1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac Delta Function Identity

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Good day.

    May I know, for Dirac Delta Function,
    Is δ(x+y)=δ(x-y)?

    3. The attempt at a solution
    Since δ(x)=δ(-x), I would say δ(x+y)=δ(x-y). Am I correct?
     
  2. jcsd
  3. Sep 23, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    No, that's not correct! The [itex]\delta[/itex] distribution (not function!) is defined by how it is acting on a function in the sense of a linear functional, written as a formal integral,
    [tex]\int_{\mathbb{R}} \mathrm{d} x \delta(x) f(x)=f(0),[/tex]
    where [itex]f[/itex] is a function of an appropriate space of functions (e.g., the Schwartz space of quickly vanishing functions or the space of functions with compact support, etc.).

    You can prove properties of the distribution by formally handling this kind of integral as if the [itex]\delta[/itex] distribution were a usual function. E.g., you have
    [tex]\int_{\mathbb{R}} \mathrm{d} x \delta(x-y) f(x) = \int_{\mathbb{R}} \mathrm{d} x' \delta(x') f(x'+y)=f(0+y)=f(y).[/tex]
    Now you can evaluate yourself, what [itex]\delta(x+y)[/itex] must be and also prove the (correct) statement [itex]\delta(-x)=\delta(x)[/itex].
     
  4. Sep 23, 2013 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    In order for two delta functions to be equal, their singularities must occur at the same place IDENTICALLY.

    Thus, you must ask yourself:
    Is x+y=x-y an identity or an equation?
     
  5. Sep 23, 2013 #4
    Hi vanhees71 and arildno,

    Thanks a lot for your prompt reply!

    I see. So ∫dx δ(x+y)f(x) will equal to f(-y),

    if f(x) is an even function, then
    ∫dx δ(x+y)f(x) = f(y)

    if f(x) is an odd functoin, then
    ∫dx δ(x+y)f(x) = -f(y)

    Am I correct? I am sorry if I still not able to get the concept, do allow me to make some mistakes so I can learn from it. While for arildno's question, I am sorry that I am not able to answer as it's little deep for me (I can't understand what equation and identity are in this context). I will dig into that.
     
  6. Sep 24, 2013 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Yes, that's correct. So you see that in general [itex]\delta(x+y) \neq \delta(x-y)[/itex].
     
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: Dirac Delta Function Identity
Loading...