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:
- 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.)
- The continuum limit $a \to 0$ therefore corresponds to the usual removal of the cutoff $\Lambda \to \infty$ seen in introductory textbook renormalization.
- 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.