Dirac Delta function

1. Oct 6, 2008

maximummman

hello every body

i am a new M.S student

and i cant understand the Dirac delta function can any one simply describe it to me in order to simplify it.

thank you

2. Oct 6, 2008

Fredrik

Staff Emeritus
If you want the actual definition, you should start by taking a look at the Wikipedia article about distributions, and then ask in one of the math forums of this site. (Unless someone like Hurkyl shows up in this thread). I think most physicists don't really know the definition of the Dirac delta distribution. It's actually not even a function, at least not a function from $\mathbb R$ to $\mathbb R$.

You can get pretty far just knowing those things that you have probably already heard, those things that sound crazy: It's a "function" that is 0 everywhere except at 0 where it's $\infty$. If you integrate it over any interval that includes 0, the result is 1.

If the presence of the delta function bothers you when you see an expression like $g(x,y)=f(x)\delta(x-y)$, just think of it as a strange way to write

$$\int_{-\infty}^{\infty} g(x,y) dx = f(y)[/itex] If you want to go beyond that, you need to learn about distributions. 3. Oct 6, 2008 HallsofIvy First of all, the "dirac delta function" is not a true function. It is an example of what mathematicians refer to as a "distribution" or "generalzed function". It is better to think of such things as "operators" ON functions: If f(x) is any function the dirac delta functions applied to it gives f(0): that is normally represented by $\int f(x)\delta(x)dx= f(0)$. One good way to think of a delta function is as the limit of a sequence of functions. Let f1(x)= 0 if x< -1, 1/2 if $-1\le x\le 1$, 0 if x> 1. f2[/sup](x)= 0 if x< -1/2, 1 if $-1/2\le x\le 1/2$, 0 if x> 1/2. In general, fn= 0 is x< -1/n, n/2 if $-1/n\le x\le 1/n$, 0 if x> 1/n. Notice that the integral of each of those, from $-\infty$ to $\infty$ is just (n/2)(1/n- (-1/n))= (n/2)(2/n)= 1. The Dirac delta function is the "limit" of those. I put "limit" in quotes because, in the usual sense, the functions {fn(x)} have no limit. We think of those as defining the Dirac delta function which, as I said, is not a true function. 4. Oct 6, 2008 maximummman Notice that the integral of each of those, from -[tex]\infty$$ to $$\infty$$is just (n/2)(1/n- (-1/n))= (n/2)(2/n)= 1.

who did you do this integration?

Last edited: Oct 7, 2008
5. Oct 7, 2008

6. Oct 7, 2008

Staff: Mentor

I'm sorry, I don't understand this question. I guess English isn't your first language, but try again using different words, or more words.

7. Oct 7, 2008

Fredrik

Staff Emeritus
He means "How did you do the integration?".

8. Oct 7, 2008

Staff: Mentor

It's amazing how much difference it makes when you switch two letters! :uhh:

9. Oct 7, 2008

samalkhaiat

10. Oct 10, 2008

maximummman

i am sorry for my spilling mistake

i am Egyptian

i mean that " could you describe for me the steps of this integration?"

i wish my spelling in this time is correct

and thank you :D:D:D:):):)

11. Oct 10, 2008

Lojzek

He was integrating a constant function (which has the shape of a rectangle), so the integral is just the area of the rectangle (=width*high=n/2*(2/n)).
Instead of a formal definition I will tell you a practical conception of delta function:
It is a function that is nonzero on a very small interval, which is so short that it can be aproximated by zero. Delta function is also so high that it can be aproximated by infinity (on the thin interval where it is nonzero) . The only exception to those two rules is a case when the area under the curve of delta function is in question. The area is defined to be 1. Delta function can have any shape (as long as it meets the 3 previous requirements), the easiest way is to imagine a very high and very thin rectangle: for example one with the hight of n and the width of 1/n for a very high n.

12. Oct 10, 2008

ice109

someone who is good at math tell me if the the dirac delta is a limit of sequence of a functions and which space is the "smallest" such that the limit converges?

13. Oct 10, 2008

Hurkyl

Staff Emeritus
Yes.... but only if you first reinterpret the functions as being distributions. (Assuming you are using Schwartz distributions as the mathematical foundation) Without such a reinterpretation, the limit (obviously!) doesn't exist.

The space that contains nothing except the functions in your sequence and the dirac delta.

14. Oct 10, 2008

ice109

a distribution being any function which i can multiply with a well behaved functin?
cute. how about a nontrivial space then. my point is does can we define the dirac delta and other such strange functions similarly to how we define real numbers.

15. Oct 10, 2008

Hurkyl

Staff Emeritus
More accurately, (in the usual notation) one that you can "multiply by a well-behaved function and then integrate".

Of course, for the usual definitions of 'well-behaved function', it is straightforward enough to define a product of the form
{distribution} * {test function} = {distribution}​
Because you can integrate {distribution} * {test function} * {test function}

The space of Schwartz distributions is very good for the purposes of quantum physics -- in particular, Fourier transforms always exist for Schwartz distributions.

16. Oct 10, 2008

HallsofIvy

The Polish Mathematician Mikusinki defined distributions in terms of sequences of function- the sequences themselves do not converge to functions but you can then define distributions as "equivalence classes" in much the same way you can define the real numbers as equivalence classes of sequences of rational numbers. He then showed that they have exactly the same properties as Schwarze's definition.

17. Oct 10, 2008

ice109

o rly? then i proclaim myself a genius for coming up with that