# Dirac delta function

1. Nov 29, 2006

### touqra

I found this equation in a field theory book, which I can't figure how it was derived:

$$\delta(x-a) \delta(x-a) = \delta(0) \delta(x-a)$$

2. Nov 29, 2006

### OlderDan

From the definition of the delta function

$$\int_{ - \infty }^\infty {f(x)\delta (x - a)dx = f(a)}$$

you get

$$\int_{ - \infty }^\infty {\delta (x - a)\delta (x - a)dx = \delta (a - a)} = \delta (0)$$

I expect this is part of the answer.

3. Nov 29, 2006

### Hurkyl

Staff Emeritus
Ugh. If that's supposed to be the dirac delta distribution, then both sides of that equation are gibberish. What is the context in which you saw it?

4. Nov 30, 2006

### arildno

From what I know (very slightly!), products of distributions like the delta "function" cannot be defined in a consistent manner.

5. Nov 30, 2006

### touqra

I was reading a section dealing with cross sections and scattering. He calculated some amplitude, A (first order) for a Feynman diagram which contains four 4-momentum delta functions. And with that amplitude, we need to get this invariant amplitude, iM which is the square of A.
Squaring A yields us 8 delta functions.
He states that 8 delta functions is bad news, and basically he gave a simple example, which was the one I posted in this forum.

It's from Gauge Theories in Particle Physics Volume I by I J R Aitchison, page 152.

6. Nov 30, 2006

### Tomsk

Equations such as $$f(x) \delta(x-a) = f(a)$$ are supposed to be read as $$\int_{-\infty}^{\infty} f(x) \delta(x-a) dx= f(a)$$, I think dropping the integral sign is just some sort of convention, not one I'm a fan of though... I think people keep swapping limits and integral signs too, I think things like that should be made more consistent, as you can't do stuff like that in general.

But yeah don't forget the integral sign!

7. Nov 30, 2006

### Meir Achuz

delta(0) can sometimes be taken to mean the volume of space (with appropriate 2pi factors) in relating delta function normalization with box normalization.