Name/significance of integral of the square of a probability density function Background/Motivation
Given a probability density function $f(x)$, the mean of the corresponding random variable is the $x$-coordinate of the centroid of the region under the graph of $f$. I wondered: "does the $y$-coordinate have any significance in probability?". Of course, the $y$-coordinate is given by $\frac12\int_{-\infty}^\infty f^2$, so without worrying about the $1/2$, I wondered if the integral of $f^2$ itself has any significance. An internet search brought up only this physics forum thread and this yahoo answer, neither of which seemed to have any useful information. 
My thoughts
One thing I thought about was the discrete case. Then $\Sigma_n p(n)^2$ is a probability, analogous to the probability of rolling doubles. Unfortunately, $\int f^2$ can easily be more than 1, so it does not have such a nice interpretation.
I also considered uniform distributions. A uniform distribution of probability density equal to $p$ has square-integral $(1/p)*p^2=p$, so that maybe this square integral is something like "the density of the uniform distribution $f$ is most like". Flipping this idea on its head, a uniform distribution over an interval of length $1/p$ has square-integral $p$, so that the reciprocal of the square-integral is like the length of the uniform distribution with the same$\ldots$"clumpiness"? (It reminded me of curvature, being the reciprocal of the radius of the circle with the same curviness.) But I don't know if these interpretations are useful in any way.
The question
Does the integral of the square of a pdf (or half of it) have a name (aside from "the square of the $L^2$ norm")? Is it used for anything? Is there a better angle from which to think about it?
 A: 
One thing I thought about was the discrete case.

Actually that works here too.  One just has to be careful of dimensions.
Notice that
$$I = \int_{\mathbb R} (f(x))^2 \, d x$$
has dimensions of $[x]^{-1}$.  Therefore, in order to interpret this as a (dimensionless) probability, it's necessary to multiply it by something else with dimensions of $[x]$.  Call this $\delta x$, and assume it's small.  Then $I \delta x$ can be interpreted as the (approximate) probability of "rolling doubles" but with a tolerance of $\delta x$.
Why is this so?  As it turns out, $I \delta x$ is just the limiting case of this integral:
$$\iint_{\mathbb R^2} f(x) f(x') \Delta(x - x') \, d x' \, d x$$
where $\Delta$ is some sort of distribution function sharply peaked around zero (e.g. a very narrow Gaussian).  If the width of $\Delta$ is narrow enough and $f$ is sufficiently smooth, then one may simply approximate the inner integral as a product of the width and the height.
Of course, even $I \delta x$ can exceed one – that's simply because the approximation has broken down by that point.
A: This quantity (actually its reciprocal) shows up sometimes in Quantum Mechanics and is known as the Süssmann measure.
It is used to define a "width" to probability density functions over $\mathbb{R}^n$. This can be useful, as other quantities that are related to the "width", like the variance, need not to exists necessarily (for example Lorentz distributions). On the other hand, ever smooth integrable function over $\mathbb{R}$ is also square integrable, so the Süssmann measure exists for all of these examples.
The intuition is, that $\int_\mathbb{R^n} d^nx p(x)^2$ can be considered the average height $h = \langle p\rangle$ of the distribution $p$ and if we want to replace $p$ by some piecewise constant distribution $q(x) = h 1_A(x)$ with that height, than the support of $q$ has to have the size $|A| = 1/\int_\mathbb{R^n} d^nx p(x)^2$ of the Süssmann measure.
A: 
I wondered if the integral of $f^2$ itself has any significance. 

As you say, this is the $L^2$ norm of $f$ in the space $L^2(λ)$, where $λ$ is the Lebesgue measure. This is also $E(f(X))$, where the random variable $X$ has density $f$. As such, this is the $L^1$ norm of $f$ in the space $L^1(μ)$ where $μ$ is the measure with density $f$.

Is it used for anything?

Not that I would be aware of.
