# Dirac Delta Function question(s)

1. Apr 21, 2008

### Crazy Tosser

OK, so my basic understanding of Dirac Delta Function is that it shows the probability of finding a point at $$(p,q)$$ at time $$t$$. Dirac Delta is 0 everywhere except for $$(p_{0},q_{0})$$.

So my question comes
Is it possible that a point enters the $$(p_{0},q_{0})$$ and stays there (for some period of time)? Then dirac's delta would have a chunk instead of one point missing, which does not match it's definition.

Correct me if I am wrong please

Last edited: Apr 21, 2008
2. Apr 21, 2008

### Andy Resnick

My understanding for the delta function is that it represents a point source. A point source of *what* depends on the application. But, when solving (for example) an inhomogeneous ODE, the use of Green's functions and the Delta function are a simple way of building up a complete solution.

3. Apr 21, 2008

### lzkelley

The probability of finding a particle at some point will be described by some sort of probability distribution function. The dirac delta function (d(x)) is a function that is zero everywhere, except at 0 (or a shifted origin) where its infinite. It ussually describes a point source, as aforementioned. For instance, in the classical point-particle case the density could be described by a delta function (i.e. zero everywhere except infinite at the point where the particle is).
You can also create a delta function that is zero everywhere, and finite at a certain point... this occassionally pops-up in a prob. dist. function... can't think of a particle case though.

4. Apr 21, 2008

### rbj

problem is that one integrates to nothing. from a physical POV, it's virtually nothing.

5. Apr 21, 2008

### lzkelley

rbj, thats really not true at all. the integral of a deltafunction is always finite, which you can see in practice - but fundamentally its true by definition.
At the same time its far from "virtually nothing" because a single source (described by a delta function) can account for all of the phenomena in a given region.

6. Apr 21, 2008

### nicksauce

Actually it is true. If f(x) = {0; x!=x0, C; x = x0} for some real number C, then the integral from -infinity to infinity of f(x) is 0.

7. Apr 21, 2008

### lzkelley

8. Apr 21, 2008

### Crazy Tosser

Well, from the book I read, the dirac delta function is described differently. let me type an approximation here (I have to translate it while I type):

"
Gibbs' ensemble can be represented as a 2-Dimensional cloud of points, which can be described by the function $$\rho(q,p,t)$$. It has a simple physical interpretation: it's nothing but the probability of finding a point at time $$\displaystyle{t}$$ with the coordinates $$\displaystyle{q,p}$$. Thus, the function $$\displaystyle{\rho}$$ imples being zero at all values of $$\displaystyle{q,p}$$, except for one case $$q_{0},p_{0}$$. Functions having the property that they are zero everywhere except for one point are called Dirac Delta Functions and are labeled $$\displaystyle{\delta(x)}$$. Function that we need in this case, $$\delta(x-x_{0})$$ equals zero everywhere except for $$x=x_{0}$$. Thus, for time $$\displaystyle{t}$$=0, the probabilty $$\displaystyle{\rho}$$ has the format $$\rho=\delta(q-q_{0})\delta(p-p_{0})$$.
"

And by the way

$$\displaystyle\int^{\infty}_{-\infty} \delta(x)\,dx = 1$$

Last edited: Apr 21, 2008
9. Apr 21, 2008

### lzkelley

That is definitely the dirac delta function. at the same time, rho isn't the probability, because at the point qo,po, its infinite instead of one (one is probability version of "hell yes, definitely" of course).
rho, in this case, is the probability density function; i.e. if you integrate it over some amount of space -> you find the probability of finding it in that space.
If you integrate this specific rho over any space that includes q0 p0, you get 1. i.e. if you are looking at any space that includes the point q0,p0, you are certain to find the particle.
Does that clear up things?

10. Apr 21, 2008

### Crazy Tosser

oooh
darn
I've been stupid.
Thanks, it all fits together into my world now! D=

11. Apr 22, 2008

### Andy Resnick

Delta functions of finite height (?) are used a lot in optics- the magnitude of the delta functions are correlated with amplitudes of spatial frequencies. Also, the height can be correlate with intensity.

12. Apr 22, 2008

### DavidWhitbeck

Dirac delta is not a function and thinking about it as if it's a function is completely misleading.

It is a distribution, it is defined in terms of what it does when integrated against a function. The visual representation of it being a big spike is cute but I think ultimately misleading.

13. Apr 22, 2008

### rbj

no, Iz, the function that i commented on (one that is zero everywhere, but for one point where it takes on a finite value) has integral of zero. i don't even need to look up your references (i had some role in writing one of them).

from a strict mathematical POV, i guess you're right, David, but from an engineering POV, i know for certain that you are not. in fact, ton's of textbooks are printed defining the dirac delta as such a limiting function of some "nascent delta" with unity area that is getting thinner and thinner and, in the limit, becomes virtually zero width. when used to sample decent and continuous functions, this "misleading" manner of thinking about is just fine and better, IMO (and i used to teach electrical engineering in years past) than trying to confuse a bunch of sophmores or juniors with the rigorous mathematical definition because they're gonna think of it as a big spike anyway.

i also think that where the dirac delta is used in Newtonian mechanics, the infinitely thin spike with finite and non-zero area is just fine also.

14. Apr 22, 2008

### nicksauce

15. Apr 22, 2008

### reilly

A delta function is a so-called generalized function or distribution. It is not a function by usual standards, but is the limit of normal function, when it is part of an integrand. For example, a delta function is, among many alternatives, the limit of a Gaussian probability distribution as the standard deviation approaches zero.

In quantum mechanics, physicists play fast and loose with delta functions, particularly with products of delta functions. Thus thinking in terms of Gaussian helps greatly.

There's an excellent book by Lighthill, Generalized Functions and Fourier Analysis, somewhat older. But worth the trouble to find and read -- it is at an advanced undergraduate level. He defines, as do most authors on the subject, generalized functions as the limits of ordinary functions. rbj is right except for the "misleading" part.

If you understand things at the level of Lighthill, then looking at a delta as a narrow Gaussian is quite proper, as long as it is used as part of an integrand.

In E&M, a delta function is used as a charge distribution of a point particle -- useful fiction. Further, it is almost always used as the source for Green's Functions -- the Green's Functions obtained this way are identical to those obtained in the old fashioned way, with use of jump conditions -- see any older text on E&M.

When I taught Advanced QM and E&M, I defined a delta function as the derivative of a so-called theta function --
THETA(X) = 0 if X<0; 1 if X>0;
some authors define THETA(0) =1/2.
This is an approach quite common in physics.

Regards,
Reilly Atkinson

16. Apr 22, 2008

### Andy Resnick

Just becasue it may not have orthodox mathematical properties does not make it less useful for physical representation of certain physical phenomena.

17. Apr 22, 2008

### DavidWhitbeck

Well I have a problem with both parts of what you said. I did not say that it was useless for representing physical phenomena. Nor did I say that it was unorthodox, I said that it was not a function, as in real valued function, which is not the same thing as unorthodox.

I know every one is saying that they usually learned the Dirac Delta to be defined as a limit of a sequence of functions, and that's the first time I saw it too... but it's a pretty stupid way to define it.

Simply defining it by what it does to functions makes much more sense because it's a practical definition, it's telling you exactly what it does. The Dirac Delta is a measure and saying it works by $$\int \delta(x)f(x)=f(0)$$ is much more simple and direct than considering a sequence of functions with an ill-defined limit.

18. Apr 22, 2008

### lzkelley

if the references say, explicitly that the integral of the delta function is non-zero, i think that supports my claim that the integral of the delta function is non-zero.
Additionally, its true even for a finite delta function; i.e. f(x) = C*(del(x)) [constant C times the delta function of x], then the integral is again non-zero, in this case C.
Again, this is exactly what the references show. For a more academic reference see Boas' Math Methods, or any other math methods book.

19. Apr 22, 2008

### arildno

20. Apr 22, 2008

### Hurkyl

Staff Emeritus
Functions have a value at zero.

Ordinary limits of functions do not satisfy the equation
$$\int_a^b \left( \lim_{t \rightarrow c} f_t(x) \right) g(x) \, dx = \lim_{t \rightarrow c} \left( \int_a^b f_t(x) g(x) \, dx\right)$$

What is to be gained by insisting upon having students think that a dirac delta distribution as something that has a value at zero, or is equal to an ordinary limit of functions?