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

A Fun with Deltas and ##L^2##: Physicists vs Mathematicians

  1. Apr 9, 2016 #1

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    <Moderator's note; thread forked from https://www.physicsforums.com/threads/question-regarding-the-dirac-delta-function.866220/>

    Why do you say that? The function that takes a function [itex]f(x)[/itex] and returns [itex]f(0)[/itex] is a well-defined distribution. Maybe what you mean is that if we identify two functions [itex]f \sim g[/itex] if [itex]\int |f(x) - g(x)|^2 dx = 0[/itex], then the delta-function distribution is ill-defined on the equivalence class. But it's perfectly well-defined on square-integrable functions, if we don't take equivalence classes.
     
    Last edited by a moderator: Apr 13, 2016
  2. jcsd
  3. Apr 9, 2016 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]\delta(x)[/itex] is not really a function, so [itex]\delta(0)[/itex] is not defined. The only thing that is defined about the delta function is integrals involving it:

    [itex]\int_{a}^b \delta(x) f(x) = f(0)[/itex] if the interval contains [itex]x=0[/itex], and is zero, otherwise.

    Statements involving [itex]\delta(x)[/itex] that don't explicitly involve integrals can only be made sense of by understanding them in terms of integration.
     
  4. Apr 9, 2016 #3

    rubi

    User Avatar
    Science Advisor

    It's not continuous, since ##\lVert f - g\rVert = 0##, but ##\left| F(f)-F(g)\right| \neq 0##. So it would be an element of the algebraic dual, but not the topological dual space.
     
  5. Apr 9, 2016 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    It's a little confusing. When people talk about [itex]L^2[/itex], they do mean functions modulo equivalence. But it seems to me that we can certainly talk about the set of square-integrable functions without taking equivalence relations.
     
  6. Apr 9, 2016 #5

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Viewed as functions (as opposed to equivalence classes of functions), [itex]\int |f(x) - g(x)|^2 dx = 0[/itex] does not imply that [itex]f=g[/itex], so I don't see what's the problem with having [itex]F(f) \neq F(g)[/itex].
     
  7. Apr 9, 2016 #6

    rubi

    User Avatar
    Science Advisor

    The space of square-integrable functions is usually denoted by ##\mathcal L^2## rather than ##L^2##, but it is not a Hilbert space, so it's not suitable for QM.

    Your ##F## is well-defined on ##\mathcal L^2##. It's just not continuous and distributions are usually defined to be continuous. It's okay to use discontinuous distributions, but you must be aware that many things might not work anymore as expected.
     
  8. Apr 9, 2016 #7

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    That's the thing isn't it. If ##f = g## a.e., why would we expect the integrals ##\int f(x)\delta(x)dx ## and ##\int g(x)\delta(x)dx## to be different.
    Furthermore, what will be the derivative of ##\delta## in your formalism?
    And why did you not put any continuity requirements on the definition of your distribution?

    I mean: ok, sure, your definition is consistent, but I don't see it as very useful.
     
  9. Apr 9, 2016 #8

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    That's the whole point--[itex]\int f(x) \delta(x) dx[/itex] is not actually an integral. It's just notation. That notation is defined to be [itex]f(0)[/itex].

    I don't understand the question. [itex]\delta(x)[/itex] is not a function. It doesn't have a derivative. It only makes sense in the expression [itex]\int \delta(x) f(x) dx[/itex].

    A distribution is not in general an integral. Certain distributions can be defined via an integral, but the delta function is one that can't be.

    There are no continuity requirements: [itex]F(f) \equiv f(0)[/itex]. That's well-defined for any function that has a value at [itex]x=0[/itex]
     
  10. Apr 9, 2016 #9

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You do realize that no math book adopts the definition you put down in this thread, right?
     
  11. Apr 9, 2016 #10

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, the context of this discussion is the delta function, and there is no way to make sense of a delta function in terms of continuous distributions.
     
  12. Apr 9, 2016 #11

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Why would you say that? Of course there is...
     
  13. Apr 9, 2016 #12

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

  14. Apr 9, 2016 #13

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You really don't see the difference between what you wrote and what's in the wiki link? For one, the space of test functions in the wiki link is not ##\mathcal{H}##, and second they require continuity.
     
  15. Apr 9, 2016 #14

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Could you elaborate? Because I don't know what you are talking about.

    I'm just giving the same definition as Wikipedia, which is this:

    As a distribution, the Dirac delta is a linear functional on the space of test functions and is defined by
    b1c2074d920e293f6ee23fc02130a233.png
    for every test function φ.
     
  16. Apr 9, 2016 #15

    rubi

    User Avatar
    Science Advisor

    The delta distrubution is continuous on the usual test function spaces like ##C^\infty_c## or ##\mathcal S##. You can embedd them into ##L^2## by sending each ##f## to its equivalence class. Of course, delta isn't continuous in the Hilbert space norm, but it's continuous in the topologies of the usual test function spaces.

    (Edit: Since micromass is active, I'm going to let him continue the argument.)
     
  17. Apr 9, 2016 #16

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, and what is their space of test functions? Also, read the following paragraph which deals about continuity.
     
  18. Apr 9, 2016 #17

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Look, the functional [itex]F[/itex] defined by [itex]F(f) = f(0)[/itex] is defined on the set of all functions from [itex]R[/itex] to [itex]C[/itex], regardless of whether they are continuous, or square-integrable, or whatever. You might be right that this function is only considered a "distribution" when it is restricted to a suitable collection of functions.
     
  19. Apr 9, 2016 #18

    jtbell

    User Avatar

    Staff: Mentor

    The limits of integration are any range that includes the point at which δ(x) ≠ 0.
     
  20. Apr 10, 2016 #19

    strangerep

    User Avatar
    Science Advisor

    Heh,... gotta love these physicist-vs-mathematician encounters... :headbang: :biggrin: [SCNR]

    (To be fair, the "usual topology on the space of test functions" is far from trivial, imho.) :angel:
     
  21. Apr 10, 2016 #20

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    :oldsmile:
    As a student, I could get really angry when someone dared to mention the delta function.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fun with Deltas and ##L^2##: Physicists vs Mathematicians
Loading...