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Observables vs. continuum and metric?

  1. Apr 8, 2015 #1
    Space in quantum mechanics seems to be modeled as a triplet of real numbers, i.e. a continuum. Same happens in special relativity. General relativity I do not know (nor field theories). And then we apply the Pythagorean theorem and triangle inequality and so forth...

    I have a few general questions:

    1) Is space (3d or Minkowski) the only continuum?

    If a quantum mechanical operator describing an observable has a continuous spectrum (partially at least),
    i.e. the observable is allowed to take, in principle, real values,
    then is this always an end result of the assumption that space is continuous?
    Or are there seemingly continuous observables that are not related to space?

    An example: Bound electrons in a hydrogen atom have discrete spectra, both theoretically and experimentally, but free ones have seemingly continuous, and the Coulomb law plays a role there theoretically, and the fact that space would be a continuum.

    2) Is it so that based on experiments, space in QM can always be assumed to have locally euclidean metric?

    Thanks
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2

    bhobba

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    Science Advisor
    Gold Member

    Of course not. But I am not aware of any models not using it that has been successful.

    This is tied up with what mechanics is. These days its defined by what transformation you use between reference frames which are assumed to contain a Cartesian 3 dimension coordinate system.

    To fully understand this view I suggest two books:

    First is Landau - Classical Mechanics
    Second is Ballentine - Quantum Mechanics - A Modern Development.

    The symmetry of inertial frames is the key, and those frames by definition have Cartesian coordinate systems with continuous real values.

    Thanks
    Bill
     
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