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Number of possible wavefunctions only countably infinite?

  1. May 7, 2013 #1
    Two related questions:
    (1) The wavefunction is characterised as encoding all the physical characteristics of a particle. But which ones? The quantum numbers? In that case, since each quantum number ranges over discrete values, there would seem to be only a countably (as opposed to a continuum) of possible wave functions. Is this correct? If so, since the wave function gives probability amplitudes, can one therefore say that there are an uncountable number of probabilities excluded?
    (2) Black hole entropy is considered to encode the information about the black hole. (Already wrong?) But entropy is a dimensionless number, and the amount of entropy is proportional to the area of a given black hole. So I do not see how entropy could act as an encoding for different combinations of these physical properties. If it does, what form does the encoding take? What quantities is it considered to encode: mass/energy of the black hole, its total spin, its angular momentum, its charge... anything else? Similarly to the first question, are there thus only a countable number of possible values that entropy for a black hole can take on, since none of those quantities are continuous?

    Thanks for any pointers.
     
    Last edited: May 7, 2013
  2. jcsd
  3. May 7, 2013 #2
    deep questions to which I do not know the answers. Is this just a hobby?
     
  4. May 7, 2013 #3
    Why should quantum numbers always be countable? An electron in free space can be in a state of any arbitrary (real-valued) momentum.

    The information in a black hole is stored at its surface in some crazy way - I'm no expert on that but I think the book "Black hole wars" has a readable account of that.
     
  5. May 7, 2013 #4

    Jano L.

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    The premise is quite controversial - you do not need to assume it necessarily.

    The quantum numbers occur when labeling eigenfunctions of some operator. For example, the Hamiltonian for hydrogen atom has eigenfunctions labelled by three quantum numbers ##n,l,m##. They are similar to labeling numbers for distinct modes of vibration, of say, metal ball. Each combination of quantum numbers gives one distinct eigenfunction. So there are countably many eigenfunctions, and since they from a basis, the basis itself is countable, which is important mathematical property of the Schr. equation.

    However, the admissible wave functions ##\psi## do not need to belong to this set of eigenfunctions; rather, any such function can be expanded into sum

    $$
    \psi(\mathbf r) = \sum_{n,l,m} c_{nlm} \phi_{nlm}(\mathbf r).
    $$

    From this it seems that the set of all wave functions is uncountable.
     
  6. May 7, 2013 #5
    Any function in the hilbert space can be expanded, so yes uncountably many functions have been excluded.
     
  7. May 8, 2013 #6
    The probability amplitude functions are of integrable sqare. This set is countable. There are many non integrable than integrable functions
     
  8. May 8, 2013 #7
    How exactly does that make the square integrable set countable?
     
  9. May 8, 2013 #8
    this is not true for all cases. for eg., the free particle energy eigenfunctions form a continuous spectrum.
    [itex]e^{i\frac{\sqrt{2mE}}{\hbar}x}[/itex]
     
  10. May 8, 2013 #9

    bhobba

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    Its a standard theorem of Lebesgue integration and Hilbert Spaces you will find in any text on functional analysis. In the context of QM you will find a proof of it in Von Neumanns classic Mathematical Foundations of QM. Basically it relies on the fact that the rationals are countable and dense in the reals.

    Thanks
    Bill
     
  11. May 8, 2013 #10

    bhobba

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    Yea that's true - basically the space of QM isn't a Hilbert space like some texts will tell you - it's really a Rigged Hilbert Space. Even these have a countable basis but, without going into the detail, which is in fact a bit hairy, have a 'looser' definition of convergence - called weak convergence - to get around the issue. It's the same sort of trick used in distribution theory where you can rigorously define stuff like the Dirac Delta function which is not a function in the usual sense, square integrable or otherwise. In fact Rigged Hilbert Spaces can in a sense be looked on as Hilbert spaces with distribution theory stitched on - that's what the Rigged in Rigged Hilbert Space means - it like the rigging or scaffolding on a ship - it provides the 'backing' you need to do this mathematically 'weird' sort of stuff. Here 'weird' means you have stuff a bit different than you usually have such as delta functions but it is mathematically totally rigorous.

    A word of warning though - don't get too caught up in this sort of stuff. I did in my first brushes with QM finding the math in Von Neumann's book perfectly OK but the stuff in other texts like Dirac's - well - problematical mathematically. This led me to advanced tomes like Gelfand and Vilenkin's encyclopedic and hard going Generalized Functions - even for guys like me with a background in math and functional analysis its hard going - but in fairness I have to say not impossibly so - I did it. Still it took me a long time. I would suggest the first chapter in Ballentine's - QM - A Modern Development instead - it outlines the basics of whats going on enough to quell questions about mathematically difficult issues.

    Thanks
    Bill
     
    Last edited: May 8, 2013
  12. May 8, 2013 #11

    Fredrik

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    There's a difference between "this Hilbert space is countable" and "this Hilbert space has a countable orthonormal basis". Even a 1-dimensional Hilbert space (which has an orthonormal basis containing only 1 vector) is not countable.
     
  13. May 9, 2013 #12
    Thanks to all those that posted; the comments have helped me a lot. There are still a couple of points that bother me; but first a few individual comments:
    Pseudo Epsilon: yes, this is "just a hobby": I have a mathematical background, but only minimal background in physics; I have been working my way independently through various texts on QM (I am not taking a course), but questions still remain, so I turn to this forum.
    bhobba: acting on your Ballentine recommendation, I have obtained the book and look forward to working through the first chapter as per your recommendation.
    Sonderval, my statement about quantum numbers being discrete came from http://en.wikipedia.org/wiki/Quantum_number, which wasn't including momentum as a quantum number. However, the other posts pointed out that the wave function is not restricted to those quantum numbers. Thanks for the recommendation on the Black Holes Wars; I shall continue to look for it (if possible, without ordering it internationally).
    The posts have shown me that there are an uncountable number of such wavefunctions. Again, thanks.
    There are some nagging doubts concerning this uncountability (limits on information content, consistency of quantization of spacetime, etc.), but I shall be working through them and will be back on this forum with new posts.
     
  14. May 9, 2013 #13
    'Rigged Hilbert Space' sounds like a misnomer. i would prefer the term physical hilbert space.
     
  15. May 9, 2013 #14

    vanhees71

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    "Rigged Hilbert Space" is a well-established term for a mathematical formulation of the theory of unbound operators in Hilbert space. As the name tells you, it's all about the Hilbert space, and the space of physical states is a Hilbert space, no more, no less. The point of the rigged-Hilbert-space formulation is to make the hand-waving way physicists treat the problem of unbound operators in quantum mechanics (non-relativistic "first-quantization" quantum mechanics!) in a mathematically rigorous way. As far as I know, it's equivalent to the older treatment with the spectral theorem by von Neumann, who established QM in a mathematically rigorous way in the early 1930ies.
     
  16. May 9, 2013 #15

    bhobba

    Staff: Mentor

    Its completely equivalent only its based on the Generalized Spectral Theorem (also called the Nuclear Spectral or Gelfand-Maurin Theorem) instead of the Spectral Theorem used by Von Neumann using the Stieltjes integral. Its proof for some reason is generally omitted from texts but ages ago I did manage to dig one up:
    http://www.math.neu.edu/~king_chris/GenEf.pdf [Broken]

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  17. May 9, 2013 #16
    okay, well i'm curious about how this name came about. The word 'Rigged' seems to imply that the concept of Hilbert space is being misused, tampered with, or wrongly used. which is not really the case. thats why i said that it sounds like a misnomer. the fact is that the concept of physical hilbert space (the term i am accustomed to) is quite natural for the description of quantum mechanics and is required to include things like wavefunction collapse.
     
  18. May 9, 2013 #17

    dextercioby

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    Hi, what you found/quoted is not the original result by Gelfand and Kostyuchenko which is presented in the 4th volume of <Generalized Functions>. It's rather a modified version due to Berezanskii (not 100% sure though; the author of the article didn't quote his bibliographical source(s)). Since I've not had access to Maurin's original work in Polish and his book containing his results in English is out of my reach, I can't comment about him.

    What I'm saying is that there are to my knowledge at least 3(*) different(?) approaches to come up with a solution for the spectral problem of a self-adjoint operator in a (complex separable) Hilbert space (of course, other than the work by MH Stone and J von Neumann millions of years ago):

    1. Gelfand-Kostyuchenko.
    2. Berezanskii.
    3. Maurin.

    (*) There might be a 4th, not sure what the work by Dutch mathematician van Eijndhoven is trully about. He wrote a book in the 1980's on a (new) mathematical fundation of Dirac's bra/ket formalism.

    https://www.amazon.com/A-Mathematic...68131967&sr=1-1-fkmr1&keywords=van+wind+dirac
     
    Last edited by a moderator: May 6, 2017
  19. May 9, 2013 #18

    dextercioby

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    This is (not) semantics, <rigged> means <fully equipped> in the sense that a <rigged Hilbert space> is an extension of the known concept of Hlbert space which is fully prepared to accomodate any self-adjoint/isometric linear operator on a Hilbert space by providing it with a solution space to its spectral equation.
     
  20. May 9, 2013 #19

    dextercioby

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    To me <physical Hilbert Space> sounds mathematically vague. Give me a source where this concept is explained accurately from a mathematician's perspective.
     
    Last edited: May 9, 2013
  21. May 9, 2013 #20

    dextercioby

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    Hi Hendrik,

    I think what you claim in the bolded part stops probably in a fundamental place:

    Let A be a self-adjoint unbounded operator on a complex separable Hilbert space H with the following spectral equation:

    [tex] Af = af [/tex] (1)

    a is a complex number (to be eventually proven as purely real), f is a vector in H (actually in D(A)[itex]\bigcap[/itex]Ran(A), D(A)[itex]\subset[/itex]H, Ran(A)[itex]\subset[/itex]H).

    Question: Does the spectral equation (1) always admit solutions* in H ? Or do we need a bigger space and an extension of A to find some solutions ?

    * to admit solutions: to find the set of all <a>'s in C and <f>'s in H satisfying (1) identically.

    Note: sorry for the multiple posting.
     
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