I'm a mathematician slowly trying to learn quantum field theory and I have a small question about renormalization, which I still have a shaky understanding of.

One common way to explain what's happening in renormalization is to pick some scheme for regularizing the apparently-infinite quantities that show up in the "naive" version of the computation. There seem to be a lot of different ways to do this regularization, but there's one scheme in particular that I've seen a lot of texts allude to in a sort of offhand way without much detail: replacing space with a finite, discrete lattice, but leaving time continuous and infinite.

I think I understand the reasons one might not want to do this: it destroys the Lorentz symmetry, probably along with whatever gauge symmetries there might have been. But it also seems like it has one advantage from a pedagogical perspective: you can express all of the quantities in the regularized theory in terms of an actual quantum-mechanical system with operators on a Hilbert space and Hamiltonians and so on.

For this reason, I've been interested to see if this can be carried out. Specifically, I suppose my questions are:

  • Does this work (that is, are all the regularized integrals associated with Feynman diagrams actually finite in this scheme) or is there some problem I'm not seeing?
  • If so, do you know of a reference that works this out carefully, say for something simple like $\phi^4$ theory?

(For what it's worth, I am aware that from the path-integral perspective, it's common to discretize both space and time in a similar way, and I've even found a nice source for that perspective in Salmhofer's book "Renormalization". But for the purposes of this question I'm specifically interested in the Hamiltonian perspective, which means I want to leave the time variable intact.)


1 Answer 1


Does this work (that is, are all the regularized integrals associated with Feynman diagrams actually finite in this scheme) or is there some problem I'm not seeing?

At least perturbatively (which is the only time you'll be computing Feynman diagrams) a spatial regulator does indeed work. I like to think about this using a smooth regulator a la Polchinski, but the details of your regulating scheme shouldn't matter. First let me introduce the "Polchinski" regulating scheme with a spatial regulator (in Polchinski's original work, he uses a regulator which regulates in both space and time). Consider the kinetic term of a free massless scalar field: $$\int dt d^{d-1}\vec{x} \left[\partial_\mu \phi \partial^\mu \phi\right] = \int dt d^{d-1}\vec{p} \left[\phi \partial_t^2\phi + \vec{p}\,^2 \phi^2\right]$$ Now let $K$ be some smooth function with the property $K(0) = 1$, $K(s) \approx 1$ for $s< 1$ and $K(s) \approx 0$ for $s > 1$. That is, $K$ is essentially some smooth approximation of a particular step function which is zero on the region $s > 1$. Now, we modify the kinetic term to be of the form: $$\int dt d^{d-1}\vec{p}\, \frac{1}{K(\vec{p}\,^2/\Lambda^2)}\left[\phi \partial_t^2 \phi + \vec{p}\,^2 \phi^2\right].$$ The presence of $K^{-1}$ causes the kinetic energy to become extremely large for any field configuration with support on momentum modes $\vec{p}$ whose magnitude is greater than some cutoff scale $\Lambda$. In the path integral formulation, this leads to a supression of all such field configurations, effectively limiting the theory to modes below the cutoff.

The reason I like this particular setup is that the implication for perturbation theory is quite clear. All we have done is modify the propagator to now take the form $$G_\Lambda(x, x') = \int d\omega d^{d-1}\vec{p}\, \frac{K(\vec{p}\,^2/\Lambda^2)}{\omega^2 - \vec{p}\,^2}e^{-i\omega(t-t') + i\vec{p}\cdot (\vec{x} - \vec{x}')}$$

When integrating over the spatial momenta, the regulating function simply kills the integral, which indeed removes any UV divergences associated to such large spatial momenta. However, from this expression we can also deduce that the unregulated $\omega$ integral is no issue. In particular, the only function of the $\omega$ integral is to pick up some pole from a given contour prescription. Using the time-ordered prescription, the integral always either picks up a pole from $\omega = \pm \sqrt{\vec{p}\,^2}$, i.e. the energy is always "on shell." So, as long as the spatial momenta is regulated, the on shell energy is also always regulated.

Now although we have argued this from a perturbative standpoint, it is reasonable to conjecture that the spatial regulator is sufficient for any Lorentz invariant QFT. The reasoning goes that energies are (on-shell) always directly related to spatial momenta via a Lorentz invariant dispersion relation, and as such, it is sufficient to only regulate in the spatial directions. Such an argument is not precise of course, and it very well may be the case that there are counter examples, although I am not aware of any concrete ones.

If so, do you know of a reference that works this out carefully, say for something simple like $𝜙^4$ theory?

Personally I don't have any explicit examples in mind. The argument I just gave above comes from this paper which only considers free fields: https://arxiv.org/abs/1609.03493.

However, it is common both in computational studies of QFT and in condensed matter contexts to think of QFTs as being on a lattice only in the spatial directions, and so I believe this is a fairly standard practice outside of high energy physics. I don't have anything concrete to say though, perhaps someone else in the answers or comments can provide some insight.

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    $\begingroup$ Thanks for this answer! That is indeed a nice setup, and I agree that it does a lot of what I want. (I also appreciate the confirmation that this idea isn't crazy.) I'm definitely still interested to see if anyone knows a source that works out the spatial lattice version of the story for a simple interacting theory. $\endgroup$ Commented Mar 17 at 20:12

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