Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A A function whose integral is either 0 or 1

  1. Mar 21, 2016 #1
    I'm trying to generalize the property of the Kronecker delta function which gives
    [tex]\sum\nolimits_{i = 0}^n {{\delta _{ij}}} = \left\{ {\begin{array}{*{20}{c}}
    1&{0 < j < n\,\,\,\,\,\,\,\,\,\,\,}\\
    0&{j < 0\,\,or\,\,n < j}
    \end{array}} \right\}\,\,.[/tex]
    The continuous case seems to be the Dirac delta function such that
    [tex]\int_R {{\rm{\delta (x - }}{{\rm{x}}_0}){\rm{dx}}} = \left\{ {\begin{array}{*{20}{c}}
    1&{{x_0} \in R}
    0&{{x_0} \notin R}
    \end{array}} \right\}\,\,.[/tex]
    But only using the Dirac delta function seems too restrictive for most applications. I'd like to keep the property of the integral being either 0 or 1, depending on whether some parameter, x0, is or is not within the limits of the integral. But I can't think of any other function for which this is true. It seems that any other continuous function defined only in R will give the same integral no matter if x0 is inside or outside R. So it seems the only way to insure that x0 is always within R is to make R be the whole real line from -∞ to +∞, in which case there is no integration to 0 since x0 is always within R.

    But I'm sure I don't know everything. And someone here might know something I don't.
  2. jcsd
  3. Mar 21, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    I don't follow. Would you have some examples ?
    Well, nobody is taking it away from you. You must mean something else than what I read from this sentence ?
    Isn't clear to me either.

    In summary: I tried to understand what your problem is but did not succeed. Can you help me ?

    I suppose you read the wikipedia lemma ?
  4. Mar 21, 2016 #3
    Ultimately, I'm hoping to be able to recognize that the wave-functions of quantum mechanics can be seen as test functions with compact support such that for values of some parameters inside the range of that support this guarantees that they can be normalized to 1, and for values outside that support the normalization must be zero. This may allow me to draw parallels between quantum mechanics and some logic that I'm considering.

    Or perhaps this is a way to prove that normalization must always be done from -∞ to +∞, (so that the parameter is always in the support) and we only normalize in smaller regions for practical reasons, such as with an infinite square well.

    So for example, I think we could always construct a formula that guarantees an integral of 1 for x0∈R, and 0 otherwise, by using the unit step function. With a step function,
    [tex]H(z) = \begin{array}{*{20}{c}}
    1&{0 < z}\\
    0&{z < 0}
    Then, [itex][H({x_0} - {x_{\min }}) - H({x_0} - {x_{\max }})] = 1[/itex] for [itex]{x_{\min }} < {x_0} < {x_{\max }}[/itex], and 0 otherwise. Then we have
    [tex]\int_R {\psi (x)} [H({x_0} - {x_{\min }}) - H({x_0} - {x_{\max }})]\,dx\,\, = \,\,\left\{ {\begin{array}{*{20}{c}}
    1&{{x_0} \in R}\\
    0&{{x_0} \notin R}
    \end{array}} \right\}[/tex]
    with [itex]R = \{ {x_{\min }} < x < {x_{\max }}\} [/itex], and only if
    [tex]\int_R {\psi (x)dx = 1\,\,.} [/tex]
    But I don't see this being used. So maybe that justifies the use of integrating throughout the entire real line.
  5. Mar 26, 2016 #4
    It seems like you don't need the integral to be 1 only when a particular point is in the range of integration for your application. Simply having an integral of 1 over an entire compact region and a value of 0 outside that region would seem to be sufficient.

    I'd look at bump functions in particular.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted