What does it mean for $P$ functions to be "more singular than a delta"?

Question

Consider the Glauber-Sudarshan $P$ representation of a state $\rho$, which is the function $\mathbb C\ni\alpha\mapsto P_\rho(\alpha)\in\mathbb R$ such that $$\rho = \int d^2\alpha \, P_\rho(\alpha) |\alpha\rangle\!\langle\alpha|.$$ Something that is often mentioned about the $P$ representation is that a state is nonclassical when $P_\rho$ is either non-positive or more singular than a Dirac delta function.

On the other hand, the $P$ representation of a coherent state $|\alpha\rangle$ is $P_\alpha(\beta)=\delta^2(\alpha-\beta)$, which does also arguably look "more singular than a $\delta$ function", in that it contains the square of a delta function, while $|\alpha\rangle$ is clearly not nonclassical.

So what gives? What exactly does "more singular than a delta" mean? The Wikipedia page gives the general expression for the $P$ function of a generic state, and this expression involves terms of the form $\big(\frac{\partial}{\partial r}\big)^\ell \delta(r)$, which I'm guessing are the source of "high degree of nonsingularity", and I'm aware that this is quite different than the $\delta^2$ term in the expression above, which is just the "standard" $\delta$ for the two-dimensional case, but I still find it quite unclear how exactly I should think about functions that are "more singular than a delta".

One possible way to understand what such functions actually represent is via some corresponding sequence of functions. For example, I can understand $\delta(x)$ as the limit of suitably normalised functions peaked around the origin, and $\delta^2(x)$ as the same thing for functions $\mathbb C\simeq \mathbb R^2\to\mathbb R$. Are there similar representations for the more nonsingular cases?

Answer 1

As you mention, the distinction here is that $\delta^2(\alpha-\beta)$ is not a derivative of a delta function, it is simply a delta function in two dimensions. Adding extra dimensions does not change the nature of the delta function in terms of how it behaves in integrals.

I will use the notation $\delta^2(\alpha-\beta)=\delta(x-x_0)\delta(y-y_0)$ because we typically abuse notation when we have functions of $\alpha$ and $\alpha^*$. We can consider $x$ and $y$ to be the real and imaginary parts of $\alpha$ and similarly $x_0$ and $y_0$ for $\beta$.

If you have any function of $\alpha$ and $\alpha^*$, ie a function of $x$ and $y$, you can integrate it with the two-dimensional delta: $$\int dx dy \,\delta(x-x_0)\delta(y-y_0)f(x,y)=f(x_0,y_0).$$ This simply uses the definition of a delta function twice, so it does not give us any more trouble than one delta function. Of course, we need $f(x,y)$ to be finite at $(x_0,y_0)$, instead of just needing a function to be finite for some value of one parameter, so perhaps that is asking for more.

Now, if we have derivatives of delta functions, we consider them in terms of chain rule / integration by parts: $$\int dx dy \,\frac{\partial^m}{\partial x^m}\delta(x-x_0)\frac{\partial^n}{\partial y^n}\delta(y-y_0)f(x,y)=(-1)^{m+n}\frac{\partial^{m+n}f(x,y)}{\partial x^m \partial y^m}\bigg{|}_{x=x_0,y=y_0}.$$ For this integral to converge, we need a lot more: we need $f(x,y)$ to be nonsingular after taking a whole bunch of derivatives! This is what it means for a function to be more singular than a delta function: it means that, when taken as a distribution and being used for integration, it asks more out of the function $f(x,y)=f(\alpha,\alpha^*)$ being integrated than what a delta function asks.

Edit: The steps with integration by parts also ask more from a function's behaviour at the limits of integration (often $x,y\to\pm\infty$). The delta function goes to zero very quickly so that is not normally an issue, but I can imagine there being functions that do not work so well once you take enough derivatives.

Further edit: One can show using integration by parts using a suitable trial function that $f(x)$ that $$\frac{\partial \delta(x)}{\partial x}=-\frac{\delta(x)}{x}.$$ This is where the singular nature of the distribution itself comes in to play: the distribution is only nonzero when $x=0$, but then there is a factor of $x$ in the denominator, which makes things singular. The more derivatives we take of a delta function, the higher the power of $x$ in the denominator.

Further edit continued: $\delta(x)$ seems singular at $x=0$, but it can be integrated and has area 1: $\int_{-\infty}^\infty dx \,\delta(x)=1$. On the other hand, $\partial_x \delta(x)$ "has no area" because $\int_{-\infty}^\infty dx \,\partial_x\delta(x)=0$. Still, if we multiply it by $x$, it regains an area of one: $\int_{-\infty}^\infty dx \,x\partial_x\delta(x)=-1$. This keeps going: for higher numbers of derivatives, derivatives of $\delta(x)$ seem to have no area, unless they are first multiplied by higher powers of $x$. So to "tame the singularity" of a derivative of a delta function, we need to do more work (multiply by $x^m$ then integrate) than to "tame the singularity" of a delta function (just integrate).

Answer 2

A footnote/ formal wisecrack, to illustrate the side issue of why derivatives of δ functions are regarded as more singular than δ functions with sequences of Gaussians of decreasing widths.

Look at the "practical definition" physicists use, $$ \delta(x) = \lim_{\epsilon\to 0} \frac{e^{-(x/\epsilon)^2}}{|\epsilon| \sqrt{\pi}}. $$ Hell, take $\epsilon$ positive. Its maximum before the limit is at the origin, $$ \frac{ 1}{\epsilon \sqrt{\pi}}, $$ quite spikey.

Now, look at $$ \partial_x \delta(x) = \lim_{\epsilon\to 0} \frac{-2x e^{-(x/\epsilon)^2}}{\epsilon^3 \sqrt{\pi}}. $$ Its maximum before the limit is to the left of the origin, at $x=-\epsilon/\sqrt 2$, $$ \sqrt{2/\pi} \frac{e^{-1/2}}{\epsilon^2}, $$ More spikey in the limit, with a minimum of the same depth symmetrically around the origin.

You may thus convince yourself that each derivative yields ever-higher peaks and dips by $/\epsilon$, leading to a more singular limit.

However, extraordinary caution is warranted. An exponential of derivatives, is a mere Lagrange shift operator, $$ \exp(a\partial_x) ~ \delta(x) = \lim_{\epsilon\to 0} \frac{e^{-((x+a)/\epsilon)^2}}{|\epsilon| \sqrt{\pi}}, $$ so the apparently runaway spikes "conspired" to simply shift the location of the singularity without altering its nature. Caveat lector.

As I indicated in my comment, Schleich's book tells one what needs to be attended to, no more and no less...