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Vertex-models and nearest-neighbor models with translation-invariant local energy functions on an infinite lattice have the troublesome feature that the sum of the local energies diverges. A standard remedy is to work on a finite lattice (e.g., imposing periodic boundary conditions). A different approach would be to multiply each local energy term by a coefficient like $e^{-|{\bf x}|^2/s^2}$ where $x$ is the location of the vertex or bond in question. (Note that for each fixed $s$ the multiplier is summable over $x$, but for each fixed $x$ the multiplier goes to 1 as the spread-parameter $s$ goes to infinity.) Given two configurations, each having infinite energy in the naive sense, one could define the energy difference between them as the limit of the difference between their $s$-regularized energies as $s \rightarrow \infty$.

Has this regularization method been successfully applied to any lattice models? Where can I read about it? (See Vorticial ground states for the O(2) rotor model for an example of a model to which this kind of regularization might be applied.)

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    $\begingroup$ Barry McCoy mentioned techniques similar to this in the Ising model context in a grad Special Topics seminar I took at SUSB in 1976 or 1977. However, we were looking for analytic solutions, and choosing a regularizer that breaks the invariance of the lagrangian under actions of the lattice translation group makes getting analytic solutions (a la Onsager's original technique) too difficult. At the time, I did not have a good grasp of renormalization group theory, so I can't say whether that notion was applied or remember if it was even mentioned. $\endgroup$
    – Mark Fischler
    Commented Oct 7, 2016 at 18:29
  • $\begingroup$ not sure I understand what "energy" means here. For the Ising model with weight $exp(\beta\sum_{x,y}J_{x,y}\sigma_x\sigma_y)$ do you mean what goes in the exponential for fixed spin configuration or do you mean expectations $\langle \sigma_{x}\sigma_{x+e}\rangle$ related to the energy field, eventually with a sum over $x$? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 13, 2016 at 16:08
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    $\begingroup$ Yes, by "energy" I mean the quantity in the exponent. (Sorry not to have responded sooner; for some reason Stack Exchange hasn't been notifying me of updates for the past month.) $\endgroup$
    – James Propp
    Commented Dec 4, 2016 at 16:08
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    $\begingroup$ usually people put the system in a box and thus set a sharp volume cut-off. What you are proposing is a bit like a smooth volume cut off with multiplication by a positive weight. So Griffiths inequalities apply and one would recover the usual Ising model when $s\rightarrow\infty$. I didn't think too hard about it but I believe that, a.s., for two independent Ising configurations in infinite volume the $s$-limits of energy differences should be infinite or something ill defined like an infinite sum of plus or minus 1's. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Dec 6, 2016 at 18:39

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As far as I know, renormalization in Physics works just the way it does in analytic number theory. Our problematic series:

$$ \sum_{n=1}^\infty a_n = \infty $$

can be regularized by adding only finitely many terms and seeing how the numbers increase when you add more stuff:

$$ \sum_{n=0}^\infty a_n e^{-ny} \asymp \frac{1}{y} \longrightarrow \sum_{n=0}^N a_n \asymp N $$

This is the Hardy-Littlewood Tauberian theorem.

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