# Dirac function

1. Nov 11, 2005

### Ratzinger

As I read in my quantum mechanics book the delta function is sometimes called the sampling function because it samples the value of the function at one point.
$$\int {\delta (x - x')} f(x')dx' = f(x)$$

But then I opened a quantum field book and I found equations like that:
$$\phi (x) = \frac{1}{{(2\pi )^{3/2} }}\int {d^4 p\delta (p^2 } - m^2 )A(p)e^{ - ip \cdot x}$$

$$(\partial _\nu \partial ^\nu + m^2 )\phi (x) = \frac{1}{{(2\pi )^{3/2} }}\int {d^4 p( - p^2 } + m^2 )\delta (p^2 - m^2 )A(p)e^{ - ip \cdot x}$$

Can someone explain me what the delta function does here? What it is sampling? How these equations work?

thank you

2. Nov 11, 2005

### Hurkyl

Staff Emeritus
The delta function does the same thing!

I don't know if "sampling" is a good adjective, but I'll run with it. In the integral

$$\int d^4 p \, \delta (p^2 - m^2 )A(p)e^{ - ip \cdot x}$$

you're "sampling" the function $A(p)e^{ - ip \cdot x}$ over the region where $p^2 - m^2 = 0$.

We can reduce this down to the known case, if we like, as follows: we can split this up into an iterated integral, with $p_0$ being the innermost. Then we have:

$$\iiint \int \delta(p_0^2 - \vec{p}\,{}^2 - m^2) A(p) e^{-ip \cdot x} \, dp_0 \, d^3 \vec{p}$$

so the innermost integral is of the form:

$$\int \delta(x^2 - a^2) f(x) \, dx$$

Now, if I make the substitution $u = x^2$, we have $du = 2 x dx$ so that $dx = du / 2 \sqrt{u}$, and the integral becomes

(yes, I know I'm being lazy with the bounds -- note that when tracing x over $(-\infty, +\infty)$, u traces over $[0, +\infty)$ twice! I really should break the integral up into two parts)

$$\int \delta(u - a^2) f(\sqrt{u}) \frac{1}{2 \sqrt{u}} \, du$$

which becomes (one term for each of the times u traces over $0, +\infty)$:

$$f(a) \frac{1}{2a} - f(-a) \frac{1}{-2a} = \frac{f(a) + f(-a)}{2a}$$

I'll leave it to you to work out the general case when the argument to $\delta$ is an arbitrary function of the dummy variable.

Anyways, if my intuition about these things is anywhere close to accurate, your integral for $\phi(x)$ should reduce to a surface integral over the hypersurface given by the equation $p^2 - m^2 = 0$.

Last edited: Nov 11, 2005
3. Nov 11, 2005

### Norman

notice here that the delta function has as its argument quadratic functions. It can be a little confusing about what to do with these. Well just use this formula:
$$\delta[g(x)]=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}$$
by the way, the sum is over the roots of g.
In your equation above, after you use this step everything should be clear- it functions just how you would expect it to.
and check out this page for help with dirac delta functions
http://mathworld.wolfram.com/DeltaFunction.html

4. Nov 11, 2005

### Ratzinger

thanks Hurykl, thanks Norman, thanks physicsforums, thanks the Internet

and a great weekend to everybody