In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. It seems to me that multiplying all diagrams with the same number of loops by the same regular-in-$\epsilon$ factor should not affect the final result.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(1-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop. But I do not have much confidence in this claim. Can someone point to a specific computation where this regularization fails to give the same answer as dimensional regularization?

To complete the specification, let's say that traces of Dirac matrix products have their usual 4-dimensional form and that $g_{\mu\nu}g^{\mu\nu} = n$.

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    $\begingroup$ I see one hole in my reasoning: It only applies to simple poles. If there's a 2nd order pole in $\epsilon$, this approach can change the coefficients of the first order poles. $\endgroup$
    – user1504
    Mar 25, 2018 at 1:06
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    $\begingroup$ (Related question here.) $\endgroup$
    – knzhou
    Mar 25, 2018 at 12:04
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    $\begingroup$ @user1504 If you let $\mu\to \mu\ (\Omega_d/\Omega_n)^{1/\epsilon}$ in your second integral you get the first one. As $\mu$ is arbitrary, the two schemes are equivalent, right? $\endgroup$ Mar 27, 2018 at 1:00
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    $\begingroup$ @user1504 the spinor traces are a matter of conventions. One sets $\text{tr}(1)=f(n)$ for some function such that $f(4)=4$. Some common options are $f(n)=4,\ n,\ 2^{n/2},\ \dots$ Any choice is as good as any other. Once $f$ is fixed, this determines the traces of any combination of gamma matrices. For more details, see Itzykson & Zuber, chapter 8 (in particular, the discussion around 8-11c). Cheers! $\endgroup$ Mar 27, 2018 at 1:40
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    $\begingroup$ Interesting article uploaded to the arXiv this morning: arxiv.org/abs/1803.09764 $\endgroup$ Mar 28, 2018 at 13:36

1 Answer 1


We work in Euclidean space-time.

OP is trying to compare the two prescriptions $$\tag{A} \int_{\mathbb R^{na}} f(p_1,\dots,p_a)\ \mathrm d^np_1\cdots\mathrm d^np_a\ \to\ \mu^{a\epsilon}\int_{\mathbb R^{da}} f(p_1,\dots,p_a)\ \mathrm d^dp_1\cdots\mathrm d^dp_a $$ and $$\tag{B} \int_{\mathbb R^{na}} f(p_1,\dots,p_a)\ \mathrm d^np_1\cdots\mathrm d^np_a\ \to\ \tilde\mu^{a\epsilon}\int_{\mathbb R^{na}} \frac{f(p_1,\dots,p_a)}{|p_1|^\epsilon\cdots|p_a|^\epsilon}\, \mathrm d^np_1\cdots\mathrm d^np_a $$ where $(n,a)\in\mathbb N^2$ is a pair of integers; $d\in\mathbb C$ is a complex parameter; $\epsilon:=n-d$; and $(\mu,\tilde\mu)\in\mathbb R^2$ is a pair of real parameters (mass scales).

We stress that $(\mathrm A)$ is the standard dimensional regularisation prescription. The case $(\mathrm B)$ is quite similar to Speer's analytic regularisation, but it is not equivalent.

If $f$ depends on the momenta only through their square, $$ f(p_1,\dots,p_a)=f(p_1^2,\dots,p_a^2) $$ then it's easy to check that the prescriptions above are equivalent to $$\tag{A} \int_{\mathbb R^{na}} f(p_1,\dots,p_a)\ \mathrm d^np_1\cdots\mathrm d^np_a\ \to\ \mu^{a\epsilon}\Omega_d^a\int_{\mathbb R^{a}} p_1^{d-1}\cdots p_a^{d-1}f(p_1,\dots,p_a)\ \mathrm dp_1\cdots\mathrm dp_a $$ and $$\tag{B} \int_{\mathbb R^{na}} f(p_1,\dots,p_a)\ \mathrm d^np_1\cdots\mathrm d^np_a\ \to\ \tilde\mu^{a\epsilon}\Omega_n^a\int_{\mathbb R^{a}} p_1^{n-1-\epsilon}\cdots p_a^{n-1-\epsilon}f(p_1,\dots,p_a)\, \mathrm dp_1\cdots\mathrm dp_a $$ and we see that the two prescriptions are mapped into each other under $$ \tilde\mu^\epsilon\to \frac{\Omega_d}{\Omega_n}\mu^\epsilon $$

As $\mu,\tilde\mu$ are arbitrary, the two prescriptions are equivalent.

If $f$ contains a non-trivial spinorial structure, the equivalence is still correct if we agree that traces in $d$ dimensions satisfy $$ \text{tr}(1)=2^{\lfloor n/2\rfloor} $$ for all $d\in\mathbb C$.

On the other hand, if $f$ contains non-trivial Lorentz structure (i.e., if it's not a Lorentz scalar or it depends on momenta not only through their squares, but also on the scalar products $p_i\cdot p_j$), then the correspondence breaks down. A simple way to see this is that, according to the standard dimensional regularisation prescription we have $$ p^\mu p^\nu\to\frac{\delta^{\mu\nu}}{d}p^2 $$ while in the modified prescription we have $$ p^\mu p^\nu\to\frac{\delta^{\mu\nu}}{n}p^2 $$

These are not equivalent, and cannot be made equivalent by modifying either prescription (we cannot define $p^\mu p^\nu$ to be $\frac{\delta^{\mu\nu}}{n}p^2$ in $d$ dimensions, because algebraic manipulations such as contractions and shifts in the integration variable would lead to inconsistencies). While it's true that the divergent part is the same in both approaches, the finite part need not.

More generally, the two prescriptions differ by a rational function of the form $$ \frac{P(n)}{P(d)} $$ where $P$ is a polynomial. This polynomial is not in general the same for all Feynman diagrams that contribute to a certain order in perturbation theory (see e.g. the case of scalar QED, cf. ref1, chapter 65). Therefore, one does not expect a cancellation of the $\xi$-dependent terms, and the Ward identity is violated. This cancellation holds in the case of dimensional regularisation, so the two prescriptions are not equivalent.

What's more, not only is the gauge symmetry anomalous in OP's scheme, the axial symmetry is not. Indeed, if $n$ is even, we can define the axial matrix $\gamma_5$. This matrix is traceless $\text{tr}(\gamma_5)=0$ for any number of spacetime dimensions, so the divergence of the axial current vanishes (in dimensional regularisation, this argument fails because $\gamma_5$ is ill-defined for complex $d$; but in OP's prescription the number of spacetime dimensions $n$ is fixed). As the axial anomaly does not vanish for any (even) $n$ (cf. this PSE post), the OP's regularisation is not equivalent to dimensional regularisation.

Finally, it bears mentioning that there is an intermediate prescription, $$\tag{C} \int_{\mathbb R^{na}} f(p_1,\dots,p_a)\ \mathrm d^np_1\cdots\mathrm d^np_a\ \to\ \hat\mu^{a\epsilon}\frac{\Omega^a_d}{\Omega_n^a}\int_{\mathbb R^{da}} f(p_1,\dots,p_a)\ \mathrm d^dp_1\cdots\mathrm d^dp_a $$ which is equivalent to dimensional regularisation. This prescription leaves the angular measure as that of $n$ instead of $d$, so it is rather close to what OP was after. But the integrand is evaluated at complex $d$ rather than at fixed $n$, so it is the same thing as dimensional regularisation (or rather, it is mapped into it under the aforementioned rescaling of the mass scale).


  1. Srednicki's QFT.
  • $\begingroup$ I'll try and calculate the vacuum polarisation in scalar QED in OP's regularisation. If I get anything meaningful I'll update the answer. For now, I'm not quite sure how it works because OP's prescription is not invariant under translations in momentum space, so it seems to be inconsistent (e.g., the value of a diagram depends on how we parametrise the momenta that flows through it; i.e., it depends on how we assign a momentum to each line). I'll have to think about it; it's kinda late here and I may be talking nonsense. $\endgroup$ Mar 27, 2018 at 3:13
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    $\begingroup$ ... what's more, in OP's scheme you can't even combine denominators using Feynman parameters. OP's prescription doesn't seem to lead to a manageable integral at all. $\endgroup$ Mar 28, 2018 at 2:27

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