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Question about wavefunctions and their Hilbert space

  1. May 31, 2014 #1
    Maybe someone here can explain me something I never understood in QM: The wave function lives in the Hilbert space spanned by the measurement operator. Is there any mathematical relation of those spaces with each other?
     
    Last edited by a moderator: May 31, 2014
  2. jcsd
  3. Jun 1, 2014 #2

    vanhees71

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    I don't know what a measurement operator might be.

    The wave function of a single particle in non-relativistic QT is the position representation of the state vector, i.e., it realizes the Hilbert space of states as the space of square-integrable functions, [itex]\mathrm{L}^2[\mathbb{R}^3].[/itex]
    Since all separable Hilbert spaces are equivalent this is just a special realization of the abstract Hilbert space as known from the representation free Dirac formulation of quantum theory.
     
  4. Jun 1, 2014 #3
    For example the energy operator
     
  5. Jun 1, 2014 #4

    bhobba

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    They are the same space, but different basis result from different complete sets of commuting observables.

    Often, but not always, the energy operator by itself forms a complete commuting set of observables eg the hydrogen atom.

    Regardless you can expand them in other observables like momentum or position if you like. By definition a wavefunction is a state expanded in terms position eigenvectors.

    There are all sorts of subtleties associated with the above such as exactly what an eigenfunction of position is in a Hilbert space setting, you really need to go to a Rigged Hilbert space. But that is is basically it.

    Thanks
    Bill
     
  6. Jun 7, 2014 #5
    If they're the same space then it's only a dimension 2 space? How can that be, that a finite dimension space have an infinite number of elements that are orthogonal to each other?
     
  7. Jun 7, 2014 #6

    bhobba

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    Exactly in what context do you think the space has dimension 2?

    Usually the eigenvectors of the energy operator is countably infinite.

    Thanks
    Bill
     
  8. Jun 8, 2014 #7
    I mean the space generated by the eigenvectors of the spin operator (up or down).
     
  9. Jun 8, 2014 #8

    bhobba

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    Sure - but at the same time such objects have position so an infinite dimensional Hilbert space is required for that. You consider it described by the superposition two wave-functions each residing in an infinite dimensional space one for up and one for down. The space of the linear combination of two such functions is itself infinite dimensional. Its just in some experimental set-ups the position is irrelevant so you only need a two dimensional space to analyse it.

    Just as an aside the usual Hilbert space formulation has issues with things like Dirac delta functions and waves of infinite extent. One way out of the difficulty is the Rigged Hilbert Space formalism and in that view you consider the fundamental states for be finite dimensional but of unknowen dimension. However that is a whole new discussion. Start a new thread about it if you are interested.

    Thanks
    Bill
     
    Last edited: Jun 8, 2014
  10. Jun 11, 2014 #9
    Ok thanks for the explanation :smile:
     
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