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Are Hilbert spaces uniquely defined for a given system?

  1. Nov 30, 2008 #1
    Hi there!

    Repeating the question on the title: Are Hilbert spaces uniquely defined for a given system?

    I started to think about this when I was reading about Schrödinger and Heisenberg pictures/formulations. From my understanding, you can describe a system analyzing the time dependent state defined by [tex]\left\vert \psi \left( t\right) \right\rangle =U\left( t,t_{0}\right)\left\vert \psi \left( t_{0}\right) \right\rangle[/tex], while keeping the observables time-independent (wich would be the Schrödinger picture). And this is totally equivalent to description of the system by a time-independent state, and observables defined by [tex] A^{\prime }\left( t\right) =U^{\dagger }\left( t,t_{0}\right) AU\left(t,t_{0}\right) [/tex]. (Heisenberg picture) However, I don't think it's the same Hilbert space! In the first one, the elements of the space are time dependent, while in the other they're not.

    Am I making any sense? :uhh:
     
  2. jcsd
  3. Nov 30, 2008 #2

    Fredrik

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    The short answer is "yes". The Hilbert space is what defines what physical system we're talking about, so it must be uniquely defined (up to isomorphisms of course) for each physical system.

    The Hilbert space of one-particle states consists of (equivalence classes of) square integrable functions from [itex]\mathbb R^3[/itex] (not [itex]\mathbb R^4[/itex]) into [itex]\mathbb C[/itex]. Your [itex]|\psi(t)\rangle[/itex] is a member of that space, but the map [itex]t\mapsto|\psi(t)\rangle[/itex] (the "time dependent wavefunction") is a curve in that space.
     
  4. Nov 30, 2008 #3
    Right. Thanks Fredrik!
    That clear things up :)
     
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