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I have two quantities involving Dirac-Deltas, where the arguments are dependent:

$$A=\delta\left(a-\frac{b}{c}\right) \delta(a-b)$$ and $$B=\partial_a \delta\left(a-\frac{b}{c}\right) \delta(a-b)$$

As $A$ and $B$ are only nonzero when the arguments of the Deltas are zero, I tried to solve it this way: $a=b$ (from second Delta), thus $a=\frac{a}{c}$. Therefore I get $c=1$ as a restriction. I would rewrite $A$ then: $$A=\delta(c-1)$$

Is that correct? And can I handle the second quantity $B$ similarly?

Thanks!!

Peter Woolfitt
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Mario Krenn
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  • In what context are you running into products of delta functions? It will help us to understand whether you're dealing with them properly – Semiclassical Jul 25 '14 at 20:39
  • In the momentum-representation of some solutions of wave equations. That means, my a,b,c are certain momentum values in some coordinate-systems. (Or do you mean something else?) – Mario Krenn Jul 25 '14 at 20:40
  • That's fine. In 1 or 2 dimensions? – Semiclassical Jul 25 '14 at 20:41
  • the full problem is in 3 dimensions, but here I only have two of them (the other one can be factored out and is not important for the problem of the deltas I guess). – Mario Krenn Jul 25 '14 at 20:43
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    Hmm. The reason why I worry is that products of delta functions don't make much mathematical sense, and so I worry you've gotten to the wrong place. (See the question at [math.stackexchange.com/questions/12944/] for a discussion of that point.) – Semiclassical Jul 25 '14 at 20:47
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    The reason I asked about dimension is that things like $\delta(x-1)\delta(y)$ do make sense. But $\delta(x)\delta(x-1)$ doesn't. – Semiclassical Jul 25 '14 at 20:49
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    Thanks for the comment, thats interesting and useful - it means I should check very carefully how I arrived at this terms. the idea was to trasform something like $\delta(x-x0)\delta(y-y0)$ in a different coordinate-system. Maybe something went wrong there. – Mario Krenn Jul 25 '14 at 21:28

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