Square root of Dirac Delta function

[ismaili]

Homework Statement

I wonder how to deal with the square root of Dirac Delta function, $$\sqrt{\delta(x)}$$. Actually, this comes from a problem which asking readers to calculate the wave function of a free particle and of a harmonic oscillator at time $$t$$, provided that the wave function at time $$t=0$$ is given, i.e. $$\psi(x,t=0) = \sqrt{\delta(x-a)}$$.

The way to obtain $$\psi(x,t)$$ in this problem is by the integral with the propagators.
However, I have no idea how to deal with $$\sqrt{\delta(x)}$$.

Homework Equations

$$\sqrt{\delta(x)}=?$$

The Attempt at a Solution

I tried to differentiate it,
$$\frac{d}{dx}\sqrt{\delta(x)} = \frac{1}{2\sqrt{\delta(x)}}\delta'(x)<br /> = -\frac{1}{2\sqrt{\delta(x)}}\frac{\delta(x)}{x} = -\frac{1}{2}\frac{\sqrt{\delta(x)}}{x}$$
$$\Rightarrow <br /> \int \sqrt{\delta(x)}f(x)dx = -2\int xf(x)d(\sqrt{\delta(x)})$$
But this is still no good...

I also tried the other definition of Dirac Delta function,
$$\delta(x)\sim\lim_{\epsilon\rightarrow 0}e^{-x^2/\epsilon}$$
$$\Rightarrow \sqrt{\delta(x)} = \lim_{\epsilon\rightarrow 0}e^{-x^2/2\epsilon}<br /> =\lim_{\epsilon'\rightarrow 0}e^{-x^2/\epsilon'} = \delta(x) ??$$

still no good...

I also tried to calculate $$\psi(x,t)^2$$ to try to get rid of the square root,
but it seems doesn't help.

Is there anyone who has any ideas about $$\sqrt{\delta(x)}$$ ?
Any help will be appreciated, thanks.

 

[ismaili]

Homework Statement

I wonder how to deal with the square root of Dirac Delta function, $$\sqrt{\delta(x)}$$. Actually, this comes from a problem which asking readers to calculate the wave function of a free particle and of a harmonic oscillator at time $$t$$, provided that the wave function at time $$t=0$$ is given, i.e. $$\psi(x,t=0) = \sqrt{\delta(x-a)}$$.

The way to obtain $$\psi(x,t)$$ in this problem is by the integral with the propagators.
However, I have no idea how to deal with $$\sqrt{\delta(x)}$$.

Homework Equations

$$\sqrt{\delta(x)}=?$$

The Attempt at a Solution

I tried to differentiate it,
$$\frac{d}{dx}\sqrt{\delta(x)} = \frac{1}{2\sqrt{\delta(x)}}\delta'(x)<br /> = -\frac{1}{2\sqrt{\delta(x)}}\frac{\delta(x)}{x} = -\frac{1}{2}\frac{\sqrt{\delta(x)}}{x}$$
$$\Rightarrow <br /> \int \sqrt{\delta(x)}f(x)dx = -2\int xf(x)d(\sqrt{\delta(x)})$$
But this is still no good...

I also tried the other definition of Dirac Delta function,
$$\delta(x)\sim\lim_{\epsilon\rightarrow 0}e^{-x^2/\epsilon}$$
$$\Rightarrow \sqrt{\delta(x)} = \lim_{\epsilon\rightarrow 0}e^{-x^2/2\epsilon}<br /> =\lim_{\epsilon'\rightarrow 0}e^{-x^2/\epsilon'} = \delta(x) ??$$

still no good...

I also tried to calculate $$\psi(x,t)^2$$ to try to get rid of the square root,
but it seems doesn't help.

Is there anyone who has any ideas about $$\sqrt{\delta(x)}$$ ?
Any help will be appreciated, thanks.

I have an idea, but I'm not sure if it's correct.
Consider
$$f(0) = \int \delta(x)f(x)dx = \int \sqrt{\delta(x)}\sqrt{\delta(x)} f(x)dx<br /> = \int \sqrt{\delta(x)}\sqrt{\delta(x)} dg(x)$$
where
$$\frac{dg}{dx} = f$$
,then integration by part gives,
$$f(0) = -2\int g(x)\frac{\delta'(x)}{2\sqrt{\delta(x)}} dx <br /> = \int g(x) \frac{\sqrt{\delta(x)}}{x} dx$$

Now, let $$g(x) = xh(x)$$, we have,
$$f(0) = \int h(x)\sqrt{\delta(x)} dx$$
,i.e.

$$\int h(x)\sqrt{\delta(x)} dx = \left[h(x) + xh'(x)\right]_{x=0}$$

I'm not sure if I were correct.
Any help would be appreciated!

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Gosh, I found one mistake just now,
I calculated the integration by part wrong,
I omitted one factor of $$\sqrt{\delta(x)}$$
So the final formula is incorrect.