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It seems to me that there are fairly good reasons to assume that quantum theories need to rely in their formulation on infinite-dimensional spaces (cf. Why do we need infinite-dimensional Hilbert spaces in physics?, Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics?). @Arnold Neumaier wrote in another thread:

With a finite-dimensional Hilbert space, the whole apparatus of practical QM is lost. Very little is left - no continuous spectra, no scattering theory, no S-matrix, no cross sections. No Dirac equation, no relativity theory, no relation between symmetry and conservation laws, no quantum fields.

It appears, however, that in some contexts, e.g. quantum gravity, people have been making claims that `the Hilbert space of quantum gravity in asymptotically de Sitter space time has a finite dimension $N$'. Here's Ed Witten:

We discuss some general properties of quantum gravity in de Sitter space. It has been argued that the Hilbert space is of finite dimension. (http://cds.cern.ch/record/504347/files/0106109.pdf)

How is it possible to reconcile those two statements? Are the respective contexts vastly different in some sense?

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  • $\begingroup$ This may just be linked to the discreteness of space - if you have a discrete spacetime with de Sitter metric (which is closed), there will only be a finite numbers of points in space, which may only require a finite number of dimensions for the Hilbert space. $\endgroup$
    – Slereah
    Commented Jul 25, 2018 at 14:42
  • $\begingroup$ The idea, I think, is that the dimension of the Hilbert space corresponds to the entropy of de Sitter space (cf. cds.cern.ch/record/447660/files/0007146.pdf). I don't quite see how this answers my initial questions, though... $\endgroup$
    – skids
    Commented Jul 25, 2018 at 15:01
  • $\begingroup$ I think the point is that there shouldn't be much physical difference between an infinite-dimensional Hilbert space and a finite (but very, very, very large) dimensional Hilbert space. For a position-space analogy, treating space as a continuum would be a good approximation if space were discrete on an extremely, extremely fine-grained scale. This is what we do in fluid mechanics all the time - we make continuum approximations of materials that are made of (a huge number of) discrete particles. $\endgroup$
    – probably_someone
    Commented Jul 25, 2018 at 15:26
  • $\begingroup$ But things do seem to be different in finite-dimensional spaces, no matter how large they are. So for instance, you do not get uncertainty principles, as there are no unbounded operators in finitely many dimensions; in what sense exactly could one approximate uncertainty principles by a single large finite-dimensional Hilbert space? $\endgroup$
    – skids
    Commented Jul 25, 2018 at 15:41
  • $\begingroup$ You definitely get uncertainty principles in finite-dimensional Hilbert spaces. Anytime you have two non-commuting Hermitian operators, the product of the standard deviations of their expectation values is bounded below by the Robertson uncertainty relation (which is just a generalization of the Heisenberg uncertainty principle). For example, in the two-dimensional spin space of a spin-1/2 particle, the spin operators along the $z$ direction and the $x$ direction do not commute with each other, so there is an uncertainty principle relating the spins measured along the $x$ and $z$ directions. $\endgroup$
    – probably_someone
    Commented Jul 25, 2018 at 15:52

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