Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Hilbert space in Everettian QM

  1. Sep 13, 2017 #1
    Is it assumed that Hilbert space is an infinite manifold that the non-collapsing wave function occupies in Everettian QM?

    Thank you.
     
  2. jcsd
  3. Sep 13, 2017 #2
    Sorry, I think I phrased the question wrong. I was concerned with whether Hilbert space is infinite dimensional in Everettian QM? Don't know if that is different than the OP question, but at least more clear.
     
  4. Sep 13, 2017 #3
    Not to self-bump, but if the assumption that the wavefunction occupies a state space in infinite dimensional Hilbert space, then doesn't that smell of metaphysics?
     
  5. Sep 14, 2017 #4

    Demystifier

    User Avatar
    Science Advisor

    Wave function lives in an infinite dimensional Hilbert space in any QM, not just Everettian QM. The question (somewhat metaphysical if you like) is whether this wave function is ontic or epistemic.
     
  6. Sep 14, 2017 #5
    The Hilbert space can have finite or infinite number of dimensions. I think most physicists would concede that any real Quantum system is probably limited to a finite (though excessively large) number of states because of the finite size of the universe and Planck limits.
     
  7. Sep 14, 2017 #6
    So, my next question is that if Hilbert space is infinite dimensional in QM, does that contribute to quantum indeterminacy? Is it pertinent to quantum indeterminacy or just a confounding factor?
     
  8. Sep 14, 2017 #7

    Nugatory

    User Avatar

    Staff: Mentor

    The indeterminacy is inherent in the Born rule, which works similarly no matter how many or few dimensions are in the Hilbert space.
     
  9. Sep 14, 2017 #8
    Can you cite the source for this?

    Cheers
     
  10. Sep 14, 2017 #9

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Indeed.

    That's one reason I like the Rigged Hilbert Space formulation.

    From the start you recognize the states you work with are not physically releasable - its just introduced for mathematical convenience.

    Thanks
    Bill
     
  11. Sep 15, 2017 #10
    Doesn't that imply metaphysics or some variety of Platonism for or given the existence of the wavefunction in infinite dimensional Hilbert space?
     
  12. Sep 15, 2017 #11

    Nugatory

    User Avatar

    Staff: Mentor

    No. It's just math, and math routinely works with things that aren't physically realizable.
     
  13. Sep 15, 2017 #12

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Physics is a mathematical model. In applied math in general, not just physics, some mathematical assumptions that often have zero impact on anything actually measurable are routinely made. This is just another case. As an example in modeling a hammer strike you often use a Dirac Delta function. Of course the actual strike isn't infinitely high and of zero duration - it's just very large and of such a short duration its fine for most purposes to model it that way.

    Same in QM. You can think of the output of some observation as some kind of digital readout. It can be a very long string of numbers - but not actually infinite. So what you do is a mathematical trick - you take all the row vectors of finite length. These are the physically realizable outcomes. You then take what's called the dual and that's the space you work in in QM. The original space is called a test space. By doing that you can apply the methods of the calculus etc and chose whatever subset of that dual is most convenient. Actually its a bit trickier than that - you chose a subset with nice mathematical properties of that big dual as your test space and take the dual of that - the choice is purely done for technical mathematical convenience.

    It is often said in physics its best to avoid words like reality etc in discussing physics. If pushed to it I define reality as what our theories describe. The above shows its more subtle than that glib remark - but the whole issue is a bit of a morass best avoided - philosophers argue about it all the time - the problem is they get nowhere but physics actually makes progress.

    Thanks
    Bill
     
  14. Sep 15, 2017 #13
    The Everett interpretation doesn't make any a priori assumptions about the dimensionality of the Hilbert space. The number of dimensions will depend on the specific model one is using to describe the universe. For example, if one supposes that we live in a de Sitter universe (and that the big bang is therefore just the most recent low entropy fluctuation), then chances are that the associated Hilbert space is finite dimensional, since a de Sitter universe has a finite maximum entropy (proportional to the surface area of the cosmic horizon).
     
  15. Sep 17, 2017 #14
    My friend, who is a physicist, had this to say to me not long ago, when I confronted him on Hilbert space ~

    ''If the representation is to be unitary, it is infinite-dimensional by the compactness of Lorentz group. These representations exist. In this case sigma matrices Sigma_munu defined as their commutators would be hermitian matrices and the formula would make sense as quantum expectation value. It would represent the curvature tensor as quantum operator. The problem with infinite-D representations of Lorentz group is that they are not encountered in particle physics, where representations are finite-D (spinors, vectors, tensors) and these representations do not allow Hilbert spzce interpretation. Poincare invariance saves the situation: one can have instead of finite-D representations of Lorentz group representations of entire Poincare group by quantum fields associated with finite-D representations of Lorentz group and everything is OK. But the problem with GRT is of course that the Poincare invariance is lost.''


    and he further spoke about the subject


    ''The infinite-D representation follows only the condition that Lorentz transformation are represented unitarily. This guarantees the curvature tensor is Hermitian operator. It is essential that one has only Lorentz invariance. If one has Poincare invariance, finite-D representations which indeed correspond to physical particles, are enough. And the problem is that GRT does not allow Poincare invariance except as approximate symmetry.''


    Which does have some interesting information in it.
     
  16. Sep 17, 2017 #15

    tom.stoer

    User Avatar
    Science Advisor

    It doesn't imply Platonism, but Everett's idea makes sense or is the result of a platonist view of physics.

    For an instrumentalist there is no reason to worry about collapse or no collapse, what a collapse means or what the wave function "really describes"; all she cares is if the wave function can be used to calculate possible outcomes, probabilities and expectation values for measurements. For a Platonist - who understands the mathematical formalism as something that encodes reality behind mere observations, i.e. something that really exists out there, for somebody who thinks beyond empirism - the collapse contradicts the unitary time evolution and cannot be real in the same sense as this time evolution. Therefore she will avoid the collapse at all cost, including the acceptance of the reality of the branch structure of the wave function predicted by unitary quantum dynamics, especially due to decoherence.

    Many proponents of the Everett interpretation seem to be Platonists; some do explain this in detail, e.g. Deutsch.
     
    Last edited: Sep 18, 2017
  17. Sep 18, 2017 #16

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    He meant "non-compactness" of the Lorentz group (I'd rather say Poincare group since the spatial translations also play a crucial role).
     
  18. Sep 18, 2017 #17
    I'm a big fan of Max Tegmark and Deutsch. I see the instrumentality of using mathematics in physics as, as close a proof of Platonism as one can get.

    If we take something like the Church-Turing-Deutsch principle, then there's little way to prove it to be true ad hoc. Then, there are some hard limits that contribute to such a situation, like Godel's incompleteness theorem, which is just a personal opinion although I'm interested in what you think about the CTD principle and what kind of relation it may or may not have with Godel's incompleteness theorems.
     
  19. Sep 19, 2017 #18
    Could you please provide links to the reference you are alluding to in your post. Thank you.
     
  20. Sep 19, 2017 #19
    Well, Ii went straight to the source and e-mailed David Deutsch and sent a Facebook message (actually, I think I'll just e-mail him too) to Max Tegmark regarding Godel and the CTD principle, and if the CTD principle is a measure of the truth of Platonism in the real world, and if there are any ways to prove that principle.

    If anyone is interested I can share their responses here (with their consent).

    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Hilbert space in Everettian QM
  1. Hilbert spaces in QM (Replies: 2)

  2. Hilbert Spaces & QM (Replies: 1)

  3. Hilbert Space (Replies: 18)

Loading...