Localizing position of electron in hydrogen atom Consider the hydrogen atom, just taking into account the electrostatic force and not magnetism nor spin. 
Is it possible to take the wave functions of the energy eigenstates of that hydrogen atom, appropriately weight each one, and then sum them up in order to get something like a Gaussian wave function of position, or at least to localize position into a single narrow range for $r$, $\theta$, and $\phi$, such that there is a $99.999$% chance that the electron is in that narrow range? 
My surmise is no, since the spherical harmonic angular momentum eigenstates do not look like they could possibly be combined in such a was as to localize the electron within the angular degrees of freedom.
 A: Indeed, it is very much possible. The trick is that however they must combine, it must be very subtle and hard to imagine immediately. But it can be done. The eigenstates form a complete orthonormal set, hence they can express "any" (with suitable caveats in the scarequotes) initial configuration, and all you have to do to find the expansion is to take inner products: any bound state $|\psi\rangle$ will satisfy
$$|\psi\rangle = \sum_{n=1}^{\infty} \sum_{l=0}^{n-1} \sum_{m=-l}^{l} \langle\psi_{n,l,m}|\psi\rangle\ |\psi_{n,l,m}\rangle$$
where $|\psi_{n,l,m}\rangle$ are the basis states for each choice of quantum numbers $n$, $l$, and $m$.
What does this look like? Well, if we go to the position basis, so that the states are now represented with wave functions $\psi(r, \theta, \phi)$, it looks as follows: the coefficient in the above is
$$\begin{align}
\langle \psi_{n,l,m}|\psi\rangle &= \iiint_\mathbb{R^3} \bar{\psi}_{n,l,m}(r, \theta, \phi)\ \psi(r, \theta, \phi)\ dV\\
&= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \bar{\psi}_{n,l,m}(r, \theta, \phi)\ \psi(r, \theta, \phi)\ r^2 \sin \theta\ dr\ d\theta\ d\phi\end{align}$$
Now suppose that we want the electron localized to a wedge $r \in [r_1, r_2]$, $\theta \in [\theta_1, \theta_2]$, $\phi \in [\phi_1, \phi_2]$. Then you can treat $\psi(r, \theta, \phi) = K$ within the wedge and as $0$ outside it, so you simply limit the bounds of integration accordingly:
$$\langle \psi_{n,l,m}|\psi\rangle = K \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} \bar{\psi}_{n,l,m}(r, \theta, \phi)\ r^2 \sin \theta\ dr\ d\theta\ d\phi$$
where the normalizing factor $K$ is equal to $(\text{volume of wedge})^{-1/2}$. Of course, working this out explicitly is likely to be extremely complicated, or one can just calculate the coefficients numerically.
A: Since you are asking about the possibility, of course, you can do that. Energy eigenstates a Hydrogen atom form a complete basis, just like the eigenstates of any Hermitian operator, and thus, they can be linearly combined to obtain any other state-vector, including a Gaussian on a sphere spiked at some particular angular value. 

For example, in the documentation of this opensource package for MATLAB, they expand a Gaussian (with $\sigma=0.4$) on a sphere in the basis of spherical harmonics and numerically the compute coefficients, see Section 3.
