Deriving the Dirac Delta Function Equation in Field Theory

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

[tex]\delta(x-a) \delta(x-a) = \delta(0) \delta(x-a)[/tex]
 
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touqra said:
I found this equation in a field theory book, which I can't figure how it was derived:

[tex]\delta(x-a) \delta(x-a) = \delta(0) \delta(x-a)[/tex]

From the definition of the delta function

[tex] \int_{ - \infty }^\infty {f(x)\delta (x - a)dx = f(a)} [/tex]

you get

[tex] \int_{ - \infty }^\infty {\delta (x - a)\delta (x - a)dx = \delta (a - a)} = \delta (0)[/tex]

I expect this is part of the answer.
 
Hurkyl said:
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?

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.
 
Equations such as [tex]f(x) \delta(x-a) = f(a)[/tex] are supposed to be read as [tex]\int_{-\infty}^{\infty} f(x) \delta(x-a) dx= f(a)[/tex], 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!