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Could you please tell me how Hilbert spaces are geometrically linked with our spacetime? Both host functions of $x$, $t$, $m$, ... and there must be a mathematical link between them? A drawing would be welcome.

Second question, related: Is there a space that englobes both?

Qmechanic
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Florent Dieterlen
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    You are essentially asking us to formulate a quantum theory of gravity, which does not exist yet ;) There's LQG which is highly speculative physically, but it is an interesting mathematical model of quantum geometry. See e.g. https://physics.stackexchange.com/q/535028/30833 – Prof. Legolasov Jul 19 '20 at 13:56

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There is no link. The dimension of the Hilbert space is determined by the number of possible outcomes of experiments measuring commuting observables.

Thus, for a single spin-1/2 system, where the number of outcomes is $2$, the dimension of the Hilbert space is $2$ and it is known one cannot expresse the spin angular momentum operators in terms of spatial coordinates.

On the other hand, for a particle trapped in a 1d harmonic oscillator, the Hilbert space is infinite dimensional.

ZeroTheHero
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  • Well, a theory of quantum gravity would have some sort of a link. I thought that was what OP was asking about. But also +1. – Prof. Legolasov Jul 19 '20 at 13:59
  • @Prof.Legolasov I see no reference to gravity in the question. – ZeroTheHero Jul 19 '20 at 14:00
  • spacetime = gravity as per General Relativity – Prof. Legolasov Jul 19 '20 at 14:00
  • @Prof.Legolasov If your reading is correct then surely someone will supply an answer that supersedes mine. – ZeroTheHero Jul 19 '20 at 14:01
  • What is the significance that "it is known one cannot expresse the spin angular momentum operators in terms of spatial coordinates."? What changes if we can express operators this way? – Charlie Jul 19 '20 at 14:25
  • @Charlie this is an example of observable that cannot be expressed in terms of the usual space and time coordinates. The number of outcomes of measuring spin and therefore the dimension of the Hilbert space is consequently independent from space and time coordinates, at least for the spin observable. – ZeroTheHero Jul 19 '20 at 14:47
  • @ZeroTheHero Is this related to the fact that the space of spin states are treated separately to the state space that includes position, energy, momentum observables etc.? And the overall description of the particles lives in a $\mathcal H \otimes \mathcal H_{spin}$ kind of contrived space? – Charlie Jul 19 '20 at 15:01
  • @Charlie if ${\cal H}$ is the Hilbert space of angular momentum states $\vert \ell,m\rangle$, then tensor product decomposes into a direct sum and the spin degree of freedom is no longer in a “separate” Hilbert space. – ZeroTheHero Jul 19 '20 at 16:21
  • @ZeroTheHero Ok thank you :) – Charlie Jul 19 '20 at 16:43
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We only know how to formulate quantum mechanics in flat (Minkowski) spacetime. For this, ignoring spin, Hilbert space uses a basis of position states on a slice of constant time.

Charles Francis
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