This is the Schroedinger equation with a particular 2D harmonic potential:
$$\begin{multline}i\hbar\frac{\partial}{\partial t}\Psi(x_1,x_2,t) = \\ \Biggl[-\frac{\hbar^2}{2m}\nabla^2 + \frac{1}{2} m\omega^2\Biggl(\biggl(x_1 – \frac{x_0(t)}{2}\biggr)^2 + \biggl(x_2 + \frac{x_0(t)}{2}\biggr)^2\Biggr)\Biggr]\Psi(x_1,x_2,t)\end{multline}$$
Can anyone please tell me what the upside down triangle means? I know its the second derivative, but since the problem I have is of variables $x_1$, $x_2$, and $t$, do I take the second derivative of time? Or does that upside down triangle only does it for $x_1$ and $x_2$?
$x_0$ is the separation distance between $x_1$ and $x_2$ and its a function of time because the two double HO potential wells are moving apart from each other.
Also, how would I go about doing separation of variables here? Would I move all the time pieces to one side and all the x pieces to the other side?
Won't I have to do separation of variables twice? Where the second time is when I have to separate $x_1$ and $x_2$?
Best Answer
The symbol $\nabla^2$ usually means the sum of second derivatives if you are in a cartesian-like system. Thus for your case $$\nabla^2=\frac{\partial^2}{\partial x^2_1}+\frac{\partial^2}{\partial x^2_2}.$$ To do separation of variables, you usually postulate separate dependences on each of the variables, $$\Psi(x_1,x_2,t)=T(t)\psi_1(x_1)\psi_2(x_2).$$ However, the time-dependence of the well centres will make matters rather more difficult, and as it stands all you can hope to do is a separation of the type $$\Psi(x_1,x_2,t)=\psi_1(x_1,t)\psi_2(x_2,t).$$
I don't quite understand, though, your comment that
If $x_0=x_2-x_1$, then it is time-independent. If it's the well centres that move apart from each other, then all you can hope for is the above.