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
