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Def.: Continuously Indexed Vector Basis?

  1. Apr 16, 2004 #1
    While reading wikipedia I found the following quote:

    "...When this basis is indexed by a discrete (finite or countable) set, the vector representation is just the column of numbers familiar to high-school students. Vector bases may also be continuously indexed. When a quantum mechanical state vector is represented with respect to such a continuous basis it is called a wavefunction."

    I know that the basis of a vector space is the set of vectors that spans a space.

    I think that indexing a vector by a discrete set means providing a coefficient for each spanning vector, iff the discrete set has the same number of elements as the number of elements in the spanning set. This is where we get the finite nature of the discrete set, i.e. the number of elements is finite, as is the dimension of the vector space, which equals the number of spanning vectors.

    Is this correct? If not, what is the correct interpretation?

    Finally, what is a "continuously indexed basis"? Is this an infinite dimensional vector--i.e. a vector in R^n, or Hilbert space? (Wouldn't that rather be an infinitely indexed basis?) Or is this a vector space where a vector is defined by a single continuous function, rather than a series of coordinates? i.e. where you have instead of discrete dimensions, a continuum of dimensions--i.e. instead of dimension 2, 3, 4, you might have dimension 1, 1.75, Pi, and so on and everything in between continuously. I really don't understand what is meant by a continuously indexed basis. What is it?

    -Mark
     
  2. jcsd
  3. Apr 16, 2004 #2

    ahrkron

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    The same way you can build any 3D vector out of the 3 linearly indep vectors on a chosen basis, you can represent any periodic function as a sum of sines and cosines, with frequencies that are integer multiples of a given basic frequency. Those sines and cosines constitute a basis in the infinite-dimensional vector space of periodic functions [for experts, yes, I'm being a little sloppy here]. The way you construct any function is by adding together all the basis vectors, each multiplied by a real coefficient.

    In this case, the vector space is infinite-dimensional, but the basis vectors are still a countable set (the basis vectors are sin(x), sin(2x), sin(3x),...). If instead of integer multiples, you use any multiple, you get a continuously indexed basis.
     
  4. Apr 16, 2004 #3

    HallsofIvy

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    First, while "finite" certainly implies "discrete", "discrete" does not necessarily mean "finite"- it can mean "countably infinite". An example is the set of all infinite sequences: (a,b,c,d,...) with scalar multiplication and addition defined, as usual, coordinatewise. Another example would be the set of all polynomials.

    "continously indexed" certainly would have to be infinite dimensional- and it probably would be impossible to write down an explicit formula for the basis.
    An example is the vector space of all real number OVER THE RATIONALS.
    That is, we have all real numbers and real number addition, but "scalar multiplication" is allowed only with rational numbers. Since the rationals are countable and the reals are not, any basis must be uncountable. Theoretically, it should be possible to give a function that, to every x from 0 to 1, f(x) is a basis number but I don't believe anyone has ever written such a function explicitely.
     
  5. Apr 16, 2004 #4
    You guys are SO cool!
     
  6. Apr 16, 2004 #5
    oh yes, I forgot to mention, thankyou very much, that just transformed my whole take on vector spaces.

    -Mark
     
  7. Apr 16, 2004 #6

    NateTG

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    While Halls is completely correct, this passage could be slightly misleading since there can also be discrete uncountable sets.

    It's not really that hard to come up with an example of a continously indexed basis for a vector space:

    Consider the set of functions from the reals to the reals as the vector space, then the usual basis is the set of characteristic functions :
    [tex]\vec{e}_x(y)= \{1 \iff x=y ,0 \iff x \neq y\}[/tex]
    (sorry my latex is not very good, and I'm too lazy to track down the correct construct right now.)

    You can see that any function [tex]\vec{g}:\Re \rightarrow \Re[/tex] can be expressed as a unique sum:
    [tex]\vec{g} = \sum_{x \in \Re} g(x) \vec{e}_x[/tex]. The notation here is a bit messy, but the example I'm trying to describe is relatively simple.

    As pointed out above, another commonly used basis is the frequecy spectrum - for example [tex]\vec{e}_x(y) = \cos(x * y)[/tex] or something similar. In practice, Fourier transformations are translations of a function from the characteristic basis to a frequency basis.
     
  8. Apr 22, 2004 #7
    Your characteristic function there would appear, in a 2 dimensional vector space, where x, runs horizontally, and y runs vertically, as a diagonal line from the lower left, to the upper right--i.e. a plot of x=y.

    We're looking at the characteristic function as another variable, z, then really this is like a 3 space function where z = f(x,y) and z can only take on the values of 0,1.

    This reminds me of a square Identity matrix, except that, instead of having a finite, discrete number of rows and columns, it is "continuously indexed"--i.e. we're in a continuous space (the reals) and it is infinite (the index goes all the way to +- infinity.) The identity matrix and the characteristic function are both 2 dimensional, where each location corresponds to a value (i.e. a point in a third variable.)

    If I am correct here, there is a relationship here between these concepts: The identity matrix, delta-functions, and symplectic manifolds.
    Can you comment on the nature of this? Is a symplectic manifold like a continuously indexed "identity matrix like" organism for calculations in hilbert space? Am I totally confused?

    -Mark
     
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