Square root of the delta function

[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?

 

[PeroK]

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

 

[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.

 

[jk22]

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

 

[zwierz]

∫∞−∞f(x)√δ(x−a)dx​

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

How is this integral evaluated?

not every sequence of mathematical symbols makes sense

 

[jk22]

You could aslo think it in the functional sense : $\int\delta(x-y)f(y)dy=f(x)=:\Delta_x(f)$

Then the root of delta would be a semi evaluation of f , which you could compute by semi integrating, using Cauchy's formula for repeated integration extended to real values of fractional calculus :

$\sqrt{\Delta_x}(f)=\frac{1}{\Gamma(1/2)}\lim_{y\rightarrow\infty}\int^y\delta(x-y)f(t)(y-t)^{-1/2}dt$

 

[Keith_McClary]

https://mathoverflow.net/questions/235827/square-root-of-dirac-delta-function
https://physics.stackexchange.com/q...s-the-square-root-of-the-dirac-delta-function

 

[jk22]

Error : my last formula is wrong since it contains a function depending on the limits of integration hence Cauchy's formula for repeated integral cannot be so easily used

 

[anuttarasammyak]

I have never met such a formula before. If it can exist, I suppose
\int f(x)\sqrt{\delta(x-a)}dx=\int \frac{f(x)}{\sqrt{\delta(x-a)}}\delta(x-a)dx=\frac{f(a)}{\sqrt{\delta(0)}}=0

 

[wrobel]

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

and then check that what it is happened does not depend on a sequence

 

[jasonRF]

EDIT: I just realized that this thread is three years old...

To me it seems like $\sqrt{\delta(t)}$ probably doesn't exist. Here is my thinking (this is not a proof).

First, as suggested by PeroK, start with one particular sequence of finite spikes that converges to a delta function. Let $p_n(t) = n$ for $-\frac{1}{2n}\leq t\leq\frac{1}{2n}$, and zero everywhere else. If $\sqrt{\delta(t)}$ exists, then I would hope that the sequence $\sqrt{p_n(t)}$ would converge to it. Let $\phi$ be a continuous test function, then
$$
\begin{eqnarray*}
\lim_{n\rightarrow\infty}\int \phi(t) \sqrt{p_n(t)} & = &
\lim_{n\rightarrow\infty} \sqrt{n}\int_{-1/2n}^{1/2n} \phi(t) dt \\
\lim_{n\rightarrow\infty} \sqrt{n}\,\phi(\tau)\int_{-1/2n}^{1/2n} dt
\end{eqnarray*}
$$
for some $\tau \in [-\frac{1}{2n}, \frac{1}{2n}]$. Then obviously we find that the limit is zero. So we have $\sqrt{p_n(t)}\rightarrow 0$, even though $p_n(t)\rightarrow \delta(t)$. This suggests that either $\sqrt{\delta(t)}=0$ or it doesn't exist.

Now I want to look at the product of $\sqrt{\delta(t)}$ with itself. We usually don't talk about multiplying distributions since in general the product of two distributions is not a distribution. However, multiplication of two distributions is allowed in those cases where the product is a distribution. In this case, if $\sqrt{\delta(t)}$ is a distribution then obviously it can be multiplied by itself to give the delta distribution. Also, if $\sqrt{\delta(t)}$ is a distribution then it has a Fourier transform that we will denote by $\mathcal{F}\left[\sqrt{\delta(t)}\right]=F(\omega)$. Now we can use the convolution theorem for Fourier transforms $$
\begin{eqnarray*}
1 & = & \mathcal{F}\left[{ \delta(t)}\right] \
& = & \mathcal{F}\left[\sqrt{ \delta(t)}\sqrt{ \delta(t)}\right] \
& = & \frac{1}{2\pi}F(\omega)\ast F(\omega).
\end{eqnarray*}
$$
I don't know whether a distribution $F(\omega)$ exists that is a constant when convolved with itself. If $F(\omega)$ doesn't exist, then neither does $\sqrt{\delta(t)}$; if it does exist then clearly $F(\omega)\neq 0$. However, this seems to contradict our earlier finding that suggests $\sqrt{\delta(t)}=0$ which would imply $F(\omega)=0$.

So I suspect that $\sqrt{\delta(t)}$ doesn't exist. Or if it does exist, it doesn't mean what the notation suggests it should mean.

jason

 

[TheDS1337]

$$\delta(x)$$ is not a function but a distribution according to Distribution Theory, I do not think of any possible case of finding it under a squareroot?

 

[wrobel]

About multiplication of generalized functions see Colombeau theory of generalized functions

 

[jasonRF]

About multiplication of generalized functions see Colombeau theory of generalized functions

Do you know of any references on Colombeau theory that are appropriate for non-mathematicians? It looks intriguing, but the references that popped up with a quick google search seem to require a little more mathematical knowledge/sophistication than I have. There are many "simple" treatments of the Schwartz theory of distributions - such as the book by Strichartz (where I first learned it). Is there anything equivalent for the Colombeau theory, or does it inherently require more tools to properly explain?

thanks,

Jason

 

[anuttarasammyak]

Supplement to post #9

As eigenfunction of x, $\delta(x)$ have the relation
f(x) \delta(x) = f(0) \delta(x)

If we allow f be not only normal function but also distribution $\delta^{a}(x)$
\delta^{a}(x)\delta(x)=\delta^{a}(0)\delta(x)

Let a=-1/2
\delta^{1/2}(x)=\delta^{-1/2}(0)\delta(x)
so it is delta function times the coefficient $\delta^{-1/2}(0)=0$. So
\delta^{1/2}(x)=0 in the similar sense with
x\delta(x)=0
when multiplied by normal function f of f(0) $\neq \infty$ and integrated.

 

[wrobel]

Do you know of any references on Colombeau theory that are appropriate for non-mathematicians? It looks intriguing, but the references that popped up with a quick google s

Long ago I saw the original Colombeau's monography The new generalized functions and multiplications of distributions. I believe it is appropriate.