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  Dirac Delta of a Dirac Delta.

+ 1 like - 0 dislike
493 views

After watching a graduate course in Quantum Physics 2, and after hearing the lecturer saying a lot of times Dirac Delta.

It had occurred to me, can you have such a thing as a Dirac Delta of a Dirac Delta, etc?

Something like: $\delta(\delta(x-x_0))$, and $\delta^{n}(x-x_0):=\delta^{n-1}(\delta(x-x_0))$?

Does it have applications in physics?

How would you rigorously define such a composition of Dirac Deltas?

Another thing, would such a thing converge and in which sense would it converge if we let $n\to \infty$?

asked May 2, 2023 in Mathematics by MathematicalPhysicist (200 points) [ revision history ]

What should be the intended meaning of such a strange construct? A function in the dual of the space of smooth real-valued function on the space of distributions?

I myself am not sure if it has any physical application. I guess at the moment it's a pure mathematical ill-defined construct that needs quite a lot of work to make it rigorous.

I guess from your response that you aren't aware of any applications as of yet.

Thank you for your response, I need to find time to work on this.

In order to speak seriously of such a construction, you have to encounter it in some real calculations. Otherwise your question is an idle play with no applications.

I agree with Vladimir Kalitvianski that such an expression, should it indeed arise, is best disentangled in the context of the specific calculation wherein it occurs. I suspect that such expressions (or questions) at least in part are born out of writing $\delta$ as a function.

If such an expression has to be resolved, it may help to redo the relevant calculations using approximations to the "$\delta$-function", like $g$ a normalised Gaussian (or similarly behaved function) and considering sequences of functions $g_\varepsilon(x)=(1/\varepsilon)g(x/\varepsilon)$ with $\varepsilon\rightarrow0$ at the end of the calculation. With repeated $\delta$s there could be several independent parameters instead of only $\varepsilon$, and you have control over the order in which the limits are taken. 

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