Is Quantum Coulomb still singular? A single free charge (e.g. electron) $q$ fixed at the coordinate origin has the well-known Coulomb/electric potential
$$\phi(\vec r) = \frac q{4\pi\epsilon_0}\frac 1r \tag{A}$$
where $r=|\vec r|$ of course. However, according to Quantum Mechanics, a particle of mass $m$ cannot be truly fixed, best you can get is a Gaussian Wave Packet
$$\psi(\vec r, t) = \left({a \over a + i\hbar t/m}\right)^{3/2} \exp\left(- \frac{r^2}{2(a + i\hbar t/m)} \right) \tag{B}$$
with $a=2\Delta x(t=0)$. So the electric potential should become an expectation value smeared by that Gaussian, and my question is, does this smooth out the singularity at $r=0$?
 A: Note that due to the time dependency of $\psi(\vec r, t)$ the retarted potential has to be considered. So we're looking for
$$\begin{align*}
  \langle\phi(\vec r, t)\rangle &= \int_{\mathbb R^3}d^3x \frac q{4\pi\epsilon_0}\frac{\Big|\psi(\vec x, t - \frac1c|\vec x - \vec r|))\Big|^2}{|\vec x - \vec r|} \quad\Bigg|\quad \vec x \to \vec x + \vec r
\\ &= \int_{\mathbb R^3}d^3x \frac q{4\pi\epsilon_0}\frac{\Big|\psi(\vec x + \vec r, \overbrace{t - \frac xc}^{=:t_c(x)}))\Big|^2}{x}
\\ &\stackrel{(B)}= \int_{\mathbb R^3} d^3x \underbrace{
  \frac q{4\pi\epsilon_0}\overbrace{\Bigg|{a \over a + i\hbar t_c/m}}^{=\frac{a}{\sqrt{a^2+\hbar^2t_c^2/m^2}}=\frac{\sqrt{2a}}{\sigma}}\Bigg|^3}_{=:N}
  \frac{\exp\overbrace{\left(- \Re \frac{(\vec x + \vec r)^2}{(a + i\hbar t_c/m)} \right)}^{=-\frac{a(\vec x + \vec r)^2}{a^2+\hbar^2t_c^2/m^2} =: -\frac{(\vec x + \vec r)^2}{2\sigma^2(t_c)}}}{x}
\end{align*}$$
The $x$-dependency of $t_c$ makes this a very nasty integral, so let's take the non-relativistic limit $c\to\infty$ such that $t_c(x) \to t$ (or alternatively, assume $m\to\infty$). Then we're looking for
$$\begin{align*}
  \langle\phi(\vec r,t)\rangle &= N\int_{\mathbb R^3}d^3x \frac{\exp\left(-\frac{(\vec x + \vec r)^2}{2\sigma^2}\right)}{x}. \tag{tk.C}\label{tk.C}
\end{align*}$$
Note how for $\vec r=0$ we can use spherical coordinates to obtain
$$\begin{align*}
  \langle\phi(0,t)\rangle &= 4\pi N\int_0^\infty \rho^2 d\rho \frac{\exp\left(-\frac{\rho^2}{2\sigma^2}\right)}{\rho}
  \\ &= -4\pi\sigma^2 N\int_0^\infty d\rho\, \partial_\rho \exp\left(-\frac{\rho^2}{2\sigma^2}\right)
  \\ &= 4\pi\sigma^2 N = \frac{q\sqrt{2a}^3}{\epsilon_0\sigma}
\end{align*}$$
which for $\sigma\neq0$ (that would yield a truly localized particle which would however diffuse infinitely the very next instant) is finite, thus the answer to the question is, considering uncertainty, there is no longer a singularity.
On paper, I actually calculated $\eqref{tk.C}$ to end up with a correction factor $\mathrm{erf}\left(\frac r{\sqrt{2\sigma^2}}\right)$ to the classical Coulomb potential (by using the Fourier transform w.r.t. $\vec r$, swapping integrals and integrating over $\sigma^2$), but that's too tedious to typeset for now, and there's a sign error in it somewhere...
A: I am supposed to answer your question to post here, but, really there is no way to answer your question.  To answer a question you could have asked:  Yes, electrons have to have smeared momentum.  But, that is not the "real" issue.  You could always put them in a potential well.  If non-relativistic quantum mechanics "worked" then you could localize electrons / charge as well as you wanted, and so have as divergent a potential as you wanted.  But, there are two objections to this, special relativistic quantum mechanics and general relativity.  If you localize an electron into a small enough region - so small that its binding energy $E_b$ is close to its rest mass, $-E_b\sim m_ec^2$, you need to think of it as a relativistic quantum field, and as a mixture of electrons and positrons.  These are treated using quantum electrodynamics, and (well) it would be reasonable in that context to claim that they have infinite (differently infinite, but still infinite) potential at the origin.  
In general relativity, on the other hand, you would find if (say) you took a classical shell with a mass and a charge and a radius, and you slowly decreased the radius, then the potential, electric field and everything becomes very large, with very large energy or mass.  When that mass becomes large enough, the system becomes a charged black hole.  There is then no "origin" - just the surface of the black hole.  The potential at the surface of the black hole is finite.  And the smallest possible mass is huge, compared to an electron mass.
And, in reality, we have don't finally know.  We don't know how to quantum mechanics and general relativity work together.  So, finally we don't know.
