Several accounts of QFT allege that using a hard momentum cutoff $p^2<\Lambda^2$ breaks Lorentz invariance. For instance, see Schwartz's book, p833, or Weinberg p14, or answers here. But I don't understand the claim: it seems that under a sensible interpretation, computations with a hard cutoff are perfectly Lorentz-invariant.

Even when calculating results for a Lorentzian QFT, the momentum cutoff is generally understood to be taken in the Euclidean QFT, as $p^2 = p_0^2 + \vec{p}^2 < \Lambda^2$. The resulting cutoff theory is $SO(d)$ invariant, i.e. the momentum-space correlation functions are invariant under Euclidean rotations. And when we use the $SO(d)$-invariant Euclidean theory to perform calculations in Lorentzian signature, we obtain Lorentz-invariant results, right?

Concretely, if we want to compute some time-ordered momentum-space correlation function in Lorentzian signature, but with a momentum-space cutoff, the standard procedure is to first compute the same momentum-space correlation function in Euclidean signature with a Euclidean momentum cutoff. That produces some $SO(d)$-invariant function of the Euclidean momenta. The result is then cast back into Lorentzian signature by replacing $p_0$ with $-ip_0$. The final Lorentzian result actually looks the same as the Euclidean result, as long as you write both using some implicit metric, i.e. as a function of $p^2, p \cdot q$, etc. And so it is manifestly Lorentz-invariant.

Have I correctly concluded that the Lorentzian signature results computed with a hard cutoff are Lorentz-invariant? When these sources claim the hard cutoff is not Lorentz-invariant, are they actually referring to using a Lorentzian cutoff $p^2 = p_0^2 - \vec{p}^2 < \Lambda^2$? But note the latter is not standard practice, and does not yield convergent loop integrals; besides, Schwartz explicitly writes $k_E < \Lambda$, i.e. refers to a Euclidean cutoff. Do they simply mean that the Euclidean-signature results are Euclidean-invariant rather than Lorentz-invariant? (But that would be a strange usage of "Lorentz invariance," since the same results, when analytically continued to Lorentzian signature, would be Lorentz invariant.)

Update: Relating Euclidean momentum cutoffs to Euclidean lattice discretization

As commented by "ACuriousMind," a Euclidean momentum cutoff bears some relation to Euclidean lattice discretization. But I don't think this relation merits the notion that momentum cutoffs break Euclidean invariance.

On one hand, lattice discretization does break both rotation and translation invariance. Relatedly, functions on an infinite $d$-dimensional lattice can be described by continuous momenta within a finite $d$-dimensional cube of Fourier space, but this subset of momenta is not closed under rotation.

On the other hand, imposing a momentum cutoff (restricting to momentum within a Euclidean ball) is not equivalent to the above lattice picture. Consider the space of functions on $\mathbb{R}^d$ whose Fourier transform is supported on $p^2 < \Lambda^2$. I emphasize that we can consider these functions on the spacetime continuum $\mathbb{R}^d$, despite the momentum cutoff. These are the functions by which we are allowed to smear our fields when computing correlation functions in the momentum-cutoff theory. This space of functions is closed under the full group of Euclidean isometries, even translations.

So in the cutoff theory, where the correlation functions must be restricted to smeared fields with bounded momentum support, it's still sensible to talk about Euclidean-invariant correlation functions, even if we can't ask about correlation functions of fields like $\langle \phi(x) \phi(y)\rangle$ that are not smeared in that way. And again, when we analytically continue these correlation functions to Lorentzian signature, it seems we obtain Poincare-invariant results.

  • $\begingroup$ I think the answer is just Yes: your reasoning is correct. Restricting the integration over momenta to $p_0^2-\vec{p}^2<\Lambda^2$ preserves Lorentz invariance. But, as you noted, the problem is this kind of cutoff simply is not a cutoff and you still have divergences at large momenta. Also note that, because of infrared problems, one would rather need to exclude instead of keep regions where $p_0^2-\vec{p}^2<\Lambda^2$. This is similar to RG around the Fermi surface in solid state physics, which is easier because the surface is compact whereas a mass hyperboloid is not. $\endgroup$ Mar 15, 2021 at 18:41
  • $\begingroup$ @Abdelmalek, you refer to a cutoff on Lorentzian momentum; I want to additionally suggest that when we instead cut off Euclidean momentum (as is usually done) and continue the results back to Lorentzian, we obtain finite Lorentz-invariant results. Is that right? $\endgroup$ Mar 15, 2021 at 18:50
  • $\begingroup$ Well, one usually does not analytically continue from Euclidean to Minkowski before removing the cutoff. I am not even sure one can do this analytic continuation when the cutoff is still there, especially in Fourier, because it destroys unitarity, i.e., Osterwalder-Schrader positivity. $\endgroup$ Mar 15, 2021 at 18:55
  • $\begingroup$ @Abdelmalek, when working perturbatively, it seems straightforward to analytically continue Euclidean momentum-space correlations to Lorentzian signature. On the other hand, are you suggesting that the non-perturbative Euclidean-cutoff results (defined through a rigorous Euclidean-cutoff functional integral as you referred to below) might not be analytic, so that we could not continue them before removing the cutoff? $\endgroup$ Mar 15, 2021 at 20:15
  • $\begingroup$ Yes, but don't take it to the bank. It's just a quick answer. Nonperturbative analytic continuation is quite subtle. In the usual OS reconstruction procedure, you use OS positivity to construct a Hilbert space. Then you translate and get a contraction semigroup which then gives a self-adjoint Hamiltonian $H$. Then analytic continuation involves replacing $e^{-tH}$ by $e^{-itH}$. With a rotation invariant Fourier cutoff, you lose the very first ingredient: OS positivity. $\endgroup$ Mar 15, 2021 at 20:34

2 Answers 2


The idea that no-one seems to explicitly spell out is that a hard cut-off in momentum space discretizes position space (the Fourier transform of functions on a finite interval $[0,\Lambda]$ lives on the discrete lattice of modes on that interval), and a discrete lattice in space is evidently not invariant under continuous spatial rotations, so both the Euclidean and the Minkowskian theories are no longer Euclidean/Lorentz-invariant "in spirit", even if the correlation functions in momentum space might be.

Note that this is also a very questionable argument - Fourier transforms only work so neatly in 1 dimension, and in general the interior of the unit sphere is not an Abelian group like the interval (if you map additions "over the edge" back from the start, i.e. make it periodic), so there exists no natural notion of Fourier transform on it.

  • $\begingroup$ Thanks for the point on lattices. I updated my question to explain why I still don't think cutoffs break Euclidean/Lorentz invariance, neither in their final answers nor even in spirit, though the "in spirit" part may be subjective. $\endgroup$ Mar 14, 2021 at 21:43

I think I agree with your analysis. Perhaps the best way to understand what's going on is to distinguish between two "philosophical" interpretations of cutoff regularization:

  • Cutoff as a formal/unphysical regulator. We define the Feynman integral as the analytical continuation of the regulated Euclidean integral. This is a well-defined prescription that, indeed, is finite and Lorentz invariant. But it has no physical meaning, much like other perturbative regulators, such as dimensional regularization or Pauli-Villars. It is just a formal tool, defined perturbatively.

  • Cutoff as a physical regulator. Here we must indeed follow ACM's adivice and obtain the cutoff integrals by discretizing spacetime. This is a physical, non-perturbative definition. We define the full theory on a lattice; if we expand it in Feynman diagrams we observe that the momentum integrals are effectively cutoff into the Brillouin zone. But we can also study the theory non-perturbatively. Note that this is not true in the previous bullet point, because the definition is only given at the level of Feynman diagrams.

Note that the "physical" point of view is what appears in, for example, Wilsonian renormalization, where the high momentum modes are interpreted as unobservable short distance physics. Or in any other situation where we give an interpretation to the regulator, such as in effective field theory. The "unphysical" point of view has nothing to say about these topics, in the same way you would not use dimensional regularization to argue that physics at different scales decouple, etc.

The "physical" cutoff is a modification of the theory; the "unphysical" one is a tool/prescription for how to compute the loop expansion. The former yields a well-defined, complete theory; but this theory is not Poincaré invariant. The latter does not define a full theory, but only its perturbative expansion; but this expansion is perfectly Poincaré invariant.

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    $\begingroup$ I may have misunderstood but this sounds like: Fourier cutoff is purely perturbative while lattice cutoffs are nonperturbative. In the Euclidean situation, the functional integral is as rigorously and nonperturbatively defined with the Fourier cutoff as it is with the lattice cutoff. $\endgroup$ Mar 15, 2021 at 18:38
  • $\begingroup$ When you refer to the physical view, do you imagine a spacetime discretization of the path integral in Lorentzian signature? If so, I'm not sure how results in that setup would compare to those we obtain from the (analytic continuation of) results obtained with a Euclidean momentum cutoff. $\endgroup$ Mar 15, 2021 at 19:12
  • $\begingroup$ Abdelmalek, how does one nonperturbatively define a Euclidean functional integral with momentum cutoff? It would seem you also need an additional regulator corresponding to a spatial volume-cutoff (IR cutoff), or a momentum discretization, and such a regulator might break Euclidean invariance. (I checked your nice ref. arxiv.org/pdf/1311.4897.pdf but it seems to skirt the infinite-volume issue.) $\endgroup$ Mar 15, 2021 at 19:39
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    $\begingroup$ That is true. You need to confine the integral in the exponential to a finite volume. To connect to what I was saying, one indeed needs to take the limit of infinite volume with fixed momentum cutoff. However, while technical, this is easier to do than removing a UV cutoff. For the $\phi^4$ model at weak coupling, the infinite volume limit can be done using cluster expansion techniques. For large coupling, I haven't seen it done with Fourier cutoffs, but it is OK with lattice cutoffs, although still technical. $\endgroup$ Mar 15, 2021 at 19:59

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