Question regarding the Dirac delta function

[Kevin McHugh]

Given the definition:

δ(x) = 0 for all x ≠ 0
∞ for x = 0

∫_-∞^∞δ(x)dx = 1

I don't understand how the integral can equal unity. The integral from -∞ to zero is zero, and the integral from 0 to ∞ is zero. How can one integrate the discontinuity at zero to equal one? What am I missing? TIA for your insight.

 

[jtbell]

The Dirac delta isn't really a function in the strict mathematical sense. One way to think of it is as the limit of a sequence of functions that becomes more and more sharply peaked. For example, consider the function f(x) = 0 for x < -a; 1/2a for -a ≤ x ≤ +a; 0 for x > +a. Its graph is a rectangle with area (integral) = 1. Now imagine making a smaller and smaller. The rectangle becomes a narrower and narrower and taller and taller "spike." Now take the limit as a --> 0.

It turns out you can make this work in a mathematically well-defined way, so long as you keep this "function" inside an integral, and you can derive properties like $$\int^{+\infty}_{-\infty} f(x)\delta(x) dx = f(0)$$ and $$\int^{+\infty}_{-\infty} f(x)\delta(x-a) dx = f(a)$$ Maybe a mathematician will come along and make this description more precise, but I think this is the way most physicists think of it.

 

[stevendaryl]

Given the definition:

δ(x) = 0 for all x ≠ 0
∞ for x = 0

∫_-∞^∞δ(x)dx = 1

I don't understand how the integral can equal unity. The integral from -∞ to zero is zero, and the integral from 0 to ∞ is zero. How can one integrate the discontinuity at zero to equal one? What am I missing? TIA for your insight.

There are two different ways to think of the delta function: (1) As a limit of functions, and (2) as a distribution.

As a limit of functions:
One approach to dealing with the delta function that almost always works is to think of it as an ordinary function that has a parameter, $W$ that in some sense measures how sharp the function is. The delta function is sort of (in a sense to be explained) a limit as $W \rightarrow 0$

Some examples:

1. $\delta(x) \approx \frac{1}{W \sqrt{\pi}} e^{-\frac{x^2}{W^2}}$
2. $\delta(x) \approx \frac{1}{\pi} \frac{sin(\frac{x}{W})}{x}$
3. $\delta(x) \approx \frac{1}{2W}$, if $-W \leq x \leq +W$), and zero otherwise.

In all three cases, the function tends to zero when $|x| \gg W$, and the integral over all space is equal to 1.

These functions don't actually have a limit as $W \rightarrow 0$, but in most cases of interest, integrals involving those functions do have limits as $W \rightarrow 0$, so you can solve the problem under the assumption that $W$ is small, but nonzero, and then let $W \rightarrow 0$ at the end.

As a distribution:
This is the mathematically rigorous way to reason about delta functions.

Let $\mathcal{H}$ be a set of square-integrable complex-valued functions of the real numbers. Then we can say that $F$ is a "distribution" for $\mathcal{H}$ if

1. $F$ takes any element of $\mathcal{H}$ and returns a complex number. That is, if $f$ is a function in $\mathcal{H}$, then $F(f)$ is a complex number.
2. $F$ is linear: For any two functions $f(x)$ and $g(x)$, and for any complex number $\alpha$, if $h(x) = f(x) + \alpha g(x)$, then $F(h) = F(f) + \alpha F(g)$.

So any function satisfying 1 and 2 is a distribution. We can then use integrals to define some particular distributions: For any function $\phi$ in $H$, there is a corresponding distribution $F_\phi$ defined via:

$F_\phi(f) = \int_{-\infty}^{+\infty} \phi^*(x) f(x) dx$ (where $^*$ means complex-conjugate)

But there are distributions that don't correspond to any actual function $\phi$. For example:

$F(f) \equiv f(0)$

That's a distribution, since it takes any function and returns a complex number, and it's linear. But it doesn't correspond to $F_\phi$ for any actual function $\phi$. But we can pretend that it is of that form, by writing: $F(f) = \int_{-\infty}^{+\infty} \delta(x) f(x) dx$. The right-hand side is just suggestive notation; it's not really an integral, although for many purposes, it works as if it were an integral.

 

[micromass]

@stevendaryl Why would you take your test space $\mathcal{H}$ as the square integrable functions? That doesn't work at all. The Dirac delta function would be ill-defined. The distributions you get by taking $\mathcal{H} = L^2$ are just the elements of $L^2$ again.

 

[Kevin McHugh]

Thanks Steven, I did read the wiki definitions before I asked the question. I think where i am having difficulty is at x = 0. By definition, f(x) is infinity at zero. I see that as a discontinuity. There are no limits of integration. Where is the Δx? The limit of a function view implies there is a Δx→ 0. (Or the W parameter in the example). I might be being pig headed about this.

 

[Mark Harder]

Yeah, you are being a tad stubborn Kevin. The delta function is not defined as infinity at x=0 and zero elsewhere. Like you say, a function defined that way is a discontinuity, and it's integral is not defined.
Instead, one definition is that the area under the delta function, ∫ δ(x) dx = 1. (Imagine the integral is from -∞ to ∞. I can't get the limits to print right.)
Another definition is that:

_-∞∫^∞ f(x) δ(x) dx = f(0) for any suitable f(x).

The way I see it, according to the 1st definition, ∫ δ(x) dx = ∫ 1⋅δ(x) dx = 1. The latter puts the integral in the form required by the 2nd definition as long as f(x) = f(0) = 1 for all x. In other words, ∫ f(x)⋅δ(x) dx = f(0) = 1. The 2 definitions are equivalent.

 

[strangerep]

Note to potential responders: more advanced aspects of this topic have been forked off into this new thread.

 

[vanhees71]

Yeah, you are being a tad stubborn Kevin. The delta function is not defined as infinity at x=0 and zero elsewhere. Like you say, a function defined that way is a discontinuity, and it's integral is not defined.
Instead, one definition is that the area under the delta function, ∫ δ(x) dx = 1. (Imagine the integral is from -∞ to ∞. I can't get the limits to print right.)
Another definition is that:

_-∞∫^∞ f(x) δ(x) dx = f(0) for any suitable f(x).

The way I see it, according to the 1st definition, ∫ δ(x) dx = ∫ 1⋅δ(x) dx = 1. The latter puts the integral in the form required by the 2nd definition as long as f(x) = f(0) = 1 for all x. In other words, ∫ f(x)⋅δ(x) dx = f(0) = 1. The 2 definitions are equivalent.

The first definition above is not correct since the constant function is not an appropriate test function. The second one is correct. You always need to use an appropriate test function with the $\delta$ distribution under the integral.