u/1strategist1

▲ 20 r/Physics

Is the difficulty of constructing QFTs in ≥4 dimensions related to existence of wick powers?

From what I've seen looking at constructive QFT

  • Building QFTs nonperturbatively in 2D isn't too hard.

  • Building them in 3D is quite difficult.

  • Building them in 4+ dimensions is Millenium prize-level. (No interacting theories constructed yet)

I recently also learned that wick powers of a Euclidean scalar field follow a similar pattern.

  • All wick powers exist in 2D.

  • Only :ϕ^(2): is well-defined in 3D.

  • No Wick powers are well-defined in 4+ dimensions.

I know the answer is probably just that the issue is the increasing irregularity of the field in higher dimensions causes both the difficulties in construction, and the failure of Wick power definitions.

However, it does seem weirdly coincidental that the dimension where mass terms in a Hamiltonian can no longer be defined exclusively via wick powers is exactly the same dimension where we run into so much trouble trying to construct QFTs. Is there any relation between those two, or am I seeing connections where there are none?

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u/1strategist1 — 6 days ago
▲ 38 r/math

What happens when you drop countable additivity of a measure for countable additivity on compact sets?

Uniform probability distributions over the real numbers can't be defined within standard measure theory because measures need to be countably additive.

Dropping countable additivity for just finite additivity, you lose a lot of nice properties. Among others, I've heard that integrals end up reducing to just Riemann integrals again.

A different modification you could consider is dropping countable additivity for finite additivity, but maintaining countable additivity whenever all the sets being unioned are contained within some compact set. This should still allow you to define uniform probability measures, but it has more structure than just finite additivity.

Does anyone know of any research or discussions on this topic? What happens to integrals in this context? Presumably integrals over compact sets would be equivalent to regular Lebesgue integrals, but how about over the full space? Do integrable functions still form some nice Banach space?

Does anyone see any obvious issues with this kind of structure, or know of similar structures?

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u/1strategist1 — 8 days ago