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}

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    $\begingroup$ I've tried square roots of families of functions converging to $\delta$ and always found the square roots to converge to identical $0$. The value is probably non-unique, but I think it might be zero quite generally. $\endgroup$
    – Void
    Sep 15, 2014 at 16:56
  • $\begingroup$ In complete agreement with Void's answer. Take it as the square root of the normalized Gaussian with vanishing width: the width increases by sqrt of 2, but the normalization is attenuated so, also by dimensional analysis, fails to normalize the integral to one. The limits go to 0, that is, the integral vanishes when you take your f (x)=1. Of course you may take a properly normalized spiked Gaussian as a wavefunction. $\endgroup$ Apr 14, 2016 at 16:03
  • $\begingroup$ The momentum-space wavefunction corresponding to an (unnormalizable) plane wave exp( ikx ) is δ(p-k) with similarly terrible normalization. You might, in a moment of playfulness take the square root of the latter as the wavefunction itself, which appears normalized, only if it is vanishing as above. $\endgroup$ Apr 14, 2016 at 23:03
  • 1
    $\begingroup$ Related question on MO.SE: mathoverflow.net/q/235827/13917 $\endgroup$
    – Qmechanic
    Apr 21, 2016 at 18:22

5 Answers 5


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.

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    $\begingroup$ I know a little about Banach algebras and spaces of distributions, but I don't see how a Dirac distribution is in a Banach algebra. What spaces does the operator map from and to? How do you multiply dirac distributions? Could you elaborate more? Thanks. $\endgroup$
    – Nick Alger
    Sep 15, 2014 at 23:47
  • $\begingroup$ An example of what you are talking about is the exponential of an operator, well known to physicists. Here the evaluation with an operator as an argument is defined plugging it into the the power series of the function, which will converge under suitable circumstances. For a function of a complex variable that is holomorphic in a point $a$, the Cauchy integral formula holds. This again can be used as a way to evaluate a function in an operator under suitable conditions. However, as @NickAlger pointed out, unfortunately we are not in this situation... $\endgroup$
    – doetoe
    Sep 16, 2014 at 8:14
  • $\begingroup$ ... Our Dirac distribution is a linear functional on a function space, not an operator from the space to itself. Rather than a matrix, in the finite dimensional case it would be a row vector. $\endgroup$
    – doetoe
    Sep 16, 2014 at 8:15
  • $\begingroup$ Can you share more light on the banach algebra? Do you think this works for every distribution? Shouldnt this need to have 'solved' the distribution multiplication problem first? $\endgroup$
    – lalala
    Jun 24, 2021 at 7:55
  • $\begingroup$ So, according to all that you said, is the value of square root of delta function zero or not? $\endgroup$
    – Prem
    Dec 6, 2022 at 21:31

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.


I'll try to answer the question of what the square root of a delta function is by telling you what it can do in the real world.

I have used the idea of $\sqrt \delta $ to define power distributions of functions of infinite energy (the integral of the square over all reals), but finite power (energy per unit time). I'm sure others have too, but I have been unable to find any examples so far. An example of such a function is $\sin \omega t$. Its Fourier transform is two sharp deltas. The energy spectrum is undefined, since it requires squaring the Fourier transform. But it is really obvious what the power spectrum should look like—two deltas at $\pm \omega$. After much mathing and fiddling with limits, I always end up with this answer:

$\rho(\omega) = \frac 1 \tau |\sqrt \delta * F|^2,$

where $F$ is the Fourier transform of the function whose power spectrum it is I'm trying to find and $\tau$ is the morally superior circle constant. So in this usage, $\sqrt \delta$ used as a convolution kernel on an energy spectrum turns it into a power spectrum. Try it on $\sin \omega t$ and you'll end up with twin delta peaks and the right amount of total power. Try it on Gaussian white noise and you'll get a flat (but locally finite) power density out to infinity. It also works on a pulse train and on arbitrary periodic signals.

I have also much less fruitfully used the square root of the Dirac delta to define a square-integrable function concentrated at a point. Such a function could be used to describe the state of a quantum particle located at a point, even though it isn't strictly possible to localise a quantum particle in this way.


The way I would define $\delta^{-n}$ is by its nth power. $$ \int f(x) \left( \delta^{-n}(x) \right)^n \mathrm{d} x = f(0) $$

In this way you can take the wave function of a particle to be $ \psi(x) = \sqrt{\delta(x)} $ such that $$ \int \psi(x)^* \psi(x) \mathrm{d} x =1. $$ However, the state this represents has “infinite” kinetic energy, such that it can not represent a physically realizable particle. The integral $ \int \psi(x)^* H \psi(x) \mathrm{d} x $ is not convergent.


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