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Inner Product (Infinite Dimensional)

  1. Nov 17, 2009 #1
    I am trying to understand the definition of the inner product of two functions on an interval. I know that the form of a scalar product in finite dimensional space is given by

    [tex]\vec{\phi} \bullet \vec{\psi}= \sum_{k} \phi_{k} \psi_{k}[/tex]

    and in infinite dimensional space

    [tex]\langle \phi , \psi \rangle = \int_{a}^{b} \phi(x) \psi(x) dx[/tex]

    If we expand out the first definition we get

    [tex]\vec{ \phi} \bullet \vec{ \psi} = \phi_{1} \psi_{1} + \phi_{2} \psi_{2} + \phi_{3} \psi_{3} ...[/tex]

    and for the second we get

    [tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]

    What is up with the factor of dx multiplying everything is this last expansion? It has to be there because that is what an integral does, it takes the function evaluated at that specific x and multiplies it by dx, but what purpose does the dx factor serve in the definition of the inner product? It is pretty obvious that apart from the factor of dx it looks identical to the finite dimensional definition. So what gives?


    Edit: I edited this to make my question more clear. And I'd like to add that it seems to me that

    [tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...= \vec{\phi} \bullet \vec{\psi}dx[/tex]

    Where [tex]\vec{\phi} \bullet \vec{\psi}[/tex] is supposed to just be a recognition of the form of the initial definition.
     
    Last edited: Nov 17, 2009
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  3. Nov 17, 2009 #2

    Hurkyl

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    They seem to be approximating the integral with a Riemann sum (the rectangle rule). IMO, dx is a poor name for the variable that expresses the step. Also, IMO, = is a poor symbol for "is approximately".
     
  4. Nov 17, 2009 #3
    Yeah, but I see the point of the argument, aside from the factor of dx.

    That essentially

    [tex]\langle \phi , \psi \rangle = dx(\vec{\phi} \bullet \vec{\psi)}[/tex]

    is true, and then you just have this dot product on the right but there is still some crazy factor of dx.
     
  5. Nov 17, 2009 #4

    Hurkyl

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    Oh, by the way, an inner product on a real vector space is merely any real-valued function of two vectors variables:
    <x+y, z> = <x,z> + <y,z>
    a<x,y> = <ax, y>
    <x,y> = <y,x>
    <x,x> is nonnegative
    <x,x> = 0 if and only if x = 0

    The forms you have described are valid only for very specific representations of your vector space!

    It is true that for any finite dimensional real vector space, there is a way to present vectors so that your formula holds. But that only works in that specific presentation! The only kind of presentation that works is to use coordinates relative to an orthonormal basis

    This can be directly extended to infinite dimensions1 -- but not as an integral!

    That integral works only in a different kind of presentation where vectors are written as functions on the interval [a,b].


    1: more precisely, countably infinite dimensional real Hilbert spaces
     
  6. Nov 17, 2009 #5

    Hurkyl

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    The dot product is crazier. It doesn't actually make any sense here.
     
  7. Nov 17, 2009 #6
    Well, let me tell you what my real desire is. I would like to have an intuitive reason for the definition given, rather than just accepting it as a definition. There has got to be some sort of, possibly 'loose', derivation of it.


    EDIT: I think I should edit the first post to maintain clarity in the discussion. I'll change the first definition to

    [tex]\vec{\phi} \bullet \vec{\psi} = \sum_{k}\phi_{k}\psi_{k}[/tex]
     
    Last edited: Nov 17, 2009
  8. Nov 17, 2009 #7
    I understand that I may be bastardizing the notation, but I still believe it is correct in the right context. If you think of the function as an infinite dimensional *vector*, that would then warrant the use of the vector symbol. And then why can't you just dot them like any other vector (thinking more along the lines of the *form* of the dot product, not necessarily what the meaning is in infinite dimensions) to get

    [tex]\vec{\phi(x)} \bullet \vec{\psi(x)}=\phi(a)\psi(a)+\phi(a+dx)\psi(a+dx)+\phi(a+2dx)\psi(a+2dx)+\phi(a+3dx)\psi(a+3dx)[/tex]

    EDIT: Sorry for the successive post.
     
    Last edited: Nov 17, 2009
  9. Nov 17, 2009 #8

    Hurkyl

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    Oh! I didn't notice the expansion was your own doing -- I thought you were citing something.


    I already gave the actual definition of inner product. Algebraically, I think the definition directly captures what is really important.

    IMO, the reason for using a function space with an inner product given by that integral is not a matter of gaining intuition about inner products. The reason is to do calculations, and to let us apply our great knowledge of calculus to solve problems of linear algebra.

    (Just like in finite dimensional inner product spaces, we use orthonormal coordinates to let us apply our great knowledge of Euclidean coordinate geometry to solve linear algebra problems)
     
  10. Nov 17, 2009 #9
    Yeah, the argument is all mine. I am trying to find the reason behind the definition. You don't just define things like

    [tex]Work=\int_{\vec{a}}^{\vec{b}}\vec{F}\bullet d\vec{r}[/tex]

    They have a reason, you know?


    Edit: And to comment on the topic of the usefulness of the abstract definiton, I am not calling that into question at all. It is very useful to think of 'lengths', 'magnitudes', and what not of functions, it is great! I already know all of that from finite dimensional stuff.
     
  11. Nov 17, 2009 #10

    Hurkyl

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    Hrm. Are you mixing up the following two terms?
    1. vector meaning an element of a vector space
    2. vector meaning a list of scalar values (I'll call these "coordinate vectors")

    I've always been using vector in the former sense....

    Dot products only make sense for coordinate vectors, and even then only when the components can be summed over....

    (e.g. even if you rewrote your function as a "list" of values at each point, there are too many points for infinite sums to make sense)


    What you wrote is a possibly useful heuristic -- but in the end, it needs to be translated back into something well-defined. Such "continuous linear combinations" are generally made rigorous using integrals. In fact, your expansion simply inverted that process.

    If you are willing to learn non-standard analysis, you could instead mutter something about standard parts and approximating the interval with a lattice of hyperfinitely many points.
     
  12. Nov 17, 2009 #11

    Hurkyl

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    My claim is that there isn't any deep reason. It's a form we can compute with, and every* case of interest can be rewritten to turn it into this form. In fact, some interesting cases were already in this form. I believe one of the earliest examples and most important driving forces of this setting were the study of Fourier series.


    *: at least, every case we are interested in
     
  13. Nov 17, 2009 #12
    Ok, I hear what you are saying. This problem of acceptance of mine could be traced back to reading the supplement to Chapter 2 in Gasiorowicz's Quantum Physics book.

    http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=1533&itemId=0471057002&chapterId=4825

    Where in equation 2A-6 he 'introduces' the factor of delta n. I know it is not an identical case, but if I can figure out the justification of the factor of delta n, then I bet there is an argument for a delta k in the very first definition of the dot product. And if the delta k sneaks into that definition, then the dx is perfectly explained.
     
    Last edited: Nov 18, 2009
  14. Nov 18, 2009 #13
    You seem to be irritated by the dx in the second definition. As Hurkyl said, they approximate the integral by a Riemann sum. Let me elaborate:

    An integral [tex]\int_{a}^{b} f(x) dx[/tex] can be approximated by [tex]\sum_{k=0}^{N-1} f(a+k\Delta x) \Delta x=f(a) \Delta x+ f(a+\Delta x) \Delta x + f(a+2\Delta x) \Delta x...[/tex]. Try to visualize this (see here).

    Now, to get to your definition we replace f(x) by [tex]\phi(x) \psi(x)[/tex], i.e. we let [tex]f(x)=\phi(x) \psi(x)[/tex].

    Then, [tex]\int_{a}^{b} \phi(x) \psi(x) dx[/tex] can be approximated by [tex]\phi(a) \psi(a) \Delta x + \phi(a+\Delta x) \psi(a+\Delta x) \Delta x + \phi(a+2\Delta x) \psi(a+2\Delta x) \Delta x ...[/tex]
     
  15. Nov 18, 2009 #14
    It really should be

    [tex]\langle \phi , \psi \rangle = \lim_{dx \rightarrow 0} \sum_{n=0}^{(b-a)/dx} \phi(a+n*dx) \psi(a+n*dx) dx [/tex]


    and the limit only exists if you have dx on the right side. Otherwise it diverges, you get an infinity.
     
  16. Nov 19, 2009 #15
    So, it is safe to say that the dx is a scaling factor that brings the otherwise divergent expression

    [tex]\vec{\phi(x)}\bullet\vec{\psi(x)}[/tex]

    back to a finite value.



    This leads to my bastardized notation where you define (for functions), *thanks to hamster for making it more precise*

    [tex]\lim_{\Delta x \rightarrow 0}\vec{\phi(x)} \bullet \vec{\psi(x)}=\lim_{\Delta x \rightarrow 0} \sum_{n=0}^{\frac{b-a}{\Delta x}} \phi(a+n\Delta x) \psi(a+n\Delta x)[/tex]

    Which, as the limit goes to zero, becomes

    [tex]\vec{\phi(x)} \bullet \vec{\psi(x)}= \phi(a) \psi(a) + \phi(a+dx) \psi(a+dx)+\phi(a+2dx) \psi(a+2dx)+...+\phi(b-dx)\psi(b-dx)dx+\phi(b)\psi(b)dx[/tex]

    This expression definitely diverges, though. If we multiply both sides by a delta x in the original equation, before the limit was applied, we get

    [tex]\lim_{\Delta x \rightarrow 0}\vec{\phi(x)} \bullet \vec{\psi(x)} \Delta x=\lim_{\Delta x \rightarrow 0} \sum_{0}^{\frac{b-a}{\Delta x}} \phi(a+n\Delta x) \psi(a+n\Delta x) \Delta x[/tex]

    And applying the limit

    [tex]\vec{\phi(x)} \bullet \vec{\psi(x)}dx= \phi(a) \psi(a)dx + \phi(a+dx) \psi(a+dx)dx+\phi(a+2dx) \psi(a+2dx)dx+...+\phi(b-dx)\psi(b-dx)dx+\phi(b)\psi(b)dx[/tex]

    Which can be recognized to be

    [tex]\vec{\phi(x)} \bullet \vec{\psi(x)}dx= \int_{a}^{b}\phi(x) \psi(x)dx[/tex]

    This integral does not diverge as long as the functions are well behaved in the interval. Then, we can define

    [tex]\langle \phi(x) , \psi(x) \rangle =\vec{\phi(x)} \bullet \vec{\psi(x)}dx=\int_{a}^{b}\phi(x) \psi(x)dx[/tex]
     
    Last edited: Nov 19, 2009
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