What is the square root of the Dirac Delta Function?

What is the square root of the Dirac Delta Function? Is it defined for functional integrals? Can it be used to describe quantum wave functions?

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

Answer

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.