Gravitational attraction between quantum particles Let's say we have a quantum particle with mass $m$ in a 1-Dimensional box. The potential outside the box is infinite. Say that $n=2$, so that $|\psi|^2$ will have two maxima.
How would the gravitational attraction between this particle (particle1) and another particle (particle2) work? Toward which maxima the particle2 is going to be pulled by the particle1? Or what will be the point toward which the particle2 is going to be pulled?

 A: For this answer, we will assume Newtonian gravity, and calculate the effects on particle 1 from the gravitation of particle 2 (which we will assume is a fixed point mass).
Suppose we have a 1-dimensional "particle in a box" of mass $m$ confined to the interval $[0,L]$. The energy eigenstates are labeled by $n=1,2,...$ and have energies
$$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$$
and wavefunctions
$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
Suppose now we introduce a small perturbation to that system, namely, the gravitational attraction from another particle of mass $M$ at position $x_G$, which we will restrict to lie outside the box so as to keep the perturbation small and avoid singularities. The potential introduced by this attraction at position $x$ is
$$V(x)=-\frac{GmM}{|x-x_G|}$$
We treat this as a perturbation of the original wavefunction, and proceed using first-order perturbation theory. The change in the energy of the $n=2$ state, to first order in $GmM$, is
$$\Delta E_2 = \langle\psi_2|V|\psi_2\rangle=\int_0^L -\frac{2GmM}{L}\frac{\sin^2\left(\frac{2\pi x}{L}\right)}{|x-x_G|}dx$$
This integral has no elementary expression for its solution, but it can be plotted. For example, here is the change in energy as a function of distance, assuming the attracting particle is to the left of the box:

In the above, the $y$-axis is in units of $E_0=\frac{2GmM}{L}$. As you can see, the closer the attracting particle is to the box, the more the energy of the particle in the box is lowered.
The first-order correction to the wavefunction can also be calculated using the well-known formula
\begin{align*}
\Delta\psi_2(x) &= \sum_{k\neq 2}\frac{\langle\psi_k|V|\psi_2\rangle}{E_2-E_k}|\psi_k\rangle \\ &= \sum_{k\neq 2}\frac{\int_0^L -\frac{2GmM}{L}\frac{\sin\left(\frac{2\pi x}{L}\right)\sin\left(\frac{k\pi x}{L}\right)}{|x-x_G|}}{\frac{(2^2-k^2)\pi^2\hbar^2}{2mL^2}}\sin\left(\frac{k\pi x}{L}\right)
\end{align*}
Once again, there is no closed-form solution, but here is an approximation of this state's probability $|\psi_2+\Delta\psi_2|^2$ for a particle with $x_G=-1$, $L=1$, and $GMm=.01$, with the $y$-axis in arbitrary units (where the corrections have been taken out to $k=6$):

As you can see, it's basically identical to the original wavefunction, which is a sign that we did things right, because first-order perturbation theory is only valid for very small perturbations to the wavefunction. To get a closer look on what has changed, let's take the ratio of the perturbed probability density to the original probability density $\frac{|\psi_2+\Delta\psi_2|^2}{|\psi_2|^2}$:

As you can see, the particle is now slightly more probable to be on the left side of the box than before. (Curiously, it also appears to be more probable to be on the edges of the box either way, but looking at the asmmetry in this ratio, it clearly prefers to be on the left side). Given that our attracting particle is on the left side of the box, this makes sense. So, overall, it seems that our probability distribution has shifted slightly leftward. Calculating the average position of the particle, in the above units:
$$\langle x\rangle = \frac{ \langle \psi_2+\Delta\psi_2|x|\psi_2+\Delta\psi_2\rangle}{\langle \psi_2+\Delta\psi_2|\psi_2+\Delta\psi_2\rangle} = \frac{\int_0^L (\psi_2+\Delta\psi_2)^2 x dx}{\int_0^L (\psi_2+\Delta\psi_2)^2 dx}\approx 0.499996$$
which is slightly left of the average position of the unperturbed particle, $0.5$. 
