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Inner product of Hilbert space functions

  1. Mar 15, 2010 #1
    this question is in reference to eq 3.9 and footnote 6 in griffith's intro to quantum mechanics

    consider a function f(x). the inner product <f|f> = int [ |f(x)|^2 dx] which is zero only* when f(x) = 0

    only points to footnote 6, where Griffith points out: "what about a function that is zero everywhere except a few isolated points? the integral would still vanish, although the function does not. if this bothers you, you should have been a math major"

    now i actually am a math major, and am interested in the reasoning behind this.
    does it have to do with the fact that you would be integrating the value of some function over a tiny amount of dx, which would then be somehow equated to zero in the limit that dx -> 0?

    if that is the case, how does this hold up with the fact that the integral of the kroneker delta function is one? i.e int[delta(x-a) * f(x)] = f(a)
     
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  3. Mar 15, 2010 #2

    Fredrik

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    The definition of the Lebesgue integral is very roughly that you take a small range of possible values of the functions, and multiply one of those values with the "size" of the subset of the domain where the function has a value in that range, and then you add up the results. So if two functions have the same values on, say, the rational numbers, the integral of the difference between those functions is zero (because the size of the set of rational numbers is zero).

    Because of this, you wouldn't get a Hilbert space if you take the set of square integrable functions and define a bilinear form on that space by

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

    It fails to satisfy

    [tex]\langle f,f\rangle\Rightarrow f=0[/tex]

    because we also have

    [tex]\langle g,g\rangle=0[/tex]

    where g is defined by g(x)=0 if x is irrational and g(x)=1 if x is rational. Note that this means that we didn't just fail to define a Hilbert space. We even failed to define an inner product.

    The way out of this mess is to say that two functions are equivalent if they only differ on a set of Lebesgue measure zero, and then define a vector space structure and an inner product on the set of equivalence classes (and then verify that this space is complete).

    That delta function is the Dirac delta, not the Kronecker delta, and that's a completely different story. It has nothing to do with Lebesgue integration. The actual definition of delta is

    [tex]\delta(f)=f(0)[/tex]

    The equation

    [tex]f(0)=\int \delta(x)f(x)dx[/tex]

    doesn't really mean anything on its own. The expression on the right isn't even an integral. It should be intepreted as a single symbol, defined to be equal to [itex]\delta(f)[/tex].
     
    Last edited: Mar 15, 2010
  4. Mar 17, 2010 #3
    Hello,

    You've instinctively picked out the relationship between quantum theory and an area of mathematics called functional analysis, which in turn is based on measure theory and the analysis of metric spaces... as a math major you may be on this road already.

    The relationship with quantum theory stems mostly from a particular function space called L^2(R^3), the space of all measurable functions on R^3 whose magnitude-square is Lebesgue integrable. This is a large space of functions; it includes all the continuous square-integrable functions. Yet, it is also the smallest extension of the continuous square-integrable functions which is "complete" in the sense of metric spaces. (In other words, L^2 is a Hilbert space.)

    L^2(R^3) is the state space for the single-particle non-relativistic Schrodinger equation. The extension to field theory has been studied for decades but as far as I know it has never been fully worked out.

    As @Fredrik said, Dirac delta is not a function and does not belong to L^2. Physics texts usually don't go into the details.

    Cheers

    Dave
     
  5. Mar 17, 2010 #4

    D H

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    As others have noted, Griffith was talking about Lebesgue integration. You might want to google that phrase, along with the phrases "almost everywhere" and "set of measure zero".

    As you are math major, you almost surely will be required to take a class in analysis, where you will learn the joys of measure theory, Lebesgue integration, and a host of other topics. (BTW, "almost surely" is yet another of those measure theoretic keywords; you will run into this phrase if you take a course in probability theory (highly recommended).)

    First off, that is the Dirac delta, not the Kroneker delta. Secondly, the Dirac delta "function" isn't a function. Thirdly, you have just found one of the key areas where mathematicians shake their heads with absolute disgust when they look at how physicists do math.
     
  6. Mar 17, 2010 #5

    Hurkyl

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    For a short, direct response:

    Functions, simply put, are not the right tool for this job; they contain "too much" information: two different functions can both name the same mathematical object of interest.

    (of course, functions are simply described, and can be used to name many objects of interest, so they are still worth studying)



    An improved notion that is very often more useful is what Fredrik mentioned: we say that two functions name the same object iff they differ on a set of measure zero.

    In addition to this thing with integrals, it cleans up some other quirks too -- e.g. the case of x=0 is no longer a relevant technicality when making a definition such as
    f(x) = x/x​
     
  7. Mar 17, 2010 #6

    D H

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    And in particular, functions are not quite the right tool for the rather important job of describing "reality".

    Even in the domain of classical physics, practically everything we "sense" is essentially an integrated response. I am using the word "sense" very broadly here: not only what we see, hear, feel, etc., but also what we measure with instruments. Go down to the domain of quantum mechanics, and those few things that do seem to be instantaneous or discrete measurements are in fact integrated responses.
     
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