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 x\delta(x)?

 

[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 \frac{\partial^2}{\partial x^2} of the laplacian and you see easily that it contributes 2\delta(x)\delta(y)\delta(z)

 

[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) 2\delta(x)\delta(y)\delta(z)(1+\frac{y^2+z^2}{x^2}). Now you can see that for example that \frac{y^2}{x^2}2\delta(x)\delta(y)\delta(z) is zero .

 

[davidbenari]

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