions/39802/… $\endgroup$ Jul 22, 2019 at 13:32

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