Proving some Dirac-Delta identity that uses Laplacian

[davidbenari]

Homework Statement

Prove that ##(x^2+y^2+z^2)\nabla^2[\delta(x)\delta(y)\delta(z)]=6\delta(x)\delta(y)\delta(z)##

Homework Equations

##\delta''(x)/2=\delta(x)/x^2##

The Attempt at a Solution

I have obtained this:

##6\delta(x)\delta(y)\delta(z) + (xy)^2/2\Bigg(\delta''(y)\delta''(z)+\delta''(x)\delta''(z)\Bigg)+(xz)^2/2\Bigg(\delta''(y)\delta''(z)+\delta''(x)\delta''(y)\Bigg)+(yz)^2/2\Bigg(\delta''(x)\delta''(z)+\delta''(x)\delta''(y)\Bigg)##

This looks like a huge mess, hehe, sorry.

Any ideas on how to make disappear those ugly terms and only remain with the first one?

Thanks.

 

[Delta2]

Whats the identity for [itex]x\delta(x)[/itex]?

 

[davidbenari]

##x\delta(x)=0## Is this a hint or were you asking me this?

 

[Delta2]

yes, it was meant as a hint, your original expression seems a mess indeed, just do only the term [itex]\frac{\partial^2}{\partial x^2}[/itex] of the laplacian and you see easily that it contributes [itex]2\delta(x)\delta(y)\delta(z)[/itex]

 

[davidbenari]

Oh! What a dope am I. Thanks hehe!

 

[davidbenari]

Wait, sorry. But those terms are divided by (e.g. the x term) ##x^2## So I would get in total that ##(x^2+y^2+z^2)\nabla^2 (\delta(x)\delta(y)\delta(z))=2\delta(x)\delta(y)\delta(z)(x^-2+y^-2+z^-2)(x^2+y^2+z^2)##

 

[Delta2]

yes, expand it, one term of the expansion will be (for x^-2) [itex]2\delta(x)\delta(y)\delta(z)(1+\frac{y^2+z^2}{x^2})[/itex]. Now you can see that for example that [itex]\frac{y^2}{x^2}2\delta(x)\delta(y)\delta(z)[/itex] is zero .

 

[davidbenari]

Thanks Delta^2 I think I've got it now.