# I Square root of the delta function

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1. Jan 30, 2017

### junt

Is square root of delta function a delta function again?

$$\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx$$

How is this integral evaluated?

2. Jan 30, 2017

### PeroK

You could try approximating the delta function by a sequence of finite spike functions and see what happens.

3. Mar 23, 2017

### jk22

This is actually related to Bell's theorem. It is not the solution to the OP : searching A s.t. $$\int A (a,\lambda)A (\lambda,b)d\lambda=-cos (a-b)$$

If we make the assumption $$A (a,\lambda)=f (a-\lambda)$$ and let $$u=a-\lambda$$ we get a convolution :

$$\int f (u)f (a-b-u)du=-cos (a-b)$$

Using the Fourier convolution theorem :

$$2F^2 (\omega)=-\delta (\omega+1)-\delta (\omega-1)$$

Hence the searched function f is given by

$$f (x)=\frac {i}{\sqrt {2}}\int e^{ikx}\sqrt {\delta (k+1)+\delta (k-1)}$$

Bell's theorem shows that there is no such f.

Last edited: Mar 23, 2017
4. Mar 26, 2017

### jk22

Addendum : no such f in the range -1 to 1

5. Mar 26, 2017

### zwierz

not every sequence of mathematical symbols makes sense