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The difference between a Hilbert space and an inner product space is that a Hilbert space is complete, i.e. has an uncountable number of points. Its all fine a

The difference between a Hilbert space and an inner product space is that a Hilbert space is complete, i.e. has an uncountable number of points. Its all fine and dandy to say that in the history of the universe we can only come up with a finite number of numbers or even finite number of mathematical concepts, but all the current theories of QFT are on top of Hilbert spaces.

lisper 2y ago
OK, but that is a consequence of the continuity of space, right? So you don't even need to go to QM. Basic Newtonian mechanics is all you need to get an uncountable continuum. But space might turn out to be quantized. In any case, the answer to your original question is obviously "no". Why do you think this matters?
- lanstin 2y ago
  Weirdly it's not related to regular space-time that we seem to inhabit - the Hilbert space in which QM happens does not correspond to the space of special relativity. Space and time of specific things are both the result of applying operators to the Hilbert space. The Hilbert space completeness means you have nice properties as far as orthonormal bases and that sort of thing (venturing outside my knowledge of QM here). The dimensionality of the Hilbert space is related to how many distinct quantum states can exist; it is unknown to physics if the universe has a finite dimensional Hilbert space or infinite. Since starting this conversation, I have tried to find any example of a quantum theory based on something other than the mathematics of the continuum without luck (but I'm not well positioned to find such a thing, as I would guess it would come from some constructivist mathematics thing but even regular ZFC QFT isn't formalized yet).
  - lisper 2y ago
    I'm at the hairy edges of my understanding here as well, but AFAIK infinite-dimensional Hilbert spaces arise when modeling position-momentum states because it is assumed that space is continuous.
  - btilly 2y ago
    Well, distributions like the Dirac delta qualify. But they can also be viewed as approximations of a continuum. Outside of that, maybe if you look somewhere like string theory? Still, in general, the continuum is enough. And while it takes work, I'm pretty sure that nothing relevant to physics about the continuum cannot be said within a strictly Constructivist version of the continuum. So no, there is no physical theory from which we can justify the existence of more reals than rationals. The justification must be philosophical. And the answers that we accept will strongly depend on what our notions about what existence should mean.
btilly 2y ago
Complete does not mean uncountable. Complete means that any sequence that looks like it is converging, actually is converging. Related to Hilbert spaces, consider the set of analytic functions with period 1. Under the L2 norm, this is an inner product space. It is also uncountable. However as Fourier showed, the Fourier series for the square wave does not converge within the set of analytic functions. In fact it is trying to converge to a function that not only fails to be analytic, it fails to be continuous!

lisper 2y ago

OK, but that is a consequence of the continuity of space, right? So you don't even need to go to QM. Basic Newtonian mechanics is all you need to get an uncountable continuum. But space might turn out to be quantized. In any case, the answer to your original question is obviously "no". Why do you think this matters?

lanstin 2y ago
Weirdly it's not related to regular space-time that we seem to inhabit - the Hilbert space in which QM happens does not correspond to the space of special relativity. Space and time of specific things are both the result of applying operators to the Hilbert space. The Hilbert space completeness means you have nice properties as far as orthonormal bases and that sort of thing (venturing outside my knowledge of QM here). The dimensionality of the Hilbert space is related to how many distinct quantum states can exist; it is unknown to physics if the universe has a finite dimensional Hilbert space or infinite. Since starting this conversation, I have tried to find any example of a quantum theory based on something other than the mathematics of the continuum without luck (but I'm not well positioned to find such a thing, as I would guess it would come from some constructivist mathematics thing but even regular ZFC QFT isn't formalized yet).
- lisper 2y ago
  I'm at the hairy edges of my understanding here as well, but AFAIK infinite-dimensional Hilbert spaces arise when modeling position-momentum states because it is assumed that space is continuous.
- btilly 2y ago
  Well, distributions like the Dirac delta qualify. But they can also be viewed as approximations of a continuum. Outside of that, maybe if you look somewhere like string theory? Still, in general, the continuum is enough. And while it takes work, I'm pretty sure that nothing relevant to physics about the continuum cannot be said within a strictly Constructivist version of the continuum. So no, there is no physical theory from which we can justify the existence of more reals than rationals. The justification must be philosophical. And the answers that we accept will strongly depend on what our notions about what existence should mean.

btilly 2y ago

Complete does not mean uncountable. Complete means that any sequence that looks like it is converging, actually is converging. Related to Hilbert spaces, consider the set of analytic functions with period 1. Under the L2 norm, this is an inner product space. It is also uncountable. However as Fourier showed, the Fourier series for the square wave does not converge within the set of analytic functions. In fact it is trying to converge to a function that not only fails to be analytic, it fails to be continuous!