Usually, it is heuristic to say that we can understand a QFT with a momentum cutoff $|k|<\Lambda$ by imagining that the system is living on a lattice. I would like to ask:

(1) Is there any mathematical proof saying that they are equivalent? Or, is there any proof saying that the beta functions of these two different regularizations are the same to certain order? The story is probably trivial for free theory. How about for interacting theory? I would imagine that different regularizations can give different results.

(2) When people say a certain lattice quantum field theory has a continuum limit, they mean they can put the lattice spacing to zero and choose a certain lattice bare coupling constant to make the theory well-defined. Does the continuum limit with lattice spacing zero have anything to do with a "continuous" quantum field theory with cutoff in the momentum space?


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    $\begingroup$ 1) They are not identical, if that is what you're asking. For one thing, cutoff regularisation corresponds to truncated integrals, and lattice regularisation to sums. They may (or may not) approach the same limit as you remove the regulator; but for finite regulator, the two theories are not identical. $\endgroup$ Aug 20, 2018 at 3:20
  • $\begingroup$ Thanks! Usually, in condensed matter community, when we have a lattice system, we will write down a QFT with a momentum cutoff to understand the system. For example, in Chapter 20 of Subir Suchdev's textbook "Quantum Phase Transition", he wrote down a Sine-Gordon field theory with a momentum cutoff to understand the phase transition between valence bond solid order and Neel order for spin systems on a lattice. Probably this has to be justified somehow. Do you agree with that? $\endgroup$
    – Herman Chu
    Aug 20, 2018 at 3:28
  • $\begingroup$ In addition, do you know: Where is the boundary of human's knowledge that we know when the two different regularizations agree and disagree? $\endgroup$
    – Herman Chu
    Aug 20, 2018 at 3:34

2 Answers 2


I generally tell a similar heuristic story when I present lattice QFT research to audiences who may not be familiar with lattice regularization, but probably have encountered hard momentum cutoffs in (for example) an introductory QFT course. I make the following points as a way to help such folks get oriented:

  1. The lattice discretization of space-time does introduce a UV cutoff scale proportional to the inverse lattice spacing, $\Lambda \propto 1/a$. (Careful calculations might have different conventions about factors of $2\pi$ and the like in that proportionality relation, and I'm not concerned about that.)
  2. The continuum limit $a \to 0$ therefore corresponds to the usual removal of the cutoff $\Lambda \to \infty$ seen in introductory textbook renormalization.
  3. However, the lattice cutoff is "smarter than the average cutoff" given a discretization of space-time that preserves a sufficiently large discrete subgroup of the $d$-dimensional SO($d$) rotation group (i.e., Lorentz symmetry Wick-rotated to euclidean signature). Then the recovery of the full (Wick-rotated) Poincaré invariance is guaranteed in the continuum limit with no fine-tuning. The usual hypercubic lattice is sufficiently regular for this purpose; arXiv:0804.1145 provides an example of a lattice that is not, and effectively requires fine-tuning the speed of light in order to reach the continuum limit.

Now, given that setup, I would respond to the questions as follows:

  • Before the removal of the cutoff, the two regularizations (hypercubic lattice vs. hard momentmum cutoff) preserve different symmetries and so cannot be equivalent.
  • After the removal of the cutoff, results (in the same renormalization scheme) obtained using either regularization must agree. That is, the continuum limit of the lattice regularization is the same as the "continuous" QFT defined via any other regularization scheme you care to apply.

Additions to address the follow-up questions in the comments below:

The lattice spacing and hard momentum cutoff discussed above are unphysical regulators introduced in order to carry out QFT calculations. Any such unphysical regulator must be removed in order for those calculations to produce physical results. Another example is carrying out dimensionally regularized calculations in $4 - \epsilon$ dimensions, which requires removing $\epsilon \to 0$. A nonzero lattice spacing and a finite momentum cutoff are every bit as unphysical as pretending we live in 3.999 dimensions.

'Non-renormalizable' theories, in contrast, are better to interpret as effective field theories (EFTs) that remain valid only up to some physical "breakdown scale". Many folks have a habit of also calling this a "cutoff" and relying on the EFT vs. QFT context to show whether they're referring to a physical breakdown scale vs. an unphysical regulator. I will use distinct terminology to make this more clear.

So, since here we're considering the case of unphysical QFT regulators, my answer to follow-up question (1) is indeed that we only expect two regularizations to produce the same physical predictions when both of their unphysical regulators are removed. (The "continuum limit" is precisely the removal of the regulator in a lattice calculation.) Moreover, (2) once those unphysical regulators are removed we are left with the same (renormalized) QFT: If two calculations produce different predictions for any (in-principle) observable physical quantity, then at least one of those calculations must be incorrect. Even unobservable, renormalization-scheme-dependent quantities such as beta functions must agree if the same renormalization conditions are imposed on the two calculations.

  • $\begingroup$ However, the removal of the cutoff cannot always be done, right? For "continuous" QFT, it is only true for renormalizable theory. $\endgroup$
    – Herman Chu
    Aug 28, 2018 at 0:23
  • $\begingroup$ Therefore, (1) are you suggesting that we only expect that two regularizations agree when both of them have their own continuum limits? (2) How do you reach the conclusion "After the removal of the cutoff, results (in the same renormalization scheme) obtained using either regularization must agree. "? Perhaps, you are using the point 3 that you mentioned? $\endgroup$
    – Herman Chu
    Aug 28, 2018 at 0:29
  • $\begingroup$ Since I am a condensed matter physicist, what I am interested in are theories effective upto some energy cutoff. Therefore, the EFT could be nonrenormalizable. I am wondering if the momentum cutoff and the lattice regularization result will agree if the theory is nonrenormalizable. Here the cutoff is reasonable and not unphysical. $\endgroup$
    – Herman Chu
    Aug 30, 2018 at 2:24
  • $\begingroup$ So it sounds like you're mixing up the two different uses of the word "cutoff" that I warned about in my edit. Certainly your physical scale has nothing to do with any (unphysical) regularization; these are different concepts. Of course, if you have a physical lattice (e.g., of atoms in a material), its lattice spacings correspond to momentum scales, so you can swap one for the other in the mathematics if you so desire. However, these scales are the regime where your EFT breaks down, so they shouldn't be popping up in any sensible calculations in the first place. $\endgroup$ Aug 30, 2018 at 13:22

I don't know if people who studied the question managed to concoct models where the two regularizations produce different continuum QFTs. I think that for most models of interest this pathology is not expected to occur. In perturbation theory one is dealing with finite dimensional integrals (for suitably renormalized Feynman diagrams) and there it should not be hard to prove the equivalence. As far as mathematical proofs which hold nonperturbatively there is not a lot available simply because there are rather few examples of continuum QFTs constructed, with whichever regularization. Nevertheless, in the case of scalar theories in 2d and 3d the equivalence can be proved by showing that correlation functions are the Borel sums of their perturbative series. See the two articles:

Note that following the work Hairer there has been new alternate constructions of these models with the stochastic quantization approach. See in particular this article and also that one by Mourrat and Weber, as well as this one by Barashkov and Gubinelli. I don't know for sure but it could be that the equivalence between lattice and Fourier cut-offs can be shown with this approach too. One simplifying feature of the SPDE approach is that the random solution of the relevant SPDE is a random variable on a fixed probability space (a functional of the underlying driving noise). That helps showing independence with respect to the choice of mollifier (the function used to put a Fourier cut-off).

  • $\begingroup$ "In perturbation theory one is dealing with finite dimensional integrals (for suitably renormalized Feynman diagrams) and there it should not be hard to prove the equivalence." Could you outline how to prove the equivalence? I only know one theorem in Weinberg QFT II pg. 138 stating that for dimenionsless coupling constant, different ways to do RG are equivalent unto two loops provided that RGEs are autonomous in the sense of ODE. $\endgroup$
    – Herman Chu
    Aug 28, 2018 at 1:21
  • $\begingroup$ Take a convergent 6 point function graph like a triangle with two external lines at each vertex in 4d scalar Euclidean QFT. It's a well defined integral which is the limit of Riemann sums (lattice cut-off) or similar integrals where propagators $1/(p^2+m^2)$ get multiplied by $\rho(p^2/\Lambda^2)$ with the Fourier cutoff function $rho$ being 1near zero and of fast decay. You get the same result because these are two ways of approximating the same absolutely convergent integral. For renormalized graphs something similar happens but is more complicated... $\endgroup$ Aug 29, 2018 at 21:27
  • $\begingroup$ ..since this uses Zimmermann's forest formula projecteuclid.org/euclid.cmp/1103841945 $\endgroup$ Aug 29, 2018 at 21:36
  • $\begingroup$ The thing you mentioned about the RG beta function being independent of the choice of $rho$ is a completely different matter. It boils down to $\int (f(x+b)-f(x)) dx $ over the real line being $ba$ independently of $f$ an increasing function with say $\lim_{x\rightarrow -\infty}f(x)=0$ and $\lim_{x\rightarrow \infty}f(x)=a$. Here $x$ is the log of the momentum in radial coordinates. $\endgroup$ Aug 29, 2018 at 21:39

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