Quantum correlation of two non-interaction spinless particles: exact solution of the time-dependent Schrodinger equation I would like to get some comment about the following obtained result.
I consider the one-dimensional time-dependent Schrodinger equation for two non-interaction particles with masses $m_1$ and $m_2$:
$$
i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m_1} \frac{\partial^2 \Psi}{\partial x_1^2} - \frac{\hbar^2}{2m_2} \frac{\partial^2 \Psi}{\partial x_2^2}.
$$
Let us assume that the wave-function $\Psi(x_1, x_2, t)$ has the gaussian form at the initial instant $t=0$:
$$
\Psi(x_1,x_2,0) = \frac{1}{\sqrt{2\pi \sigma_1 \sigma_2}\,(1-r^2)^{1/4}}\, \exp \left[ - \frac{1}{1-r^2} \left( \frac{x_1^2}{4\sigma_1^2} + \frac{(x_2-x_0)^2}{4\sigma_2^2} - \frac{r x_1 (x_2 - x_0)}{2\sigma_1\sigma_2}\right) + \frac{i}{\hbar}\, p_0 x_2 \right].
$$
In my understanding, it describes the situation when the first particle is placed at the origin, while the second one is moving along $x$-axis with the velocity $v_0 = p_0/m_2$. Here $\sigma_1$ and $\sigma_2$ are the initial standard deviations of the particle coordinates, and $r$ is the initial correlation coefficient, $|r|<1$. Thus, if $r \neq 0$, then we have a two-particle entangled state at $t=0$. Without loss of generality, we will assume that $r>0$.
I have exactly solved the time-dependent Schrodinger equation with the given initial condition and obtained an explicit form of the wave-function $\Psi(x_1,x_2,t)$ (the corresponding expression is rather cumbersome and I do not write it here). Then I have calculated the time-dependent correlation coefficient:
$$
r_t := \frac{\langle x_1 x_2 \rangle - \langle x_1 \rangle \langle x_2 \rangle}{\delta x_1 \delta x_2} = \frac{r \sigma_1 \sigma_2 - \frac{r \hbar^2 t^2}{4 m_1 m_2 (1-r^2) \sigma_1 \sigma_2}}{\sqrt{\sigma_1^2 + \frac{\hbar^2 t^2}{4 m_1^2 (1-r^2)\sigma_1^2}} \, \sqrt{\sigma_2^2 + \frac{\hbar^2 t^2}{4 m_2^2 (1-r^2) \sigma_2^2}}},
$$
where $\langle \hat{A} \rangle := \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \overline{\Psi(x_1, x_2, t)} \, \hat{A} \Psi(x_1, x_2, t) dx_1 dx_2$, $\delta x_i = \sqrt{\langle x_i^2 \rangle - \langle x_i \rangle^2}$, $i=1,2$. An analysis of this expression shows that: 1) the function $r_t$ has a maximum at $t=0$, which equals $r$; 2) for $t>0$ the function $r_t$ monotonically decreases and asymptotically approaches the value $r_{\infty} = -r$; 3) the function $r_t$ vanishes at the instant $t_0 = 2 \sigma_1 \sigma_2 \sqrt{m_1 m_2(1-r^2)}/\hbar$. I would like to get comments on these results in the physical context. Especially, what means the change of sign for the correlation coefficient? I would be grateful for any clarifications and references.
 A: Honestly, the best approach here is to just look, i.e., to examine the behaviour of $|\Psi(x_1,x_2,t)|^2$ (best plotted as a density plot on the $x_1,x_2$ plane) as a function of $t$. The initial correlation coefficient $r$ tells you that the wavefunction tilts diagonally up in that plane, and the change in sign tells you that this tilt is reversed at some point in the evolution.
And this is indeed what happens:

The reason for this is that, if you look at a diagonal frame, you have an uncorrelated gaussian wavepacket which is narrower in one direction (diagonal in the $x_1,x_2$ frame) and wider in the other. The wavepacket expands, like all gaussian wavepackets do, but it expands faster in the direction where it was narrower (since it has a higher momentum variance in that direction), so eventually the expansion makes that direction bigger than the orthogonal one.
As for the upwards motion, it's irrelevant for the physics of the tilt switch.
A: As the two particles are distinguishable and non interacting, they are completely uncorrelated. The two particle Schrodinger equation reduces to two free one particle equations and the wave functions at t=0 are $\psi_1(x_1,0)=\int dx_2 \Psi(x_1,x_2,0)$ and $\psi_2(x_2,0)=\int dx_1 \Psi(x_1,x_2,0)$.
