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Algebraic structure of Dirac delta functions

  1. Jun 1, 2012 #1
    OK, the Dirac delta function has the following properties:

    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1[/tex]

    [tex]\int_{ - \infty }^{ + \infty } {f({x_1})\delta ({x_1} - {x_0})d{x_1}} = f({x_0})[/tex]
    which is a convolution integral. Then if [tex]f({x_1}) = \delta (x - {x_1})[/tex]
    we get

    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})d{x_1}} = \delta (x - {x_0})[/tex]
    which is a self-convolution and is seen on the wikipedia site for the Dirac delta function. But this can be iterated to get:

    [tex]\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\delta (x - {x_2})\delta ({x_2} - {x_1})\delta ({x_1} - {x_0})d{x_2}d{x_1}} } = \int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})d{x_1}} = \delta (x - {x_0})[/tex]
    And if iterated an infinite number of times we get:

    [tex]\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {\delta (x - {x_n})\delta ({x_n} - {x_{n - 1}}) \cdot \cdot \cdot \delta ({x_1} - {x_0})d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}} } } = \delta (x - {x_0})[/tex]
    which is seen in Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 4th Edition, by Hagen Kleinert, page 91, eq 2.17

    So all the Dirac delta functions seem to form a space closed under self-convolution. And I wonder what other algebraic structures are implied by the above. And particularly I wonder if there is any structure here that proves that the Dirac deltas must be complex. Or what other structures are necessary to make the deltas complex?
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2
    I would suggest learning some functional analysis. As you are probably aware, the dirac delta is not a function, it's a continuous linear functional on some function space.

    Perhaps what you are looking for are banach algebras. I'm not sure because you wonder if the dirac delta must be complex. The definition of a dirac delta function in terms of functional analysis should clear that point up.

    You may find my post in this thread helpful.
  4. Jun 1, 2012 #3
    Yes, yes, I'm aware of the objections to multiplying distributions. But perhaps it would help to show an explicit proof of the self-convolution integral. The Chapman-Kolmogorov equation is:

    [tex]f(x|{x_0}) = \int_{ - \infty }^{ + \infty } {f(x|{x_1})} f({x_1}|{x_0})d{x_1}[/tex]
    And when

    [tex]f(x|{x_o}) = {\left( {\frac{\lambda }{{2\pi (t - {t_0})}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{\frac{{\lambda {{(x - {x_0})}^2}}}{{2(t - {t_0})}}}}[/tex]
    it can be proved that
    [tex]{\left( {\frac{\lambda }{{2\pi (t - {t_0})}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{\frac{{\lambda {{(x - {x_0})}^2}}}{{2(t - {t_0})}}}} = \int_{ - \infty }^{ + \infty } {{{\left( {\frac{\lambda }{{2\pi (t - {t_1})}}} \right)}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{\frac{{\lambda {{(x - {x_1})}^2}}}{{2(t - {t_1})}}}}} {\left( {\frac{\lambda }{{2\pi ({t_1} - {t_0})}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{\frac{{\lambda {{({x_1} - {x_0})}^2}}}{{2({t_1} - {t_0})}}}}d{x_1}[/tex]
    This is proved in The Feynman Integral and Feynman's Operational Calculus, by Gerald W. Johnson and Michael L. Lapidus, page 37 and 38. And then as [tex]t \to {t_0}[/tex] with [tex]{t_0} \le {t_1} \le t[/tex] turns the above equation into the self-convolution integral of the Dirac delta since,

    [tex]\delta (x - {x_0}) = \mathop {\lim }\limits_{t \to {t_0}} {\left( {\frac{\lambda }{{2\pi (t - {t_0})}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{\frac{{\lambda {{(x - {x_0})}^2}}}{{2(t - {t_0})}}}}[/tex]
    Or maybe I should drop the term Dirac delta and just refer to the Chapman-Kolmogorov equation. Does this avoid the multiplication of distribution issue?

    So I wonder about the algebraic structure of this Chapman-Kolmogorov/self-convolution integral. Thank you.
    Last edited: Jun 1, 2012
  5. Jan 30, 2013 #4
    I've been studying the complex numbers and their extension to hypercomplex numbers a little. There seems to be a progression from the real numbers to complex numbers to quaternions, to octonions numbers, etc. Each of these have a set of algebraic properties and with each progression one loses some algebraic property. For example the real number have the properties of magnitude and commutativity and associativity. But complex numbers lose the property of magnitude, quaternions lose commutativity, and octonions lose associativity. Beyond that you're no longer dealing with division algebras which makes them hard to work with.

    So it seems to me that the set of dirac delta functions that I use in the integrals above can be represented by complex numbers because they share the same algebraic property as complex numbers. So how do the dirac delta functions have the algebraic property of the complex numbers. The complex numbers lose the ability to state which complex number is greater or less than another. And, of course, you can't say which dirac delta function is greater or less than another dirac delta function. But the dirac delta function still has the properties of commutativity and associativity like the complex numbers. So they both share the same algebraic properties, which mean that the dirac delta function can be represented by complex numbers, at least in the context above where they seem to be compared with each other.
    Last edited: Jan 30, 2013
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