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Functions and infinite dimensional vectors

  1. Jun 11, 2009 #1
    while skimming through a linear algebra book today i read that functions were a vector space

    Can you actually describe a function [tex]f: \mathbb{R} \rightarrow \mathbb{R}[/tex] that is defined over all real numbers as a vector with uncountably infinite components?

    Like this? [tex]|f\rangle = \left [ f(x_1), f(x_2),....... \right ] [/tex], the components representing the values the function takes at each real number.

    Is this what "vector spaces" of Functions and "functionals" (functions which has real functions as arguments) all about? What do all of this have to do with quantum mechanics? (wiki qm article blabbers on about that)
     
    Last edited: Jun 11, 2009
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  3. Jun 11, 2009 #2

    Hurkyl

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    Yes. The set of all functions R->R is a (real) vector space -- the vector space axioms are almost trivial to check.

    You have to be careful about what you mean by "components", far moreso than you do in the finite dimensional case. I've even seen a textbook on infinite-dimensional linear algebra advise that you forget everything you know about finite-dimensional linear algebra, because it is just as likely to mislead you as to guide you.
     
  4. Jun 12, 2009 #3

    HallsofIvy

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    If you were to read (not just skim) a Linear Algebra text, you would find that the "vectors" they are talking about are not the same vectors you see in Calculus or Physics. The vector spaces discussed in Linear Algebra are a generalization of those. You probably will not see the term "component" except as part of a discussion of a basis for a vector space.

    As Hurkyl said, it is easy to see that if f and g are functions and a is a number then f+ g and af are also functions: The set of all functions forms a vector space with those operations.

    Since the powers of x, x0= 1, x1= x, x2, x3, etc. are easily shown to be independent, it follows that no finite collection of functions can span the whole space and so the "vector space of all functions" is infinite dimensional. That is not quite the same as talking about "components" or an "infinite number of components".

    (In all of this, I should be talking about functions defined on a particular set. I didn't just to make it easier to write.)
     
  5. Jun 12, 2009 #4
    i meant can the value of the function at each point be the basis for the vector spaces of functions? Can not every function that is defined over the whole real line be expressed as sums of scalar multiples of the indicator functions for every real number: [tex]f_i(x)[/tex] = a nonzero real number only at one real number [tex]x_i[/tex], and zero everywhere else? If we add up the functions f_i for all real numbers x_i, than we get the function back.

    can this not also be expressed as [tex]f(x) = \sum_{i\in \mathbb{R}}a_i f_i(x)[/tex]?

    The problem with this site is that the mentor guys think they helped you enough already when they first post a reply
     
    Last edited: Jun 12, 2009
  6. Jun 12, 2009 #5

    jbunniii

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    What do you mean by [tex]\sum_{i\in \mathbb{R}}[/tex]? There are uncountably many real numbers, whereas the summation symbol normally indicates at most a countably infinite sum, a notion that is well-defined in terms of sequences of partial sums.

    Also, by definition, in order for a subset B of V to be a basis for V, it must be possible to write every element of V as a FINITE linear combination of elements of V. Such a basis is called a Hamel basis. For certain types of spaces, this restriction can be relaxed to allow for countably infinite linear combinations, e.g., an orthonormal basis for a Hilbert space. But I don't know of any notion of basis that allows for uncountably infinite linear combinations, however you choose to define that.
     
  7. Jun 12, 2009 #6

    HallsofIvy

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    The answer to that is obviously "no". values of functions are numbers, not functions and so cannot give a basis for a vector space of vectors.

    Then you have misunderstood. Us "mentor guys" expect you to come back with additional questions if you do not understood.
     
  8. Jun 12, 2009 #7
    kinds of spaces in qm?

    ok...

    What kind of infinite-dimensional complex Hilbert space or functional space occurs in quantum mechanics? Wikipedia just says infinite-dimensional complex hilbert space
     
  9. Jun 12, 2009 #8

    jbunniii

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    Re: kinds of spaces in qm?

    I'm not a physicist, but the Wikipedia article mentions:

    "In the mathematically rigorous formulation of quantum mechanics, developed by Paul Dirac[35] and John von Neumann[36], the possible states of a quantum mechanical system are represented by unit vectors (called "state vectors") residing in a complex separable Hilbert space (variously called the "state space" or the "associated Hilbert space" of the system) well defined up to a complex number of norm 1 (the phase factor). In other words, the possible states are points in the projectivization of a Hilbert space, usually called the complex projective space. The exact nature of this Hilbert space is dependent on the system; for example, the state space for position and momentum states is the space of square-integrable functions, while the state space for the spin of a single proton is just the product of two complex planes."

    The space of square-integrable functions is called "L^2", and it is more or less what it sounds like: it's the set of all functions f(x) that satisfy

    [tex]\int |f(x)|^2 dx < \infty[/tex]

    (Technically, L^2 consists of equivalence classes of functions where two functions are equivalent if they differ from each other only on a set of measure zero.)

    A Hilbert space is a vector space that also has an inner product:

    [tex]<f,g> = \int f(x) g^*(x) dx[/tex]

    and the norm induced by the inner product:

    [tex]||f|| = <f,f>^{1/2}[/tex]

    One more technical detail: a Hilbert space must be a COMPLETE, meaning that it doesn't have any "gaps" in the sense that a sequence of functions that get closer and closer together must converge to a function in the space. An example of a vector space that is not complete is the set of rational numbers: a sequence of rationals that get closer and closer to sqrt(2) doesn't have a rational limit. We don't allow this stuff in Hilbert spaces.

    L^2 has an orthogonal basis, for example, the complex exponential functions.
     
  10. Jun 12, 2009 #9
    do you mean [tex]\int_{\mathbb{R}} |f(x)|^2 < \infty[/tex] ? (the blackboard bold R means integrate over entire real line)
     
    Last edited: Jun 12, 2009
  11. Jun 12, 2009 #10

    jbunniii

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    Yes, I should have been more specific. What I wrote above assumes that the integration is taken over a finite interval [a,b].

    However, it's all true if we integrate over [tex]\mathbb{R}[/tex], EXCEPT in that case the complex exponentials are not a basis: indeed, they aren't even in the space because

    [tex]\int_{\mathbb{R}} |\exp(i\omega x)|^2 dx = \int_{\mathbb{R}} (1) dx = \infty[/tex]

    By Zorn's lemma, every Hilbert space, including [tex]L^2(\mathbb{R})[/tex] has a (possibly uncountable) orthonormal basis. Unfortunately, Zorn's lemma doesn't tell you how to construct such a basis in general.

    For [tex]L^2([a,b])[/tex], there's a countable orthonormal basis, namely

    [tex]\left\{\frac{1}{\sqrt{b-a}} e^{i 2\pi n x / (b-a)}\right\}[/tex] for [tex]n \in \mathbb{Z}[/tex].

    Expressing a function in terms of this basis gives you the classical Fourier series.
     
  12. Jun 12, 2009 #11

    Fredrik

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    Re: kinds of spaces in qm?

    Von Neumann postulated that it's a separable infinite-dimensional Hilbert space over [itex]\mathbb C[/itex]. "Separable" means "contains a countable dense subset". A Hilbert space is separable if and only if it has a countable orthonormal basis. Any two separable Hilbert spaces are isomorphic to each other. So it doesn't really matter which one we use, as long as it's separable.
     
  13. Jun 12, 2009 #12

    Fredrik

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    Your choice of words inspired me to ask about something I've been wondering about for some time. Is there a reason why it's called linear algebra when the space is finite-dimensional, and functional analysis when it's infinite-dimensional? And why is "linear analysis" specifically about Fourier series and stuff? Is there something fundamentally "more linear" about Fourier series than the rest of functional analysis? Also, why functional analysis? I mean, a "functional" is specifically a linear operator into the set of scalars, and most of the linear operators we encounter in functional analysis have some other Hilbert or Banach space as their codomain.

    Sorry if it's a dumb question, but this terminology seems weird to me. :smile: It reminds me of a Dilbert strip where Dogbert had a computer generate suggestions for company names, by randomly combining words from science and technology. The result was "Uranus-Hertz".
     
  14. Jun 12, 2009 #13

    jbunniii

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    Analysis usually implies that there are limits involved, which is natural as soon as you move from the finite to the infinite.

    I guess "linear analysis" emphasizes tools that are useful in the context of linear differential equations.

    Apparently "functional analysis" originally dealt with functionals; see for example the first paragraph in the preface of this book at Google Books:

    http://tinyurl.com/l2oapx

    [I used tinyurl because the original link is ridiculously long!]
     
  15. Jun 12, 2009 #14

    Fredrik

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    Thank you, that makes sense.

    You can use url tags instead of tinyurl if you want, like this.
     
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