Deriving the Dirac Delta Function Equation in Field Theory

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Homework Help Overview

The discussion revolves around the derivation of an equation involving the Dirac delta function in the context of field theory. Participants are examining the implications and definitions related to the delta function, particularly in relation to products of distributions.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the definition of the delta function and its properties, questioning the validity of the equation presented. Some express confusion over the context in which the equation was derived, while others note the challenges in defining products of distributions consistently.

Discussion Status

The discussion is active, with participants sharing insights and raising questions about the equation's context and the nature of delta functions. Some guidance has been offered regarding the interpretation of delta functions and their properties, but there is no explicit consensus on the validity of the equation.

Contextual Notes

Participants mention that products of distributions like the delta function are problematic and that the equation may arise in specific contexts, such as scattering amplitudes in particle physics. There is also a reference to a specific textbook for further context.

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.
 
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?
 
From what I know (very slightly!), products of distributions like the delta "function" cannot be defined in a consistent manner.
 
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!
 
delta(0) can sometimes be taken to mean the volume of space (with appropriate 2pi factors) in relating delta function normalization with box normalization.
 

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