# Square root of the delta function

• I
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
Homework Helper
Gold Member
You could try approximating the delta function by a sequence of finite spike functions and see what happens.

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.

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

∫∞−∞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

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##

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

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Delta2
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
Gold Member
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

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wrobel
$$\delta(x)$$ 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
jasonRF
Gold Member
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
Gold Member
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.

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wrobel