What is the square root of the Dirac Delta Function?

You can think of the Dirac distribution as being an element in some operator valued Banach algebra. One can then use the Riesz calculus (http://en.wikipedia.org/wiki/Holomorphic_functional_calculus) to define an arbitrary holomorphic function of this Dirac delta (or more general distributions).

In particular (much like the Cauchy-Integral Formula) we have, for a Banach algebra $\mathscr{A}$ and holomorphic function $f$, for $a \in \mathscr{A}$:

$f(a) = \frac {1} {2 \pi i} \oint \frac {f(z)} {z - a} d z$ ;

This can in principle be evaluated. For example, you can represent the Dirac delta as a matrix (if you are thinking about finite dimensions). The integration is then fairly straightforward (for the square-root you would set $f(z) = z^{1/2} = e^{\frac {1} {2} \log(z)}$). You will have to be careful about branch cuts for this particular case.

As for quantum wave functions, I could not say. However, this tool is very useful to define Dirac operators. For example, you could consider the algebra $\mathscr{A}$ to consist of differential operators of some dimension. Taking the Klein-Gordon operator $\nabla^2 + m^2$ one can attempt to define square-roots of this operator by evaluating the above integral.

The space of locally integrable functions is dense in the space of distributions in the topology in which a sequence of functions $(f_n)$ converges to a distribution $T$ if the integrals $\int f_n\phi dx$ converge to $T(\phi)$ for all test functions $\phi$, typically the compactly supported smooth functions. In this sense the Dirac distribution is defined by $\delta(\phi) = \phi(0)$, often written as $\int f(x)\delta(x) dx = f(0)$.

This means that to evaluate a distribution, we could take a sequence of ordinary functions converging to it, and compute the limit. In our case we could take a sequence of positive functions converging to the Dirac delta, take their square roots, and evaluate the limits. It looks to me like this is identically 0, not very useful.

That doesn't mean that in other generalized function spaces this couldn't be given a more useful meaning, see e.g. this article (pdf) (that I didn't read), with the title "Generalized Functions for Applications" and abstract

A simple rigorous approach is given to generalized functions, suitable for applications. Here, a generalized function is defined as a genuine function on a superset of the real line, so that multiplication is unrestricted and associative, and various manipulations retain their classical meanings. The superset is simply constructed, and does not require Robinson's nonstandard real line. The generalized functions go beyond the Schwartz distributions, enabling products and square roots of delta functions to be discussed.

I don't think the square root of the Dirac delta is well defined. if you have some distribution $g$ such that $g^2 = \delta$, that requires the wavefront set of $\delta$ to be a subset of the wavefront of $g$, meaning that the square root has at least the same wavefront set as the Dirac distribution, which would prevent it from being a well defined distribution.