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.